{"id":2215,"date":"2015-04-22T21:02:17","date_gmt":"2015-04-22T21:02:17","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=2215"},"modified":"2016-10-27T16:01:31","modified_gmt":"2016-10-27T16:01:31","slug":"integrated-rate-laws","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/chapter\/integrated-rate-laws\/","title":{"raw":"Integrated Rate Laws","rendered":"Integrated Rate Laws"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this module, you will be able to:\r\n<ul>\r\n \t<li>Explain the form and function of an integrated rate law<\/li>\r\n \t<li>Perform integrated rate law calculations for zero-, first-, and second-order reactions<\/li>\r\n \t<li>Define half-life and carry out related calculations<\/li>\r\n \t<li>Identify the order of a reaction from concentration\/time data<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.\r\n\r\nUsing calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.\r\n<h2>First-Order Reactions<\/h2>\r\nAn equation relating the rate constant <em>k<\/em> to the initial concentration [<em>A<\/em>]<sub>0<\/sub> and the concentration [<em>A<\/em>]<sub>t<\/sub> present after any given time <em>t<\/em> can be derived for a first-order reaction and shown to be:\r\n<p style=\"text-align: center;\">[latex]\\text{ln}\\left(\\frac{{\\left[A\\right]}_{t}}{{\\left[A\\right]}_{0}}\\right)=-kt[\/latex]<\/p>\r\nor\r\n<p style=\"text-align: center;\">[latex]\\text{ln}\\left(\\frac{{\\left[A\\right]}_{0}}{{\\left[A\\right]}_{t}}\\right)=kt[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0The Integrated Rate Law for a First-Order Reaction<\/h3>\r\nThe rate constant for the first-order decomposition of cyclobutane, C<sub>4<\/sub>H<sub>8<\/sub> at 500 \u00b0C is 9.2 \u00d7 10<sup>\u22123<\/sup> s<sup>\u22121<\/sup>:\r\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{4}{\\text{H}}_{8}\\rightarrow{\\text{2C}}_{2}{\\text{H}}_{4}[\/latex]<\/p>\r\nHow long will it take for 80.0% of a sample of C<sub>4<\/sub>H<sub>8<\/sub> to decompose?\r\n\r\n[reveal-answer q=\"541265\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"541265\"]\r\n\r\nWe use the integrated form of the rate law to answer questions regarding time:\r\n<p style=\"text-align: center;\">[latex]\\text{ln}\\left(\\frac{{\\left[A\\right]}_{0}}{\\left[A\\right]}\\right)=kt[\/latex]<\/p>\r\nThere are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [<em>A<\/em>]<sub>0<\/sub>, [<em>A<\/em>], and <em>k<\/em>, and need to find <em>t<\/em>.\r\n\r\nThe initial concentration of C<sub>4<\/sub>H<sub>8<\/sub>, [<em>A<\/em>]<sub>0<\/sub>, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let <em>x<\/em> be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of <em>x<\/em> or 0.200<em>x.<\/em> Rearranging the rate law to isolate <em>t<\/em> and substituting the provided quantities yields:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill t&amp; =\\mathrm{ln}\\frac{\\left[x\\right]}{\\left[0.200x\\right]}\\times \\frac{1}{k}\\hfill \\\\ &amp; =\\text{ln}\\frac{0.100{\\text{mol L}}^{-1}}{0.020{\\text{mol L}}^{-1}}\\times \\frac{1}{9.2\\times {10}^{-3}{\\text{s}}^{-1}}\\hfill \\\\ &amp; =1.609\\times \\frac{1}{9.2\\times {10}^{-3}{\\text{s}}^{-1}}\\hfill \\\\ &amp; =1.7\\times {10}^{2}\\text{s}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nIodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:\r\n<p style=\"text-align: center;\">[latex]\\text{I-131}\\rightarrow\\text{Xe-131}+\\text{electron}[\/latex]<\/p>\r\nThe decay is first-order with a rate constant of 0.138 d<sup>\u22121<\/sup>. All radioactive decay is first order. How many days will it take for 90% of the iodine-131 in a 0.500 <em>M<\/em> solution of this substance to decay to Xe-131?\r\n\r\n[reveal-answer q=\"531173\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"531173\"]16.7 days[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}\\left[A\\right]&amp; =&amp; \\left(\\text{-}k\\right)\\left(t\\right)+\\text{ln}{\\left[A\\right]}_{0}\\hfill \\\\ \\hfill y&amp; =&amp; mx+b\\hfill \\end{array}[\/latex]<\/p>\r\nA plot of ln[<em>A<\/em>] versus <em>t<\/em> for a first-order reaction is a straight line with a slope of -<em>k<\/em> and an intercept of ln[<em>A<\/em>]<sub>0<\/sub>. If a set of rate data are plotted in this fashion but do <em>not<\/em> result in a straight line, the reaction is not first order in <em>A<\/em>.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Determination of Reaction Order by Graphing<\/h3>\r\nShow that the data in Figure\u00a01 can be represented by a first-order rate law by graphing ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time. Determine the rate constant for the rate of decomposition of H<sub>2<\/sub>O<sub>2<\/sub> from this data.\r\n\r\n[caption id=\"attachment_5373\" align=\"aligncenter\" width=\"906\"]<img class=\"size-full wp-image-5373\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09033022\/Table2.png\" alt=\"A table with five columns is shown. The first column is labeled, \u201cTime, h.\u201d Beneath it the numbers 0.00, 6.00, 12.00, 18.00, and 24.00 are listed. The second column is labeled, \u201c[ H subscript 2 O subscript 2 ], mol \/ L.\u201d Below, the numbers 1.000, 0.500, 0.250, 0.125, and 0.0625 are double spaced. To the right, a third column is labeled, \u201ccapital delta [ H subscript 2 O subscript 2 ], mol \/ L.\u201d Below, the numbers negative 0.500, negative 0.250, negative 0.125, and negative 0.062 are listed such that they are double spaced and offset, beginning one line below the first number listed in the column labeled, \u201c[ H subscript 2 O subscript 2 ], mol \/ L.\u201d The first two numbers in the second column have line segments extending from their right side to the left side of the first number in the third row. The second and third numbers in the second column have line segments extending from their right side to the left side of the second number in the third row. The third and fourth numbers in the second column have line segments extending from their right side to the left side of the third number in the third row. The fourth and fifth numbers in the second column have line segments extending from their right side to the left side of the fourth number in the third row. The fourth column in labeled, \u201ccapital delta t, h.\u201d Below the title, the value 6.00 is listed four times, each single-spaced. The fifth and final column is labeled \u201cRate of Decomposition, mol \/ L \/ h.\u201d Below, the following values are listed single-spaced: negative 0.0833, negative 0.0417, negative 0.0208, and negative 0.0103.\" width=\"906\" height=\"292\" \/> Figure\u00a01. The rate of decomposition of H<sub>2<\/sub>O<sub>2<\/sub> in an aqueous solution decreases as the concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases.[\/caption]\r\n[reveal-answer q=\"559973\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"559973\"]The data from Figure\u00a01 with the addition of values of ln[H<sub>2<\/sub>O<sub>2<\/sub>] are given in Figure\u00a02.[caption id=\"attachment_4849\" align=\"aligncenter\" width=\"400\"]<img class=\" wp-image-4849\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214429\/FrstOKin.jpg\" alt=\"A graph is shown with the label \u201cTime ( h )\u201d on the x-axis and \u201cl n [ H subscript 2 O subscript 2 ]\u201d on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).\" width=\"400\" height=\"286\" \/> Figure\u00a02. The linear relationship between the ln[H<sub>2<\/sub>O<sub>2<\/sub>] and time shows that the decomposition of hydrogen peroxide is a first-order reaction.[\/caption]\r\n<table id=\"fs-idp114347648\" class=\"medium unnumbered\" summary=\"This table contains four columns and six rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( h ),\u201d \u201c[ H subscript 2 O subscript 2 ] ( M ),\u201d and \u201cl n [ H subscript 2 O subscript 2 ].\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the column, \u201cTime ( h )\u201d are the numbers 0, 6.00, 12.00, 18.00, and 24.00. Under the column \u201c[ H subscript 2 O subscript 2 ] ( M ),\u201d are the numbers 1.000, 0.500, 0.250, 0.125, and 0.0625. Under the column, \u201cl n [ H subscript 2 O subscript 2 ],\u201d are the numbers: 0.0, negative 0.693, negative 1.386, negative 2.079, and negative 2.772.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Trial<\/th>\r\n<th>Time (h)<\/th>\r\n<th>[H<sub>2<\/sub>O<sub>2<\/sub>] (<em>M<\/em>)<\/th>\r\n<th>ln[H<sub>2<\/sub>O<sub>2<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>1<\/td>\r\n<td>0<\/td>\r\n<td>1.000<\/td>\r\n<td>0.0<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>6.00<\/td>\r\n<td>0.500<\/td>\r\n<td>\u20130.693<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3<\/td>\r\n<td>12.00<\/td>\r\n<td>0.250<\/td>\r\n<td>\u20131.386<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4<\/td>\r\n<td>18.00<\/td>\r\n<td>0.125<\/td>\r\n<td>\u20132.079<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>5<\/td>\r\n<td>24.00<\/td>\r\n<td>0.0625<\/td>\r\n<td>\u20132.772<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe plot of ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.\r\n\r\nThe rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time where:\r\n<p style=\"text-align: center;\">[latex]\\text{slope}=\\frac{\\text{change in }y}{\\text{change in }x}=\\frac{\\Delta y}{\\Delta x}=\\frac{\\Delta\\text{ ln}\\left[{\\text{H}}_{2}{\\text{O}}_{2}\\right]}{\\Delta t}[\/latex]<\/p>\r\nIn order to determine the slope of the line, we need two values of ln[H<sub>2<\/sub>O<sub>2<\/sub>] at different values of <em>t<\/em> (one near each end of the line is preferable). For example, the value of ln[H<sub>2<\/sub>O<sub>2<\/sub>] when <em>t<\/em> is 6.00 h is -0.693; the value when <em>t<\/em> = 12.00 h is \u22121.386:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{slope}&amp; =&amp; \\frac{-1.386-\\left(-0.693\\right)}{\\text{12.00 h}-\\text{6.00 h}}\\hfill \\\\ &amp; =&amp; \\frac{-0.693}{\\text{6.00 h}}\\hfill \\\\ &amp; =&amp; -1.155\\times {10}^{-2}{\\text{h}}^{-1}\\hfill \\\\ \\hfill k&amp; =&amp; -\\text{slope}=-\\left(-1.155\\times {10}^{-2}{\\text{h}}^{-1}\\right)=1.155\\times {10}^{-2}{\\text{h}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nGraph the following data to determine whether the reaction [latex]A\\rightarrow B+C[\/latex] is first order.\r\n<table id=\"fs-idm149704608\" class=\"medium unnumbered\" summary=\"This table has three columns and six rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and, \u201c[ A ].\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201cTime ( s )\u201d column are the numbers: 4.0, 8.0, 12.0, 16.0, and 20.0. Under the \u201c [ A ]\u201d column are the numbers: 0.220, 0.144, 0.110, 0.088, and 0.074.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Trial<\/th>\r\n<th>Time (s)<\/th>\r\n<th>[<em>A<\/em>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>1<\/td>\r\n<td>4.0<\/td>\r\n<td>0.220<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>8.0<\/td>\r\n<td>0.144<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3<\/td>\r\n<td>12.0<\/td>\r\n<td>0.110<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4<\/td>\r\n<td>16.0<\/td>\r\n<td>0.088<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>5<\/td>\r\n<td>20.0<\/td>\r\n<td>0.074<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"663896\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"663896\"]The plot of ln[<em>A<\/em>] vs. <em>t<\/em> is not a straight line. The equation is not first order:\r\n\r\n<img class=\"wp-image-2204 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212251\/CNX_Chem_12_04_CYL1_img1.jpg\" alt=\"A graph, labeled above as \u201cl n [ A ] vs. Time\u201d is shown. The x-axis is labeled, \u201cTime ( s )\u201d and the y-axis is labeled, \u201cl n [ A ].\u201d The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).\" width=\"400\" height=\"232\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Second-Order Reactions<\/h2>\r\nThe equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant\u2019s concentration and described by the differential rate law:\r\n<p style=\"text-align: center;\">[latex]\\text{Rate}=k{\\left[A\\right]}^{2}[\/latex]<\/p>\r\nFor these second-order reactions, the integrated rate law is:\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\left[A\\right]}=kt+\\frac{1}{{\\left[A\\right]}_{0}}[\/latex]<\/p>\r\nwhere the terms in the equation have their usual meanings as defined above.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0The Integrated Rate Law for a Second-Order Reaction<\/h3>\r\nThe reaction of butadiene gas (C<sub>4<\/sub>H<sub>6<\/sub>) with itself produces C<sub>8<\/sub>H<sub>12<\/sub> gas as follows:\r\n<p style=\"text-align: center;\">[latex]{\\text{2C}}_{4}{\\text{H}}_{6}\\text{(}g\\text{)}\\rightarrow{\\text{C}}_{8}{\\text{H}}_{12}\\text{(}g\\text{)}[\/latex]<\/p>\r\nThe reaction is second order with a rate constant equal to 5.76 \u00d7 10<sup>\u22122<\/sup> L\/mol\/min under certain conditions. If the initial concentration of butadiene is 0.200 <em>M<\/em>, what is the concentration remaining after 10.0 min?\r\n\r\n[reveal-answer q=\"768114\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"768114\"]\r\n\r\nWe use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have:\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\left[A\\right]}=kt+\\frac{1}{{\\left[A\\right]}_{0}}[\/latex]<\/p>\r\nWe know three variables in this equation: [<em>A<\/em>]<sub>0<\/sub> = 0.200 mol\/L, <em>k<\/em> = 5.76 \u00d7 10<sup>\u22122<\/sup> L\/mol\/min, and <em>t<\/em> = 10.0 min. Therefore, we can solve for [<em>A<\/em>], the fourth variable:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{\\left[A\\right]}&amp; =&amp; \\left(5.76\\times {10}^{-2}{\\text{L mol}}^{-1}{\\mathrm{min}}^{-1}\\right)\\left(10\\text{min}\\right)+\\frac{1}{0.200{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill \\frac{1}{\\left[A\\right]}&amp; =&amp; \\left(5.76\\times {10}^{-1}{\\text{L mol}}^{-1}\\right)+5.00{\\text{L mol}}^{-1}\\hfill \\\\ \\hfill \\frac{1}{\\left[A\\right]}&amp; =&amp; 5.58{\\text{L mol}}^{-1}\\hfill \\\\ \\hfill \\left[A\\right]&amp; =&amp; 1.79\\times {10}^{-1}{\\text{mol L}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\r\nTherefore 0.179 mol\/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol\/L that was originally present.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nIf the initial concentration of butadiene is 0.0200 <em>M<\/em>, what is the concentration remaining after 20.0 min?\r\n\r\n[reveal-answer q=\"741332\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"741332\"]0.0196 mol\/L[\/hidden-answer]\r\n\r\n<\/div>\r\nThe integrated rate law for our second-order reactions has the form of the equation of a straight line:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{\\left[A\\right]}&amp; =&amp; kt+\\frac{1}{{\\left[A\\right]}_{0}}\\hfill \\\\ \\hfill y&amp; =&amp; mx+b\\hfill \\end{array}[\/latex]<\/p>\r\nA plot of [latex]\\frac{1}{\\left[A\\right]}[\/latex] versus <em>t<\/em> for a second-order reaction is a straight line with a slope of <em>k<\/em> and an intercept of [latex]\\frac{1}{{\\left[A\\right]}_{0}}[\/latex]. If the plot is not a straight line, then the reaction is not second order.\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0Determination of Reaction Order by Graphing<\/h3>\r\nTest the data given to show whether the dimerization of C<sub>4<\/sub>H<sub>6<\/sub> is a first- or a second-order reaction.\r\n\r\n[reveal-answer q=\"186612\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"186612\"]\r\n<table id=\"fs-idm140502592\" class=\"medium unnumbered\" summary=\"This table contains three columns and six rows. The first row is a header row and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and \u201c[ C subscript 4 H subscript 6 ] ( M ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201cTime ( s )\u201d column are the numbers: 0, 1600, 3200, 4800, and 6200. Under the column \u201c[ C subscript 4 H subscript 6 ] ( M )\u201d are the numbers: 1.00 times ten to the negative 2; 5.04 times ten to the negative 3; 3.37 times ten to the negative 3; 2.53 times ten to the negative 3; and 2.08 times ten to the negative 3.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Trial<\/th>\r\n<th>Time (s)<\/th>\r\n<th>[C<sub>4<\/sub>H<sub>6<\/sub>] (<em>M<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>1<\/td>\r\n<td>0<\/td>\r\n<td>1.00 \u00d7 10<sup>\u20132<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>1600<\/td>\r\n<td>5.04 \u00d7 10<sup>\u20133<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3<\/td>\r\n<td>3200<\/td>\r\n<td>3.37 \u00d7 10<sup>\u20133<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4<\/td>\r\n<td>4800<\/td>\r\n<td>2.53 \u00d7 10<sup>\u20133<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>5<\/td>\r\n<td>6200<\/td>\r\n<td>2.08 \u00d7 10<sup>\u20133<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn order to distinguish a first-order reaction from a second-order reaction, we plot ln[C<sub>4<\/sub>H<sub>6<\/sub>] versus <em>t<\/em> and compare it with a plot of [latex]\\frac{\\text{1}}{\\left[{\\text{C}}_{4}{\\text{H}}_{6}\\right]}[\/latex] versus <em>t<\/em>. The values needed for these plots follow.\r\n<table id=\"fs-idp62232800\" class=\"medium unnumbered\" summary=\"This table contains three columns and six rows. The first row is a header row and it labels each column, \u201cTime ( s ),\u201d \u201c1 over [ C subscript 4 H subscript 6 ] ( M superscript negative 1 ),\u201d and \u201cl n [ C subscript 4 H subscript 6 ].\u201d Under the column \u201cTime ( s )\u201d are the numbers: 0, 1600, 3200, 4800, and 6200. Under the \u201c1 over [ C subscript 4 H subscript 6 ] ( M superscript negative 1 )\u201d column are the numbers: 100, 198, 296, 395, and 481. Under the \u201cl n [ C subscript 4 H subscript 6 ]\u201d column are the numbers: negative 4.605, negative 5.289, negative 5.692, negative 5.978, and negative 6.175.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Time (s)<\/th>\r\n<th>[latex]\\frac{1}{\\left[{\\text{C}}_{4}{\\text{H}}_{6}\\right]}\\left({M}^{-1}\\right)[\/latex]<\/th>\r\n<th>ln[C<sub>4<\/sub>H<sub>6<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>0<\/td>\r\n<td>100<\/td>\r\n<td>\u20134.605<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1600<\/td>\r\n<td>198<\/td>\r\n<td>\u20135.289<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3200<\/td>\r\n<td>296<\/td>\r\n<td>\u20135.692<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4800<\/td>\r\n<td>395<\/td>\r\n<td>\u20135.978<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>6200<\/td>\r\n<td>481<\/td>\r\n<td>\u20136.175<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe plots are shown in Figure\u00a03. As you can see, the plot of ln[C<sub>4<\/sub>H<sub>6<\/sub>] versus <em>t<\/em> is not linear, therefore the reaction is not first order. The plot of [latex]\\frac{1}{\\left[{\\text{C}}_{4}{\\text{H}}_{6}\\right]}[\/latex] versus <em>t<\/em> is linear, indicating that the reaction is second order.\r\n\r\n[caption id=\"attachment_2205\" align=\"alignnone\" width=\"975\"]<img class=\"size-full wp-image-2205\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212252\/CNX_Chem_12_04_2OrdKin1.jpg\" alt=\"Two graphs are shown, each with the label \u201cTime ( s )\u201d on the x-axis. The graph on the left is labeled, \u201cl n [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The graph on the right is labeled \u201c1 divided by [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).\" width=\"975\" height=\"355\" \/> Figure\u00a03. These two graphs show first- and second-order plots for the dimerization of C<sub>4<\/sub>H<sub>6<\/sub>. Since the first-order plot (left) is not linear, we know that the reaction is not first order. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.[\/caption][\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nDoes the following data fit a second-order rate law?\r\n<table id=\"fs-idm88760288\" class=\"medium unnumbered\" summary=\"This table contains three columns and seven rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and, \u201c[ A ] ( M ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, 5, and 6. Under the \u201cTime ( s )\u201d column are the numbers: 5, 10, 15, 20, 25, and 35. Under the \u201c[ A ] ( M )\u201d column are the numbers 0.952, 0.625, 0.465, 0.370, 0.308, and 0.230.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Trial<\/th>\r\n<th>Time (s)<\/th>\r\n<th>[<em>A<\/em>] (<em>M<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>1<\/td>\r\n<td>5<\/td>\r\n<td>0.952<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>10<\/td>\r\n<td>0.625<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3<\/td>\r\n<td>15<\/td>\r\n<td>0.465<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4<\/td>\r\n<td>20<\/td>\r\n<td>0.370<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>5<\/td>\r\n<td>25<\/td>\r\n<td>0.308<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>6<\/td>\r\n<td>35<\/td>\r\n<td>0.230<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"260710\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"260710\"]Yes. The plot of [latex]\\frac{1}{\\left[A\\right]}[\/latex] vs. <em>t<\/em> is linear:\r\n\r\n<img class=\"wp-image-2206 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212253\/CNX_Chem_12_04_CYL2_img1.jpg\" alt=\"A graph, with the title \u201c1 divided by [ A ] vs. Time\u201d is shown, with the label, \u201cTime ( s ),\u201d on the x-axis. The label \u201c1 divided by [ A ]\u201d appears left of the y-axis. The x-axis shows markings beginning at zero and continuing at intervals of 10 up to and including 40. The y-axis on the left shows markings beginning at 0 and increasing by intervals of 1 up to and including 5. A line with an increasing trend is drawn through six points at approximately (4, 1), (10, 1.5), (15, 2.2), (20, 2.8), (26, 3.4), and (36, 4.4).\" width=\"400\" height=\"269\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Zero-Order Reactions<\/h2>\r\nFor zero-order reactions, the differential rate law is:\r\n<p style=\"text-align: center;\">[latex]\\text{Rate}=k{\\left[A\\right]}^{0}=k[\/latex]<\/p>\r\nA zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactants.\r\n\r\nThe integrated rate law for a zero-order reaction also has the form of the equation of a straight line:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left[A\\right]&amp; =&amp; {-}kt+{\\left[A\\right]}_{0}\\hfill \\\\ \\hfill y&amp; =&amp; mx+b\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_2207\" align=\"alignright\" width=\"350\"]<img class=\" wp-image-2207\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212255\/CNX_Chem_12_04_AmDecomK1.jpg\" alt=\"A graph is shown with the label, \u201cTime ( s ),\u201d on the x-axis and, \u201c[ N H subscript 3 ] M,\u201d on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled \u201cDecomposition on W.\u201d A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled \u201cDecomposition on S i O subscript 2.\u201d\" width=\"350\" height=\"353\" \/> Figure\u00a04. The decomposition of NH<sub>3<\/sub> on a tungsten (W) surface is a zero-order reaction, whereas on a quartz (SiO<sub>2<\/sub>) surface, the reaction is first order.[\/caption]A plot of [<em>A<\/em>] versus <em>t<\/em> for a zero-order reaction is a straight line with a slope of \u2212<em>k<\/em> and an intercept of [<em>A<\/em>]<sub>0<\/sub>. Figure\u00a04 shows a plot of [NH<sub>3<\/sub>] versus <em>t<\/em> for the decomposition of ammonia on a hot tungsten wire and for the decomposition of ammonia on hot quartz (SiO<sub>2<\/sub>). The decomposition of NH<sub>3<\/sub> on hot tungsten is zero order; the plot is a straight line. The decomposition of NH<sub>3<\/sub> on hot quartz is not zero order (it is first order). From the slope of the line for the zero-order decomposition, we can determine the rate constant:\r\n<p style=\"text-align: center;\">slope = \u2212<em>k<\/em> = 1.3110<sup>\u22126<\/sup> mol\/L\/s<\/p>\r\n\r\n<h2>The Half-Life of a Reaction<\/h2>\r\nThe <strong>half-life of a reaction (<em>t<\/em><sub>1\/2<\/sub>)<\/strong> is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide in Figure\u00a01 as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases from 1.000 <em>M<\/em> to 0.500 <em>M<\/em>. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 <em>M<\/em> to 0.250 <em>M<\/em>; during the third half-life, it decreases from 0.250 <em>M<\/em> to 0.125 <em>M<\/em>. The concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.\r\n<h3>First-Order Reactions<\/h3>\r\nWe can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\hfill \\mathrm{ln}\\frac{{\\left[A\\right]}_{0}}{\\left[A\\right]}&amp; =&amp; kt\\hfill \\\\ \\hfill t&amp; =&amp; \\mathrm{ln}\\frac{{\\left[A\\right]}_{0}}{\\left[A\\right]}\\times \\frac{1}{k}\\hfill \\end{array}[\/latex]<\/p>\r\nIf we set the time <em>t<\/em> equal to the half-life, [latex]{t}_{1\\text{\/}2},[\/latex] the corresponding concentration of <em>A<\/em> at this time is equal to one-half of its initial concentration. Hence, when [latex]t={t}_{1\\text{\/}2},[\/latex] [latex]\\left[A\\right]=\\frac{1}{2}{\\left[A\\right]}_{0}[\/latex].\r\n\r\nTherefore:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {t}_{1\\text{\/}2}&amp; =\\mathrm{ln}\\frac{{\\left[A\\right]}_{0}}{\\frac{1}{2}{\\left[A\\right]}_{0}}\\times \\frac{1}{k}\\hfill \\\\ &amp; =\\mathrm{ln}2\\times \\frac{1}{k}=0.693\\times \\frac{1}{k}\\hfill \\end{array}[\/latex]<\/p>\r\nThus:\r\n<p style=\"text-align: center;\">[latex]{t}_{1\\text{\/}2}=\\frac{0.693}{k}[\/latex]<\/p>\r\nWe can see that the half-life of a first-order reaction is inversely proportional to the rate constant <em>k<\/em>. A fast reaction (shorter half-life) will have a larger <em>k<\/em>; a slow reaction (longer half-life) will have a smaller <em>k<\/em>.\r\n<div class=\"textbox examples\">\r\n<h3>Example 5:\u00a0Calculation of a First-order Rate Constant using Half-Life<\/h3>\r\nCalculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 \u00b0C, using the data given in Figure\u00a05.\r\n\r\n[caption id=\"attachment_2208\" align=\"aligncenter\" width=\"627\"]<img class=\" wp-image-2208\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212256\/CNX_Chem_12_04_HPerDcmp1.jpg\" alt=\"A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, \u201c1.000 M, 0 s, and ( 0 h ).\u201d The second beaker contains a slightly lighter green substance and is labeled below as, \u201c0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).\u201d The third beaker contains an even lighter green substance and is labeled below as, \u201c0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).\u201d The fourth beaker contains a green tinted substance and is labeled below as, \u201c0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).\u201d The fifth beaker contains a colorless substance and is labeled below as, \u201c0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).\u201d\" width=\"627\" height=\"197\" \/> Figure\u00a05. The decomposition of H<sub>2<\/sub>O<sub>2<\/sub> (2H<sub>2<\/sub>O<sub>2<\/sub> \u2192 2H<sub>2<\/sub>O + O<sub>2<\/sub>) at 40 \u00b0C is illustrated. The intensity of the color symbolizes the concentration of H<sub>2<\/sub>O<sub>2<\/sub> at the indicated times; H<sub>2<\/sub>O<sub>2<\/sub> is actually colorless.[\/caption]\r\n\r\n[reveal-answer q=\"81889\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"81889\"]\r\n\r\nThe half-life for the decomposition of H<sub>2<\/sub>O<sub>2<\/sub> is 2.16 \u00d7 10<sup>4<\/sup> s:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {t}_{1\\text{\/}2}&amp; =&amp; \\frac{0.693}{k}\\hfill \\\\ \\hfill k&amp; =&amp; \\frac{0.693}{{t}_{1\\text{\/}2}}=\\frac{0.693}{2.16\\times {10}^{4}\\text{s}}=3.21\\times {10}^{-5}{\\text{s}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nThe first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d<sup>\u22121<\/sup>. What is the half-life for this decay?\r\n\r\n[reveal-answer q=\"409771\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"409771\"]5.02 d[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Second-Order Reactions<\/h3>\r\nWe can derive the equation for calculating the half-life of a second order as follows:\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\left[A\\right]}=kt+\\frac{1}{{\\left[A\\right]}_{0}}[\/latex]<\/p>\r\nor\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\left[A\\right]}-\\frac{1}{{\\left[A\\right]}_{0}}=kt[\/latex]<\/p>\r\nIf\u00a0[latex]t={t}_{1\\text{\/}2}[\/latex],\u00a0then\u00a0[latex]\\left[A\\right]=\\frac{1}{2}{\\left[A\\right]}_{0}[\/latex],\u00a0and we can write:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{\\frac{1}{2}{\\left[A\\right]}_{0}}-\\frac{1}{{\\left[A\\right]}_{0}}&amp; =&amp; k{t}_{1\\text{\/}2}\\hfill \\\\ \\hfill 2{\\left[A\\right]}_{0}-\\frac{1}{{\\left[A\\right]}_{0}}&amp; =&amp; k{t}_{1\\text{\/}2}\\hfill \\\\ \\hfill \\frac{1}{{\\left[A\\right]}_{0}}&amp; =&amp; k{t}_{1\\text{\/}2}\\hfill \\end{array}[\/latex]<\/p>\r\nThus:\r\n<p style=\"text-align: center;\">[latex]{t}_{1\\text{\/}2}=\\frac{1}{k{\\left[A\\right]}_{0}}[\/latex]<\/p>\r\nFor a second-order reaction, [latex]{t}_{1\\text{\/}2}[\/latex] is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.\r\n<h3>Zero-Order Reactions<\/h3>\r\nWe can derive an equation for calculating the half-life of a zero order reaction as follows:\r\n<p style=\"text-align: center;\">[latex]\\left[A\\right]={-}kt+{\\left[A\\right]}_{0}[\/latex]<\/p>\r\nWhen half of the initial amount of reactant has been consumed [latex]t={t}_{1\\text{\/}2}[\/latex] and [latex]\\left[A\\right]=\\frac{{\\left[A\\right]}_{0}}{2}.[\/latex] Thus:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{{\\left[\\text{A}\\right]}_{0}}{2}&amp; =&amp; {-}k{t}_{1\\text{\/}2}+{\\left[\\text{A}\\right]}_{0}\\hfill \\\\ \\hfill k{t}_{1\\text{\/}2}&amp; =&amp; \\frac{{\\left[\\text{A}\\right]}_{0}}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]{t}_{1\\text{\/}2}=\\text{}\\frac{{\\left[A\\right]}_{0}}{2k}[\/latex]<\/p>\r\nThe half-life of a zero-order reaction increases as the initial concentration increases.\r\n\r\nEquations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table 1.\r\n<table id=\"fs-idm117482272\" summary=\"This table contains four columns and seven rows. The first column and the first row both serve as headers. The first cell in the first column is blank and is followed by, \u201crate law,\u201d \u201cunits of rate constant,\u201d \u201cintegrated rate law,\u201d \u201cplot needed for linear fit of rate date,\u201d \u201crelationship between slope of linear plot and rate constant,\u201d and \u201chalf-life.\u201d The first row labels each column, \u201cZero-Order,\u201d \u201cFirst-Order,\u201d and \u201cSecond-Order.\u201d Under \u201cZero-order\u201d are the following: \u201crate = k,\u201d \u201cM s superscript negative 1,\u201d \u201c[ A ] = negative k t + [ A ] subscript 0,\u201d \u201c[ A ] v s. t,\u201d \u201ck = negative slope,\u201d and \u201ct subscript one half = [ A ] subscript 0 over 2 k.\u201d Under \u201cFirst-Order\u201d are the following: \u201crate = k [ A ],\u201d \u201cs superscript negative 1,\u201d \u201cl n [ A ] = negative k t plus l n [ A ] subscript 0,\u201d \u201cl n [ A ] v s. t,\u201d \u201ck = negative slope,\u201d and \u201ct subscript one half = 0.693 over k.\u201d Under \u201cSecond-Order\u201d are the following: \u201crate = k [ A ] superscript 2,\u201d \u201cM superscript negative 1 s superscript negative 1,\u201d \u201c1 over [ A ] = k t + ( 1 over [ A ] subscript 0 ),\u201d \u201c1 over [ A ] v s. t,\u201d \u201ck = positive slope,\u201d and \u201ct subscript one half \u2013 1 over [ A ] subscript 0 k.\u201d\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"4\">Table 1. Summary of Rate Laws for Zero-, First-, and Second-Order Reactions<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<th><\/th>\r\n<th>Zero-Order<\/th>\r\n<th>First-Order<\/th>\r\n<th>Second-Order<\/th>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>rate law<\/td>\r\n<td>rate = <em>k<\/em><\/td>\r\n<td>rate = <em>k<\/em>[A]<\/td>\r\n<td>rate = <em>k<\/em>[A]<sup>2<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>units of rate constant<\/td>\r\n<td><em>M<\/em> s<sup>\u22121<\/sup><\/td>\r\n<td>s<sup>\u22121<\/sup><\/td>\r\n<td><em>M<\/em><sup>\u22121<\/sup> s<sup>\u22121<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>integrated rate law<\/td>\r\n<td>[<em>A<\/em>] = \u2212<em>kt<\/em> + [<em>A<\/em>]0<\/td>\r\n<td>ln [<em>A<\/em>] = \u2212<em>kt<\/em> + ln[<em>A<\/em>]0<\/td>\r\n<td>[latex]\\frac{1}{\\left[A\\right]}=kt+\\left(\\frac{1}{{\\left[A\\right]}_{0}}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>plot needed for linear fit of rate data<\/td>\r\n<td>[<em>A<\/em>] vs. <em>t<\/em><\/td>\r\n<td>ln [<em>A<\/em>] vs. <em>t<\/em><\/td>\r\n<td>[latex]\\frac{1}{\\left[A\\right]}[\/latex] vs. <em>t<\/em><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>relationship between slope of linear plot and rate constant<\/td>\r\n<td><em>k<\/em> = \u2212slope<\/td>\r\n<td><em>k<\/em> = \u2212slope<\/td>\r\n<td><em>k<\/em> = +slope<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>half-life<\/td>\r\n<td>[latex]{t}_{1\\text{\/}2}=\\frac{{\\left[A\\right]}_{0}}{2k}[\/latex]<\/td>\r\n<td>[latex]{t}_{1\\text{\/}2}=\\frac{0.693}{k}[\/latex]<\/td>\r\n<td>[latex]{t}_{1\\text{\/}2}=\\frac{1}{{\\left[A\\right]}_{0}k}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nDifferential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction.\r\n\r\nThe half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>integrated rate law for zero-order reactions: [latex]\\left[A\\right]={-}kt+{\\left[A\\right]}_{0},[\/latex] [latex]{t}_{1\\text{\/}2}=\\frac{{\\left[A\\right]}_{0}}{2k}[\/latex]<\/li>\r\n \t<li>integrated rate law for first-order reactions: [latex]\\mathrm{ln}\\left[A\\right]={-}kt+{\\left[A\\right]}_{0},\\text{}{t}_{1\\text{\/}2}=\\frac{0.693}{k}[\/latex]<\/li>\r\n \t<li>integrated rate law for second-order reactions: [latex]\\frac{1}{\\left[A\\right]}=kt+\\frac{1}{{\\left[A\\right]}_{0}},[\/latex] [latex]{t}_{1\\text{\/}2}=\\frac{1}{{\\left[A\\right]}_{0}k}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of <em>A<\/em> at varying times.<\/li>\r\n \t<li>Use the data provided to graphically determine the order and rate constant of the following reaction: [latex]{\\text{SO}}_{2}{\\text{Cl}}_{2}\\rightarrow{\\text{SO}}_{2}+{\\text{Cl}}_{2}[\/latex]\r\n<table id=\"fs-idp120830736\" class=\"medium unnumbered\" summary=\"This table contains two columns and eight rows. The first column is labeled, \u201cTime ( s ),\u201d and the second column is labeled, \u201c[ S O subscript 2 C l subscript 2 ] ( M ).\u201d Under the \u201cTime ( s )\u201d column are the numbers: 0; 5.00 times ten to the third power; 1.00 times ten to the fourth power; 1.50 times ten to the fourth power; 2.50 times ten to the fourth power; 3.00 times ten to the fourth power; and 4.00 times ten to the fourth power. Under the \u201c[ S O subscript 2 C l subscript 2 ] ( M )\u201d column are the numbers: 0.100, 0.0896, 0.0802, 0.0719, 0.0577, 0.0517, and 0.0415.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Time (s)<\/th>\r\n<th>[SO<sub>2<\/sub>Cl<sub>2<\/sub>] (<em>M<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>0<\/td>\r\n<td>0.100<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>5.00 \u00d7 10<sup>3<\/sup><\/td>\r\n<td>0.0896<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.00 \u00d7 10<sup>4<\/sup><\/td>\r\n<td>0.0802<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.50 \u00d7 10<sup>4<\/sup><\/td>\r\n<td>0.0719<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.50 \u00d7 10<sup>4<\/sup><\/td>\r\n<td>0.0577<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3.00 \u00d7 10<sup>4<\/sup><\/td>\r\n<td>0.0517<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4.00 \u00d7 10<sup>4<\/sup><\/td>\r\n<td>0.0415<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>Use the data provided in a graphical method to determine the order and rate constant of the following reaction:[latex]2P\\rightarrow Q+W[\/latex]\r\n<table id=\"fs-idm142901952\" class=\"medium unnumbered\" summary=\"This table contains two columns and six rows. The first row is a header row, and it labels each column, \u201cTime ( s ), \u201c and \u201c[ P ] ( M ).\u201d Under the \u201cTime ( s )\u201d column are the numbers: 9.0, 13.0, 18.0, 22.0, and 25.0. Under the \u201c[ P ] ( M )\u201d column are the numbers: 1.077 times ten to the negative 3; 1.068 times ten to the negative 3; 1.055 times ten to the negative 3; 1.046 times ten to the negative 3; and 1.039 times ten to the negative 3.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Time (s)<\/th>\r\n<th>[P] (<em>M<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>9.0<\/td>\r\n<td>1.077 \u00d7 10<sup>-3<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>13.0<\/td>\r\n<td>1.068 \u00d7 10<sup>-3<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>18.0<\/td>\r\n<td>1.055 \u00d7 10<sup>-3<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>22.0<\/td>\r\n<td>1.046 \u00d7 10<sup>-3<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>25.0<\/td>\r\n<td>1.039 \u00d7 10<sup>-3<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>Pure ozone decomposes slowly to oxygen, [latex]{\\text{2O}}_{3}\\left(g\\right)\\rightarrow{\\text{3O}}_{2}\\left(g\\right).[\/latex] Use the data provided in a graphical method and determine the order and rate constant of the reaction.\r\n<table id=\"fs-idp72766816\" class=\"medium unnumbered\" summary=\"This table has two columns and eight rows. The first row is a header row, and it labels each column, \u201cTime ( h ),\u201d and, \u201c[ O subscript 3 ] ( M ).\u201d Under the \u201cTime ( h )\u201d column are the numbers: 0; 2.0 times ten to the third power; 7.6 times ten to the third power; 1.00 times ten to the fourth power; 1.23 times ten to the fourth power; 1.43 times ten to the fourth power; 1.70 times ten to the fourth power. Under the \u201c[ O subscript 3 ] ( M )\u201d column are the numbers: 1.00 times ten to the negative 5; 4.98 times ten to the negative 6; 2.07 times ten to the negative 6; 1.66 times ten to the negative 6; 1.39 times ten to the negative 6; 1.22 times ten to the negative 6; and 1.05 times ten to the negative 6.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Time (h)<\/th>\r\n<th>[O<sub>3<\/sub>] (<em>M<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>0<\/td>\r\n<td>1.00 \u00d7 10<sup>\u22125<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.0 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<td>4.98 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>7.6 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<td>2.07 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.00 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<td>1.66 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.23 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<td>1.39 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.43 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<td>1.22 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.70 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<td>1.05 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>From the given data, use a graphical method to determine the order and rate constant of the following reaction:[latex]2X\\rightarrow Y+Z[\/latex]\r\n<table id=\"fs-idm133654768\" class=\"medium unnumbered\" summary=\"This table contains two columns and nine rows. The first row is a header row, and it labels each column, \u201cTime ( s ),\u201d and \u201c[ X ] ( M ).\u201d Under the \u201cTime ( s )\u201d column are the numbers: 5.0, 10.0, 15.0, 20.0, 25.0, 30.0, 35.0, and 40.0. Under the \u201c[ X ] ( M )\u201d column are the numbers; 0.0990, 0.497, 0.0332, 0.0249, 0.0200, 0.0166, 0.0143, and 0.0125.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Time (s)<\/th>\r\n<th>[<em>X<\/em>] (<em>M<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>5.0<\/td>\r\n<td>0.0990<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>10.0<\/td>\r\n<td>0.0497<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>15.0<\/td>\r\n<td>0.0332<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>20.0<\/td>\r\n<td>0.0249<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>25.0<\/td>\r\n<td>0.0200<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>30.0<\/td>\r\n<td>0.0166<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>35.0<\/td>\r\n<td>0.0143<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>40.0<\/td>\r\n<td>0.0125<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>What is the half-life for the first-order decay of phosphorus-32 [latex]\\left({}_{15}^{32}\\text{P}\\rightarrow{}_{16}^{32}\\text{S}+{\\text{e}}^{-}\\right)[\/latex] The rate constant for the decay is 4.85 \u00d7 10<sup>\u22122<\/sup> day <sup>\u22121<\/sup>.<\/li>\r\n \t<li>What is the half-life for the first-order decay of carbon-14? [latex]\\left({}_{14}^{6}\\text{C}\\rightarrow{}_{14}^{7}\\text{N}+{\\text{e}}^{-}\\right)[\/latex] The rate constant for the decay is 1.21 \u00d7 10<sup>\u22124<\/sup> year<sup>\u22121<\/sup>.<\/li>\r\n \t<li>What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 <em>M<\/em>? The rate constant for this second-order reaction is 8.0 \u00d7 10<sup>-8<\/sup> L\/mol\/s.<\/li>\r\n \t<li>What is the half-life for the decomposition of O<sub>3<\/sub> when the concentration of O<sub>3<\/sub> is 2.35 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>? The rate constant for this second-order reaction is 50.4 L\/mol\/h.<\/li>\r\n \t<li>The reaction of compound <em>A<\/em> to give compounds <em>C<\/em> and <em>D<\/em> was found to be second-order in <em>A<\/em>. The rate constant for the reaction was determined to be 2.42 L\/mol\/s. If the initial concentration is 0.500 mol\/L, what is the value of t<sub>1\/2<\/sub>?<\/li>\r\n \t<li>The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol\/L. How long will it take for the concentration to drop to 0.0300 mol\/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?<\/li>\r\n \t<li>Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 \u00d7 10<sup>4<\/sup> g\/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 \u00b5g (0.15 \u00d7 10<sup>\u22126<\/sup> g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.\r\n<table id=\"fs-idp79245440\" class=\"medium unnumbered\" summary=\"This table contains two columns and four rows. The first row is a header row, and it labels each column, \u201c[ Penicillin ] ( M ),\u201d and, \u201cRate ( mol \/ L \/ min ).\u201d Under the \u201c[ Penicillin ] ( M )\u201d column are the numbers: 2.0 times ten to the negative six; 3.0 times ten to the negative six; and 4.0 times ten to the negative 6. Under the \u201cRate ( mol \/ L \/ min )\u201d column are the numbers: 1.0 times ten to the negative ten; 1.5 times ten to the negative 10; and 2.0 times ten to the negative 10.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[Penicillin] (<em>M<\/em>)<\/th>\r\n<th>Rate (mol\/L\/min)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>2.0 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td>1.0 \u00d7 10<sup>\u221210<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3.0 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td>1.5 \u00d7 10<sup>\u221210<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4.0 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td>2.0 \u00d7 10<sup>\u221210<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?<\/li>\r\n \t<li>There are two molecules with the formula C<sub>3<\/sub>H<sub>6<\/sub>. Propene, CH<sub>3<\/sub>CH=CH<sub>2,<\/sub> is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:\r\n<img class=\"alignnone size-full wp-image-2213\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212304\/CNX_Chem_12_04_Cycloprop_img1.jpg\" alt=\"A structural formula for cyclopropane is shown. Three C H subscript 2 groups are positioned as vertices of an equilateral triangle connected with single bonds represented by line segments.\" width=\"272\" height=\"180\" \/>\r\nWhen heated to 499 \u00b0C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of [latex]5.95\\times {10}^{-4}{s}^{-1}[\/latex].\u00a0What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499.5 \u00b0C?<\/li>\r\n \t<li>Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is [latex]{}_{9}^{18}\\text{F}\\rightarrow{}_{8}^{18}\\text{O}+{\\text{e}}^{+}.[\/latex] ) Physicians use <sup>18<\/sup>F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>What is the rate constant for the decomposition of fluorine-18?<\/li>\r\n \t<li>If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?<\/li>\r\n \t<li>How long does it take for 99.99% of the <sup>18<\/sup>F to decay?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for [latex]\\frac{1}{64}[\/latex] of the initial dose to remain in the athlete\u2019s body?<\/li>\r\n \t<li>Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.<\/li>\r\n \t<li>Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 \u00b0C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:\r\n<table id=\"fs-idp121895360\" class=\"medium unnumbered\" summary=\"This table contains three columns and nine rows. The first row is a header row, and it labels each column, \u201cInitial [ C subscript 3 H subscript 5 N subscript 3 O subscript 9 ] ( M ),\u201d \u201ct ( s ),\u201d and \u201c% Decomposed.\u201d Under the \u201cInitial [ C subscript 3 H subscript 5 N subscript 3 O subscript 9 ] ( M )\u201d column are the numbers: 4.88, 3.52, 2.29, 1.81, 5.33, 4.05, 2.95, and 1.72. Under the \u201ct ( s )\u201d column are the numbers: 300, 300, 300, 300, 180, 180, 180, and 180. Under the \u201c% Decomposed\u201d column are the numbers: 52.0, 52.9, 53.2, 53.9, 34.6, 35.9, 36.0, and 35.4.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Initial [C<sub>3<\/sub>H<sub>5<\/sub>N<sub>3<\/sub>O<sub>9<\/sub>] (M)<\/th>\r\n<th><em>t<\/em> (s)<\/th>\r\n<th>% Decomposed<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>4.88<\/td>\r\n<td>300<\/td>\r\n<td>52.0<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3.52<\/td>\r\n<td>300<\/td>\r\n<td>52.9<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.29<\/td>\r\n<td>300<\/td>\r\n<td>53.2<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.81<\/td>\r\n<td>300<\/td>\r\n<td>53.9<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>5.33<\/td>\r\n<td>180<\/td>\r\n<td>34.6<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4.05<\/td>\r\n<td>180<\/td>\r\n<td>35.9<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.95<\/td>\r\n<td>180<\/td>\r\n<td>36.0<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.72<\/td>\r\n<td>180<\/td>\r\n<td>35.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene (CH<sub>2<\/sub>=CH\u2013CH=CH<sub>2<\/sub>) has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:\r\n<img class=\"alignnone size-full wp-image-2214\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212305\/CNX_Chem_12_04_ExSolutio2_img1.jpg\" alt=\"A structural formula for cyclobutene is shown. The figure has two C H subscript 2 groups as the upper two vertices of a square structure. These groups are connected by a single, short line segment. Line segments extend below each of these C H subscript 2 groups to C H groups positioned at the lower two vertices of the square structure. The C H groups are connected with a double line segment indicating a double bond.\" width=\"220\" height=\"156\" \/>\r\nThe isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as 2.0 \u00d7 10<sup>\u22124<\/sup> s<sup>\u22121<\/sup> at 150\u00b0C in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150\u00b0C with an initial pressure of 55 torr.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"658620\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"658620\"]\r\n\r\n2.\u00a0Plotting a graph of ln[SO<sub>2<\/sub>Cl<sub>2<\/sub>] versus <em>t<\/em> reveals a linear trend; therefore we know this is a first-order reaction:\r\n\r\n<img class=\"alignnone size-full wp-image-5382\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09033827\/Graph.png\" alt=\"Graph of ln SO2Cl2 M and time. A line connects the following points: coordinates negative 2.303, zero; coordinates negative 2.412, 5 times 10 to the power of three; coordinates negative 2.523, 1 times 10 to the power of 4; coordinates negative 2.632, 1.5 times 10 to the power of 4; coordinates negative 2.852, 2.5 times 10 to the power of 4; coordinates negative 2.962 times 3 times 10 to the power of 4; coordinates negative 3.182, 4 times 10 to the power of 4.\" width=\"471\" height=\"351\" \/>\r\n\r\nThe value of <em>k<\/em> is found from the slope of the line since [latex]\\mathrm{ln}\\left[A\\right]={-}kt+\\mathrm{ln}{\\left[A\\right]}_{0}[\/latex] is in the form of a straight line, y = <em>mx<\/em> + <em>b<\/em>.\r\n\r\n[latex]\\frac{\\Delta y}{\\Delta x}=\\frac{\\text{ln}0.0896-\\text{ln}0.100}{5.00\\times {10}^{3}-0\\text{s}}=\\frac{-0.1098}{5.00\\times {10}^{3}\\text{s}}=-2.20\\times {10}^{5}{\\text{s}}^{-1}[\/latex]\r\n\r\n4.\u00a0To distinguish a first-order reaction from a second-order reaction, we plot ln[<em>P<\/em>] against <em>t<\/em> and compare that plot with a plot of [latex]\\frac{1}{\\left[P\\right]}[\/latex] versus <em>t<\/em>. The values needed for these plots are abbreviated to include only the data needed for a second-order plot, as the data do not seem to support a first-order reaction:\r\n<table id=\"fs-idm111039952\" class=\"medium unnumbered\" summary=\"This table contains one column and eight rows. The first row is a header row and it labels the column, \u201c1 over [ O subscript 3 ] ( M superscript negative 1 ).\u201d Under this column are the numbers: 1.00 times ten to the fifth power; 2.01 times ten to the fifth power; 4.83 times ten to the fifth power; 6.02 times ten to the fifth power; 7.19 times ten to the fifth power; 8.20 times ten to the fifth power; 9.52 times ten to the fifth power.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]\\frac{1}{\\left[{\\text{O}}_{3}\\right]}\\text{(}{M}^{-1}\\text{)}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>1.00 \u00d7 10<sup>5<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.01 \u00d7 10<sup>5<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4.83 \u00d7 10<sup>5<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>6.02 \u00d7 10<sup>5<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>7.19 \u00d7 10<sup>5<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>8.20 \u00d7 10<sup>5<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>9.52 \u00d7 10<sup>5<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe plot is nicely linear, so the reaction is second order.\r\n<img class=\"alignnone size-full wp-image-5383\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09033910\/Graph2.png\" alt=\"Graph of 1 over O3 M and time in hours. The rise is delta 1 over O3. The run is delta t.\" width=\"316\" height=\"426\" \/>\r\n[latex]\\text{slope}=\\frac{9.52\\times {10}^{5}-1.00\\times {10}^{5}}{17\\times {10}^{3}-0}=50.1{\\text{L mol}}^{-1}{\\text{h}}^{-1}[\/latex]\r\n\r\n6. \u00a0The half-life is [latex]{t}_{1\\text{\/}2}=\\frac{0.693}{k},[\/latex] where <em>k<\/em> is the rate constant:\r\n\r\n[latex]k=\\frac{0.693}{{t}_{1\\text{\/}2}}=4.85\\times {10}^{-2}{\\text{d}}^{-1}[\/latex]\r\n\r\n[latex]{t}_{1\\text{\/}2}=\\frac{0.693}{4.85\\times {10}^{-2}{\\text{d}}^{-1}}=\\text{14.3 d}[\/latex]\r\n\r\n8.\u00a0In a second-order reaction, the rate is concentration-dependent, [latex]{t}_{1\\text{\/}2}=\\frac{1}{k{\\left[A\\right]}_{0}}.[\/latex]\r\n\r\n[latex]{t}_{1\\text{\/}2}=\\frac{1}{k{\\left[A\\right]}_{0}}=\\frac{1}{8.0\\times {10}^{-8}{\\text{L mol}}^{-1}{\\text{s}}^{-1}\\left[0.15M\\right]}=8.3\\times {10}^{7}\\text{s}[\/latex]\r\n\r\n10.\u00a0For a second-order reaction, the half-life is concentration-dependent:\r\n\r\n[latex]{t}_{1\\text{\/}2}=\\frac{1}{k{\\left[A\\right]}_{0}}=\\frac{1}{2.42{\\text{L mol}}^{-1}{\\text{s}}^{-1}\\times 0.500{\\text{mol L}}^{-1}}=0.826\\text{s}[\/latex]\r\n\r\n12.\u00a0The reaction is first order with respect to penicillinase, and the rate doubles as [penicillin] doubles. Thus the rate equation is:\r\n\r\nrate = <em>k<\/em>[penicillinase][penicillin]\r\n\r\nUsing the data in the first row,\r\n\r\n[latex]k=\\frac{1.0\\times {10}^{-10}{\\text{mol L}}^{-1}{\\text{min}}^{-1}}{\\left(\\frac{0.15\\times {10}^{-6}{\\text{g L}}^{-1}}{3.0\\times {10}^{4}{\\text{g mol}}^{-1}}\\right)\\left(2.0\\times {10}^{-6}{\\text{mol L}}^{-1}\\right)}=1.0\\times {10}^{7}{\\text{mol}}^{-1}{\\text{min}}^{-1}[\/latex]\r\n\r\n14.\u00a0The unit of the provided rate constant indicates the reaction is first order. Using the integrated form of the first-order rate law, we have:\r\n\r\n[latex]\\text{ln}\\frac{{\\left[A\\right]}_{0}}{\\left[A\\right]}{-}kt[\/latex]\r\n\r\nLet [<em>A<\/em>]<sub>0<\/sub> = 1 , then:\r\n\r\n[latex]\\text{ln}\\frac{\\left[1\\right]}{\\left[A\\right]}=5.95\\times {10}^{-4}{\\cancel{\\text{s}}}^{-1}\\times 0.75\\cancel{\\text{h}}\\times 60\\frac{\\cancel{\\text{min}}}{\\cancel{\\text{h}}}\\times 60\\frac{\\cancel{\\text{s}}}{\\cancel{\\text{min}}}=1.606[\/latex]\r\n\r\nConvert 1.606, a natural log, to the corresponding number by taking the <em>e<\/em>s of both sides:\r\n\r\n[latex]\\frac{1}{\\left[\\text{A}\\right]}=4.98\\text{[A]}=0.20,\\text{so 20% remains}.[\/latex]\r\n\r\n16.\u00a0[latex]\\frac{1}{64}=\\frac{1}{{2}^{x}}[\/latex] where <em>x<\/em> represents the number of half-life periods <em>x<\/em> = 6, so (6)(42) = 252 days.\r\n\r\n18.\u00a0From the first-order rate law, calculate the value of [<em>A<\/em>], [latex]\\text{ln}\\left(\\frac{{\\left[\\text{A}\\right]}_{0}}{\\left[\\text{A}\\right]}\\right),[\/latex] and <em>k<\/em>. The values are tabulated:\r\n<table id=\"fs-idp20979328\" class=\"medium unnumbered\" summary=\"This table has five columns and nine rows. The first row is a header row, and it labels each column: \u201c[ A ] subscript 0 ( M ),\u201d \u201c[ A ] ( M ),\u201d \u201cl n ( [ A ] subscript 0 over [ A ] ),\u201d \u201ct ( s ),\u201d and \u201ck times 10 to the third power ( s superscript negative 1 ).\u201d Under the \u201c[ A ] subscript 0 ( M )\u201d column are the numbers: 4.88, 3.52, 2.29, 1.81, 5.33, 4.05, 2.95, and 1.72. Under the \u201c[ A ] ( M )\u201d column are the numbers: 2.34, 1.66, 1.07, 0.834, 3.49, 2.61, 1.89, and 1.11. Under the \u201cl n ( [ A ] subscript 0 over [ A ] )\u201d column are the numbers: 0.734, 0.752, 0.761, 0.775, 0.423, 0.439, 0.445, and 0.438. Under the \u201ct ( s )\u201d column are the numbers: 300, 300, 300, 300, 180, 180, 180, and 180. Under the \u201ck times 10 to the third power ( s superscript negative 1 )\u201d columns are the numbers: 2.45, 2.51, 2.54, 2.58, 2.35, 2.44, 2.47, and 2.43.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[A]<sub>0<\/sub> (<em>M<\/em>)<\/th>\r\n<th>[A] (<em>M<\/em>)<\/th>\r\n<th>[latex]\\text{ln}\\left(\\frac{{\\left[\\text{A}\\right]}_{0}}{\\left[\\text{A}\\right]}\\right)[\/latex]<\/th>\r\n<th><em>t<\/em> (s)<\/th>\r\n<th><em>k<\/em> \u00d7 10<sup>3<\/sup> (s<sup>\u22121<\/sup>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>4.88<\/td>\r\n<td>2.34<\/td>\r\n<td>0.734<\/td>\r\n<td>300<\/td>\r\n<td>2.45<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3.52<\/td>\r\n<td>1.66<\/td>\r\n<td>0.752<\/td>\r\n<td>300<\/td>\r\n<td>2.51<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.29<\/td>\r\n<td>1.07<\/td>\r\n<td>0.761<\/td>\r\n<td>300<\/td>\r\n<td>2.54<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.81<\/td>\r\n<td>0.834<\/td>\r\n<td>0.775<\/td>\r\n<td>300<\/td>\r\n<td>2.58<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>5.33<\/td>\r\n<td>3.49<\/td>\r\n<td>0.423<\/td>\r\n<td>180<\/td>\r\n<td>2.35<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4.05<\/td>\r\n<td>2.61<\/td>\r\n<td>0.439<\/td>\r\n<td>180<\/td>\r\n<td>2.44<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.95<\/td>\r\n<td>1.89<\/td>\r\n<td>0.445<\/td>\r\n<td>180<\/td>\r\n<td>2.47<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>1.72<\/td>\r\n<td>1.11<\/td>\r\n<td>0.438<\/td>\r\n<td>180<\/td>\r\n<td>2.43<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>half-life of a reaction (<em>t<\/em><sub>l\/2<\/sub>): <\/strong>time required for half of a given amount of reactant to be consumed\r\n\r\n<strong>integrated rate law: <\/strong>equation that relates the concentration of a reactant to elapsed time of reaction","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this module, you will be able to:<\/p>\n<ul>\n<li>Explain the form and function of an integrated rate law<\/li>\n<li>Perform integrated rate law calculations for zero-, first-, and second-order reactions<\/li>\n<li>Define half-life and carry out related calculations<\/li>\n<li>Identify the order of a reaction from concentration\/time data<\/li>\n<\/ul>\n<\/div>\n<p>The rate laws we have seen thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.<\/p>\n<p>Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.<\/p>\n<h2>First-Order Reactions<\/h2>\n<p>An equation relating the rate constant <em>k<\/em> to the initial concentration [<em>A<\/em>]<sub>0<\/sub> and the concentration [<em>A<\/em>]<sub>t<\/sub> present after any given time <em>t<\/em> can be derived for a first-order reaction and shown to be:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{ln}\\left(\\frac{{\\left[A\\right]}_{t}}{{\\left[A\\right]}_{0}}\\right)=-kt[\/latex]<\/p>\n<p>or<\/p>\n<p style=\"text-align: center;\">[latex]\\text{ln}\\left(\\frac{{\\left[A\\right]}_{0}}{{\\left[A\\right]}_{t}}\\right)=kt[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0The Integrated Rate Law for a First-Order Reaction<\/h3>\n<p>The rate constant for the first-order decomposition of cyclobutane, C<sub>4<\/sub>H<sub>8<\/sub> at 500 \u00b0C is 9.2 \u00d7 10<sup>\u22123<\/sup> s<sup>\u22121<\/sup>:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{4}{\\text{H}}_{8}\\rightarrow{\\text{2C}}_{2}{\\text{H}}_{4}[\/latex]<\/p>\n<p>How long will it take for 80.0% of a sample of C<sub>4<\/sub>H<sub>8<\/sub> to decompose?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q541265\">Show Answer<\/span><\/p>\n<div id=\"q541265\" class=\"hidden-answer\" style=\"display: none\">\n<p>We use the integrated form of the rate law to answer questions regarding time:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{ln}\\left(\\frac{{\\left[A\\right]}_{0}}{\\left[A\\right]}\\right)=kt[\/latex]<\/p>\n<p>There are four variables in the rate law, so if we know three of them, we can determine the fourth. In this case we know [<em>A<\/em>]<sub>0<\/sub>, [<em>A<\/em>], and <em>k<\/em>, and need to find <em>t<\/em>.<\/p>\n<p>The initial concentration of C<sub>4<\/sub>H<sub>8<\/sub>, [<em>A<\/em>]<sub>0<\/sub>, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let <em>x<\/em> be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of <em>x<\/em> or 0.200<em>x.<\/em> Rearranging the rate law to isolate <em>t<\/em> and substituting the provided quantities yields:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill t& =\\mathrm{ln}\\frac{\\left[x\\right]}{\\left[0.200x\\right]}\\times \\frac{1}{k}\\hfill \\\\ & =\\text{ln}\\frac{0.100{\\text{mol L}}^{-1}}{0.020{\\text{mol L}}^{-1}}\\times \\frac{1}{9.2\\times {10}^{-3}{\\text{s}}^{-1}}\\hfill \\\\ & =1.609\\times \\frac{1}{9.2\\times {10}^{-3}{\\text{s}}^{-1}}\\hfill \\\\ & =1.7\\times {10}^{2}\\text{s}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{I-131}\\rightarrow\\text{Xe-131}+\\text{electron}[\/latex]<\/p>\n<p>The decay is first-order with a rate constant of 0.138 d<sup>\u22121<\/sup>. All radioactive decay is first order. How many days will it take for 90% of the iodine-131 in a 0.500 <em>M<\/em> solution of this substance to decay to Xe-131?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q531173\">Show Answer<\/span><\/p>\n<div id=\"q531173\" class=\"hidden-answer\" style=\"display: none\">16.7 days<\/div>\n<\/div>\n<\/div>\n<p>We can use integrated rate laws with experimental data that consist of time and concentration information to determine the order and rate constant of a reaction. The integrated rate law can be rearranged to a standard linear equation format:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}\\left[A\\right]& =& \\left(\\text{-}k\\right)\\left(t\\right)+\\text{ln}{\\left[A\\right]}_{0}\\hfill \\\\ \\hfill y& =& mx+b\\hfill \\end{array}[\/latex]<\/p>\n<p>A plot of ln[<em>A<\/em>] versus <em>t<\/em> for a first-order reaction is a straight line with a slope of &#8211;<em>k<\/em> and an intercept of ln[<em>A<\/em>]<sub>0<\/sub>. If a set of rate data are plotted in this fashion but do <em>not<\/em> result in a straight line, the reaction is not first order in <em>A<\/em>.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Determination of Reaction Order by Graphing<\/h3>\n<p>Show that the data in Figure\u00a01 can be represented by a first-order rate law by graphing ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time. Determine the rate constant for the rate of decomposition of H<sub>2<\/sub>O<sub>2<\/sub> from this data.<\/p>\n<div id=\"attachment_5373\" style=\"width: 916px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5373\" class=\"size-full wp-image-5373\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09033022\/Table2.png\" alt=\"A table with five columns is shown. The first column is labeled, \u201cTime, h.\u201d Beneath it the numbers 0.00, 6.00, 12.00, 18.00, and 24.00 are listed. The second column is labeled, \u201c[ H subscript 2 O subscript 2 ], mol \/ L.\u201d Below, the numbers 1.000, 0.500, 0.250, 0.125, and 0.0625 are double spaced. To the right, a third column is labeled, \u201ccapital delta [ H subscript 2 O subscript 2 ], mol \/ L.\u201d Below, the numbers negative 0.500, negative 0.250, negative 0.125, and negative 0.062 are listed such that they are double spaced and offset, beginning one line below the first number listed in the column labeled, \u201c[ H subscript 2 O subscript 2 ], mol \/ L.\u201d The first two numbers in the second column have line segments extending from their right side to the left side of the first number in the third row. The second and third numbers in the second column have line segments extending from their right side to the left side of the second number in the third row. The third and fourth numbers in the second column have line segments extending from their right side to the left side of the third number in the third row. The fourth and fifth numbers in the second column have line segments extending from their right side to the left side of the fourth number in the third row. The fourth column in labeled, \u201ccapital delta t, h.\u201d Below the title, the value 6.00 is listed four times, each single-spaced. The fifth and final column is labeled \u201cRate of Decomposition, mol \/ L \/ h.\u201d Below, the following values are listed single-spaced: negative 0.0833, negative 0.0417, negative 0.0208, and negative 0.0103.\" width=\"906\" height=\"292\" \/><\/p>\n<p id=\"caption-attachment-5373\" class=\"wp-caption-text\">Figure\u00a01. The rate of decomposition of H<sub>2<\/sub>O<sub>2<\/sub> in an aqueous solution decreases as the concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q559973\">Show Answer<\/span><\/p>\n<div id=\"q559973\" class=\"hidden-answer\" style=\"display: none\">The data from Figure\u00a01 with the addition of values of ln[H<sub>2<\/sub>O<sub>2<\/sub>] are given in Figure\u00a02.<\/p>\n<div id=\"attachment_4849\" style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4849\" class=\"wp-image-4849\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214429\/FrstOKin.jpg\" alt=\"A graph is shown with the label \u201cTime ( h )\u201d on the x-axis and \u201cl n &#091; H subscript 2 O subscript 2 &#093;\u201d on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).\" width=\"400\" height=\"286\" \/><\/p>\n<p id=\"caption-attachment-4849\" class=\"wp-caption-text\">Figure\u00a02. The linear relationship between the ln[H<sub>2<\/sub>O<sub>2<\/sub>] and time shows that the decomposition of hydrogen peroxide is a first-order reaction.<\/p>\n<\/div>\n<table id=\"fs-idp114347648\" class=\"medium unnumbered\" summary=\"This table contains four columns and six rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( h ),\u201d \u201c&#091; H subscript 2 O subscript 2 &#093; ( M ),\u201d and \u201cl n &#091; H subscript 2 O subscript 2 &#093;.\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the column, \u201cTime ( h )\u201d are the numbers 0, 6.00, 12.00, 18.00, and 24.00. Under the column \u201c&#091; H subscript 2 O subscript 2 &#093; ( M ),\u201d are the numbers 1.000, 0.500, 0.250, 0.125, and 0.0625. Under the column, \u201cl n &#091; H subscript 2 O subscript 2 &#093;,\u201d are the numbers: 0.0, negative 0.693, negative 1.386, negative 2.079, and negative 2.772.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Trial<\/th>\n<th>Time (h)<\/th>\n<th>[H<sub>2<\/sub>O<sub>2<\/sub>] (<em>M<\/em>)<\/th>\n<th>ln[H<sub>2<\/sub>O<sub>2<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>1<\/td>\n<td>0<\/td>\n<td>1.000<\/td>\n<td>0.0<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>6.00<\/td>\n<td>0.500<\/td>\n<td>\u20130.693<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3<\/td>\n<td>12.00<\/td>\n<td>0.250<\/td>\n<td>\u20131.386<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4<\/td>\n<td>18.00<\/td>\n<td>0.125<\/td>\n<td>\u20132.079<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>5<\/td>\n<td>24.00<\/td>\n<td>0.0625<\/td>\n<td>\u20132.772<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The plot of ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.<\/p>\n<p>The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time where:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{slope}=\\frac{\\text{change in }y}{\\text{change in }x}=\\frac{\\Delta y}{\\Delta x}=\\frac{\\Delta\\text{ ln}\\left[{\\text{H}}_{2}{\\text{O}}_{2}\\right]}{\\Delta t}[\/latex]<\/p>\n<p>In order to determine the slope of the line, we need two values of ln[H<sub>2<\/sub>O<sub>2<\/sub>] at different values of <em>t<\/em> (one near each end of the line is preferable). For example, the value of ln[H<sub>2<\/sub>O<sub>2<\/sub>] when <em>t<\/em> is 6.00 h is -0.693; the value when <em>t<\/em> = 12.00 h is \u22121.386:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{slope}& =& \\frac{-1.386-\\left(-0.693\\right)}{\\text{12.00 h}-\\text{6.00 h}}\\hfill \\\\ & =& \\frac{-0.693}{\\text{6.00 h}}\\hfill \\\\ & =& -1.155\\times {10}^{-2}{\\text{h}}^{-1}\\hfill \\\\ \\hfill k& =& -\\text{slope}=-\\left(-1.155\\times {10}^{-2}{\\text{h}}^{-1}\\right)=1.155\\times {10}^{-2}{\\text{h}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Graph the following data to determine whether the reaction [latex]A\\rightarrow B+C[\/latex] is first order.<\/p>\n<table id=\"fs-idm149704608\" class=\"medium unnumbered\" summary=\"This table has three columns and six rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and, \u201c[ A ].\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201cTime ( s )\u201d column are the numbers: 4.0, 8.0, 12.0, 16.0, and 20.0. Under the \u201c [ A ]\u201d column are the numbers: 0.220, 0.144, 0.110, 0.088, and 0.074.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Trial<\/th>\n<th>Time (s)<\/th>\n<th>[<em>A<\/em>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>1<\/td>\n<td>4.0<\/td>\n<td>0.220<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>8.0<\/td>\n<td>0.144<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3<\/td>\n<td>12.0<\/td>\n<td>0.110<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4<\/td>\n<td>16.0<\/td>\n<td>0.088<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>5<\/td>\n<td>20.0<\/td>\n<td>0.074<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q663896\">Show Answer<\/span><\/p>\n<div id=\"q663896\" class=\"hidden-answer\" style=\"display: none\">The plot of ln[<em>A<\/em>] vs. <em>t<\/em> is not a straight line. The equation is not first order:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2204 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212251\/CNX_Chem_12_04_CYL1_img1.jpg\" alt=\"A graph, labeled above as \u201cl n [ A ] vs. Time\u201d is shown. The x-axis is labeled, \u201cTime ( s )\u201d and the y-axis is labeled, \u201cl n [ A ].\u201d The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).\" width=\"400\" height=\"232\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Second-Order Reactions<\/h2>\n<p>The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant\u2019s concentration and described by the differential rate law:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Rate}=k{\\left[A\\right]}^{2}[\/latex]<\/p>\n<p>For these second-order reactions, the integrated rate law is:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\left[A\\right]}=kt+\\frac{1}{{\\left[A\\right]}_{0}}[\/latex]<\/p>\n<p>where the terms in the equation have their usual meanings as defined above.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0The Integrated Rate Law for a Second-Order Reaction<\/h3>\n<p>The reaction of butadiene gas (C<sub>4<\/sub>H<sub>6<\/sub>) with itself produces C<sub>8<\/sub>H<sub>12<\/sub> gas as follows:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{2C}}_{4}{\\text{H}}_{6}\\text{(}g\\text{)}\\rightarrow{\\text{C}}_{8}{\\text{H}}_{12}\\text{(}g\\text{)}[\/latex]<\/p>\n<p>The reaction is second order with a rate constant equal to 5.76 \u00d7 10<sup>\u22122<\/sup> L\/mol\/min under certain conditions. If the initial concentration of butadiene is 0.200 <em>M<\/em>, what is the concentration remaining after 10.0 min?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q768114\">Show Answer<\/span><\/p>\n<div id=\"q768114\" class=\"hidden-answer\" style=\"display: none\">\n<p>We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\left[A\\right]}=kt+\\frac{1}{{\\left[A\\right]}_{0}}[\/latex]<\/p>\n<p>We know three variables in this equation: [<em>A<\/em>]<sub>0<\/sub> = 0.200 mol\/L, <em>k<\/em> = 5.76 \u00d7 10<sup>\u22122<\/sup> L\/mol\/min, and <em>t<\/em> = 10.0 min. Therefore, we can solve for [<em>A<\/em>], the fourth variable:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{\\left[A\\right]}& =& \\left(5.76\\times {10}^{-2}{\\text{L mol}}^{-1}{\\mathrm{min}}^{-1}\\right)\\left(10\\text{min}\\right)+\\frac{1}{0.200{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill \\frac{1}{\\left[A\\right]}& =& \\left(5.76\\times {10}^{-1}{\\text{L mol}}^{-1}\\right)+5.00{\\text{L mol}}^{-1}\\hfill \\\\ \\hfill \\frac{1}{\\left[A\\right]}& =& 5.58{\\text{L mol}}^{-1}\\hfill \\\\ \\hfill \\left[A\\right]& =& 1.79\\times {10}^{-1}{\\text{mol L}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore 0.179 mol\/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol\/L that was originally present.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>If the initial concentration of butadiene is 0.0200 <em>M<\/em>, what is the concentration remaining after 20.0 min?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q741332\">Show Answer<\/span><\/p>\n<div id=\"q741332\" class=\"hidden-answer\" style=\"display: none\">0.0196 mol\/L<\/div>\n<\/div>\n<\/div>\n<p>The integrated rate law for our second-order reactions has the form of the equation of a straight line:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{\\left[A\\right]}& =& kt+\\frac{1}{{\\left[A\\right]}_{0}}\\hfill \\\\ \\hfill y& =& mx+b\\hfill \\end{array}[\/latex]<\/p>\n<p>A plot of [latex]\\frac{1}{\\left[A\\right]}[\/latex] versus <em>t<\/em> for a second-order reaction is a straight line with a slope of <em>k<\/em> and an intercept of [latex]\\frac{1}{{\\left[A\\right]}_{0}}[\/latex]. If the plot is not a straight line, then the reaction is not second order.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0Determination of Reaction Order by Graphing<\/h3>\n<p>Test the data given to show whether the dimerization of C<sub>4<\/sub>H<sub>6<\/sub> is a first- or a second-order reaction.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q186612\">Show Answer<\/span><\/p>\n<div id=\"q186612\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"fs-idm140502592\" class=\"medium unnumbered\" summary=\"This table contains three columns and six rows. The first row is a header row and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and \u201c&#091; C subscript 4 H subscript 6 &#093; ( M ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201cTime ( s )\u201d column are the numbers: 0, 1600, 3200, 4800, and 6200. Under the column \u201c&#091; C subscript 4 H subscript 6 &#093; ( M )\u201d are the numbers: 1.00 times ten to the negative 2; 5.04 times ten to the negative 3; 3.37 times ten to the negative 3; 2.53 times ten to the negative 3; and 2.08 times ten to the negative 3.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Trial<\/th>\n<th>Time (s)<\/th>\n<th>[C<sub>4<\/sub>H<sub>6<\/sub>] (<em>M<\/em>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>1<\/td>\n<td>0<\/td>\n<td>1.00 \u00d7 10<sup>\u20132<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>1600<\/td>\n<td>5.04 \u00d7 10<sup>\u20133<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3<\/td>\n<td>3200<\/td>\n<td>3.37 \u00d7 10<sup>\u20133<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4<\/td>\n<td>4800<\/td>\n<td>2.53 \u00d7 10<sup>\u20133<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>5<\/td>\n<td>6200<\/td>\n<td>2.08 \u00d7 10<sup>\u20133<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In order to distinguish a first-order reaction from a second-order reaction, we plot ln[C<sub>4<\/sub>H<sub>6<\/sub>] versus <em>t<\/em> and compare it with a plot of [latex]\\frac{\\text{1}}{\\left[{\\text{C}}_{4}{\\text{H}}_{6}\\right]}[\/latex] versus <em>t<\/em>. The values needed for these plots follow.<\/p>\n<table id=\"fs-idp62232800\" class=\"medium unnumbered\" summary=\"This table contains three columns and six rows. The first row is a header row and it labels each column, \u201cTime ( s ),\u201d \u201c1 over &#091; C subscript 4 H subscript 6 &#093; ( M superscript negative 1 ),\u201d and \u201cl n &#091; C subscript 4 H subscript 6 &#093;.\u201d Under the column \u201cTime ( s )\u201d are the numbers: 0, 1600, 3200, 4800, and 6200. Under the \u201c1 over &#091; C subscript 4 H subscript 6 &#093; ( M superscript negative 1 )\u201d column are the numbers: 100, 198, 296, 395, and 481. Under the \u201cl n &#091; C subscript 4 H subscript 6 &#093;\u201d column are the numbers: negative 4.605, negative 5.289, negative 5.692, negative 5.978, and negative 6.175.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Time (s)<\/th>\n<th>[latex]\\frac{1}{\\left[{\\text{C}}_{4}{\\text{H}}_{6}\\right]}\\left({M}^{-1}\\right)[\/latex]<\/th>\n<th>ln[C<sub>4<\/sub>H<sub>6<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>0<\/td>\n<td>100<\/td>\n<td>\u20134.605<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1600<\/td>\n<td>198<\/td>\n<td>\u20135.289<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3200<\/td>\n<td>296<\/td>\n<td>\u20135.692<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4800<\/td>\n<td>395<\/td>\n<td>\u20135.978<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>6200<\/td>\n<td>481<\/td>\n<td>\u20136.175<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The plots are shown in Figure\u00a03. As you can see, the plot of ln[C<sub>4<\/sub>H<sub>6<\/sub>] versus <em>t<\/em> is not linear, therefore the reaction is not first order. The plot of [latex]\\frac{1}{\\left[{\\text{C}}_{4}{\\text{H}}_{6}\\right]}[\/latex] versus <em>t<\/em> is linear, indicating that the reaction is second order.<\/p>\n<div id=\"attachment_2205\" style=\"width: 985px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2205\" class=\"size-full wp-image-2205\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212252\/CNX_Chem_12_04_2OrdKin1.jpg\" alt=\"Two graphs are shown, each with the label \u201cTime ( s )\u201d on the x-axis. The graph on the left is labeled, \u201cl n &#091; C subscript 4 H subscript 6 &#093;,\u201d on the y-axis. The graph on the right is labeled \u201c1 divided by &#091; C subscript 4 H subscript 6 &#093;,\u201d on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).\" width=\"975\" height=\"355\" \/><\/p>\n<p id=\"caption-attachment-2205\" class=\"wp-caption-text\">Figure\u00a03. These two graphs show first- and second-order plots for the dimerization of C<sub>4<\/sub>H<sub>6<\/sub>. Since the first-order plot (left) is not linear, we know that the reaction is not first order. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Does the following data fit a second-order rate law?<\/p>\n<table id=\"fs-idm88760288\" class=\"medium unnumbered\" summary=\"This table contains three columns and seven rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and, \u201c[ A ] ( M ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, 5, and 6. Under the \u201cTime ( s )\u201d column are the numbers: 5, 10, 15, 20, 25, and 35. Under the \u201c[ A ] ( M )\u201d column are the numbers 0.952, 0.625, 0.465, 0.370, 0.308, and 0.230.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Trial<\/th>\n<th>Time (s)<\/th>\n<th>[<em>A<\/em>] (<em>M<\/em>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>1<\/td>\n<td>5<\/td>\n<td>0.952<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>10<\/td>\n<td>0.625<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3<\/td>\n<td>15<\/td>\n<td>0.465<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4<\/td>\n<td>20<\/td>\n<td>0.370<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>5<\/td>\n<td>25<\/td>\n<td>0.308<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>6<\/td>\n<td>35<\/td>\n<td>0.230<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q260710\">Show Answer<\/span><\/p>\n<div id=\"q260710\" class=\"hidden-answer\" style=\"display: none\">Yes. The plot of [latex]\\frac{1}{\\left[A\\right]}[\/latex] vs. <em>t<\/em> is linear:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2206 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212253\/CNX_Chem_12_04_CYL2_img1.jpg\" alt=\"A graph, with the title \u201c1 divided by &#091; A &#093; vs. Time\u201d is shown, with the label, \u201cTime ( s ),\u201d on the x-axis. The label \u201c1 divided by &#091; A &#093;\u201d appears left of the y-axis. The x-axis shows markings beginning at zero and continuing at intervals of 10 up to and including 40. The y-axis on the left shows markings beginning at 0 and increasing by intervals of 1 up to and including 5. A line with an increasing trend is drawn through six points at approximately (4, 1), (10, 1.5), (15, 2.2), (20, 2.8), (26, 3.4), and (36, 4.4).\" width=\"400\" height=\"269\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Zero-Order Reactions<\/h2>\n<p>For zero-order reactions, the differential rate law is:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Rate}=k{\\left[A\\right]}^{0}=k[\/latex]<\/p>\n<p>A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactants.<\/p>\n<p>The integrated rate law for a zero-order reaction also has the form of the equation of a straight line:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left[A\\right]& =& {-}kt+{\\left[A\\right]}_{0}\\hfill \\\\ \\hfill y& =& mx+b\\hfill \\end{array}[\/latex]<\/p>\n<div id=\"attachment_2207\" style=\"width: 360px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2207\" class=\"wp-image-2207\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212255\/CNX_Chem_12_04_AmDecomK1.jpg\" alt=\"A graph is shown with the label, \u201cTime ( s ),\u201d on the x-axis and, \u201c[ N H subscript 3 ] M,\u201d on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled \u201cDecomposition on W.\u201d A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled \u201cDecomposition on S i O subscript 2.\u201d\" width=\"350\" height=\"353\" \/><\/p>\n<p id=\"caption-attachment-2207\" class=\"wp-caption-text\">Figure\u00a04. The decomposition of NH<sub>3<\/sub> on a tungsten (W) surface is a zero-order reaction, whereas on a quartz (SiO<sub>2<\/sub>) surface, the reaction is first order.<\/p>\n<\/div>\n<p>A plot of [<em>A<\/em>] versus <em>t<\/em> for a zero-order reaction is a straight line with a slope of \u2212<em>k<\/em> and an intercept of [<em>A<\/em>]<sub>0<\/sub>. Figure\u00a04 shows a plot of [NH<sub>3<\/sub>] versus <em>t<\/em> for the decomposition of ammonia on a hot tungsten wire and for the decomposition of ammonia on hot quartz (SiO<sub>2<\/sub>). The decomposition of NH<sub>3<\/sub> on hot tungsten is zero order; the plot is a straight line. The decomposition of NH<sub>3<\/sub> on hot quartz is not zero order (it is first order). From the slope of the line for the zero-order decomposition, we can determine the rate constant:<\/p>\n<p style=\"text-align: center;\">slope = \u2212<em>k<\/em> = 1.3110<sup>\u22126<\/sup> mol\/L\/s<\/p>\n<h2>The Half-Life of a Reaction<\/h2>\n<p>The <strong>half-life of a reaction (<em>t<\/em><sub>1\/2<\/sub>)<\/strong> is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide in Figure\u00a01 as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases from 1.000 <em>M<\/em> to 0.500 <em>M<\/em>. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 <em>M<\/em> to 0.250 <em>M<\/em>; during the third half-life, it decreases from 0.250 <em>M<\/em> to 0.125 <em>M<\/em>. The concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.<\/p>\n<h3>First-Order Reactions<\/h3>\n<p>We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}\\hfill \\mathrm{ln}\\frac{{\\left[A\\right]}_{0}}{\\left[A\\right]}& =& kt\\hfill \\\\ \\hfill t& =& \\mathrm{ln}\\frac{{\\left[A\\right]}_{0}}{\\left[A\\right]}\\times \\frac{1}{k}\\hfill \\end{array}[\/latex]<\/p>\n<p>If we set the time <em>t<\/em> equal to the half-life, [latex]{t}_{1\\text{\/}2},[\/latex] the corresponding concentration of <em>A<\/em> at this time is equal to one-half of its initial concentration. Hence, when [latex]t={t}_{1\\text{\/}2},[\/latex] [latex]\\left[A\\right]=\\frac{1}{2}{\\left[A\\right]}_{0}[\/latex].<\/p>\n<p>Therefore:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill {t}_{1\\text{\/}2}& =\\mathrm{ln}\\frac{{\\left[A\\right]}_{0}}{\\frac{1}{2}{\\left[A\\right]}_{0}}\\times \\frac{1}{k}\\hfill \\\\ & =\\mathrm{ln}2\\times \\frac{1}{k}=0.693\\times \\frac{1}{k}\\hfill \\end{array}[\/latex]<\/p>\n<p>Thus:<\/p>\n<p style=\"text-align: center;\">[latex]{t}_{1\\text{\/}2}=\\frac{0.693}{k}[\/latex]<\/p>\n<p>We can see that the half-life of a first-order reaction is inversely proportional to the rate constant <em>k<\/em>. A fast reaction (shorter half-life) will have a larger <em>k<\/em>; a slow reaction (longer half-life) will have a smaller <em>k<\/em>.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 5:\u00a0Calculation of a First-order Rate Constant using Half-Life<\/h3>\n<p>Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 \u00b0C, using the data given in Figure\u00a05.<\/p>\n<div id=\"attachment_2208\" style=\"width: 637px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2208\" class=\"wp-image-2208\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212256\/CNX_Chem_12_04_HPerDcmp1.jpg\" alt=\"A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, \u201c1.000 M, 0 s, and ( 0 h ).\u201d The second beaker contains a slightly lighter green substance and is labeled below as, \u201c0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).\u201d The third beaker contains an even lighter green substance and is labeled below as, \u201c0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).\u201d The fourth beaker contains a green tinted substance and is labeled below as, \u201c0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).\u201d The fifth beaker contains a colorless substance and is labeled below as, \u201c0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).\u201d\" width=\"627\" height=\"197\" \/><\/p>\n<p id=\"caption-attachment-2208\" class=\"wp-caption-text\">Figure\u00a05. The decomposition of H<sub>2<\/sub>O<sub>2<\/sub> (2H<sub>2<\/sub>O<sub>2<\/sub> \u2192 2H<sub>2<\/sub>O + O<sub>2<\/sub>) at 40 \u00b0C is illustrated. The intensity of the color symbolizes the concentration of H<sub>2<\/sub>O<sub>2<\/sub> at the indicated times; H<sub>2<\/sub>O<sub>2<\/sub> is actually colorless.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q81889\">Show Answer<\/span><\/p>\n<div id=\"q81889\" class=\"hidden-answer\" style=\"display: none\">\n<p>The half-life for the decomposition of H<sub>2<\/sub>O<sub>2<\/sub> is 2.16 \u00d7 10<sup>4<\/sup> s:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {t}_{1\\text{\/}2}& =& \\frac{0.693}{k}\\hfill \\\\ \\hfill k& =& \\frac{0.693}{{t}_{1\\text{\/}2}}=\\frac{0.693}{2.16\\times {10}^{4}\\text{s}}=3.21\\times {10}^{-5}{\\text{s}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d<sup>\u22121<\/sup>. What is the half-life for this decay?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q409771\">Show Answer<\/span><\/p>\n<div id=\"q409771\" class=\"hidden-answer\" style=\"display: none\">5.02 d<\/div>\n<\/div>\n<\/div>\n<h3>Second-Order Reactions<\/h3>\n<p>We can derive the equation for calculating the half-life of a second order as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\left[A\\right]}=kt+\\frac{1}{{\\left[A\\right]}_{0}}[\/latex]<\/p>\n<p>or<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\left[A\\right]}-\\frac{1}{{\\left[A\\right]}_{0}}=kt[\/latex]<\/p>\n<p>If\u00a0[latex]t={t}_{1\\text{\/}2}[\/latex],\u00a0then\u00a0[latex]\\left[A\\right]=\\frac{1}{2}{\\left[A\\right]}_{0}[\/latex],\u00a0and we can write:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{\\frac{1}{2}{\\left[A\\right]}_{0}}-\\frac{1}{{\\left[A\\right]}_{0}}& =& k{t}_{1\\text{\/}2}\\hfill \\\\ \\hfill 2{\\left[A\\right]}_{0}-\\frac{1}{{\\left[A\\right]}_{0}}& =& k{t}_{1\\text{\/}2}\\hfill \\\\ \\hfill \\frac{1}{{\\left[A\\right]}_{0}}& =& k{t}_{1\\text{\/}2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Thus:<\/p>\n<p style=\"text-align: center;\">[latex]{t}_{1\\text{\/}2}=\\frac{1}{k{\\left[A\\right]}_{0}}[\/latex]<\/p>\n<p>For a second-order reaction, [latex]{t}_{1\\text{\/}2}[\/latex] is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.<\/p>\n<h3>Zero-Order Reactions<\/h3>\n<p>We can derive an equation for calculating the half-life of a zero order reaction as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[A\\right]={-}kt+{\\left[A\\right]}_{0}[\/latex]<\/p>\n<p>When half of the initial amount of reactant has been consumed [latex]t={t}_{1\\text{\/}2}[\/latex] and [latex]\\left[A\\right]=\\frac{{\\left[A\\right]}_{0}}{2}.[\/latex] Thus:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{{\\left[\\text{A}\\right]}_{0}}{2}& =& {-}k{t}_{1\\text{\/}2}+{\\left[\\text{A}\\right]}_{0}\\hfill \\\\ \\hfill k{t}_{1\\text{\/}2}& =& \\frac{{\\left[\\text{A}\\right]}_{0}}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]{t}_{1\\text{\/}2}=\\text{}\\frac{{\\left[A\\right]}_{0}}{2k}[\/latex]<\/p>\n<p>The half-life of a zero-order reaction increases as the initial concentration increases.<\/p>\n<p>Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table 1.<\/p>\n<table id=\"fs-idm117482272\" summary=\"This table contains four columns and seven rows. The first column and the first row both serve as headers. The first cell in the first column is blank and is followed by, \u201crate law,\u201d \u201cunits of rate constant,\u201d \u201cintegrated rate law,\u201d \u201cplot needed for linear fit of rate date,\u201d \u201crelationship between slope of linear plot and rate constant,\u201d and \u201chalf-life.\u201d The first row labels each column, \u201cZero-Order,\u201d \u201cFirst-Order,\u201d and \u201cSecond-Order.\u201d Under \u201cZero-order\u201d are the following: \u201crate = k,\u201d \u201cM s superscript negative 1,\u201d \u201c[ A ] = negative k t + [ A ] subscript 0,\u201d \u201c[ A ] v s. t,\u201d \u201ck = negative slope,\u201d and \u201ct subscript one half = [ A ] subscript 0 over 2 k.\u201d Under \u201cFirst-Order\u201d are the following: \u201crate = k [ A ],\u201d \u201cs superscript negative 1,\u201d \u201cl n [ A ] = negative k t plus l n [ A ] subscript 0,\u201d \u201cl n [ A ] v s. t,\u201d \u201ck = negative slope,\u201d and \u201ct subscript one half = 0.693 over k.\u201d Under \u201cSecond-Order\u201d are the following: \u201crate = k [ A ] superscript 2,\u201d \u201cM superscript negative 1 s superscript negative 1,\u201d \u201c1 over [ A ] = k t + ( 1 over [ A ] subscript 0 ),\u201d \u201c1 over [ A ] v s. t,\u201d \u201ck = positive slope,\u201d and \u201ct subscript one half \u2013 1 over [ A ] subscript 0 k.\u201d\">\n<thead>\n<tr>\n<th colspan=\"4\">Table 1. Summary of Rate Laws for Zero-, First-, and Second-Order Reactions<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<th><\/th>\n<th>Zero-Order<\/th>\n<th>First-Order<\/th>\n<th>Second-Order<\/th>\n<\/tr>\n<tr valign=\"top\">\n<td>rate law<\/td>\n<td>rate = <em>k<\/em><\/td>\n<td>rate = <em>k<\/em>[A]<\/td>\n<td>rate = <em>k<\/em>[A]<sup>2<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>units of rate constant<\/td>\n<td><em>M<\/em> s<sup>\u22121<\/sup><\/td>\n<td>s<sup>\u22121<\/sup><\/td>\n<td><em>M<\/em><sup>\u22121<\/sup> s<sup>\u22121<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>integrated rate law<\/td>\n<td>[<em>A<\/em>] = \u2212<em>kt<\/em> + [<em>A<\/em>]0<\/td>\n<td>ln [<em>A<\/em>] = \u2212<em>kt<\/em> + ln[<em>A<\/em>]0<\/td>\n<td>[latex]\\frac{1}{\\left[A\\right]}=kt+\\left(\\frac{1}{{\\left[A\\right]}_{0}}\\right)[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>plot needed for linear fit of rate data<\/td>\n<td>[<em>A<\/em>] vs. <em>t<\/em><\/td>\n<td>ln [<em>A<\/em>] vs. <em>t<\/em><\/td>\n<td>[latex]\\frac{1}{\\left[A\\right]}[\/latex] vs. <em>t<\/em><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>relationship between slope of linear plot and rate constant<\/td>\n<td><em>k<\/em> = \u2212slope<\/td>\n<td><em>k<\/em> = \u2212slope<\/td>\n<td><em>k<\/em> = +slope<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>half-life<\/td>\n<td>[latex]{t}_{1\\text{\/}2}=\\frac{{\\left[A\\right]}_{0}}{2k}[\/latex]<\/td>\n<td>[latex]{t}_{1\\text{\/}2}=\\frac{0.693}{k}[\/latex]<\/td>\n<td>[latex]{t}_{1\\text{\/}2}=\\frac{1}{{\\left[A\\right]}_{0}k}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>Differential rate laws can be determined by the method of initial rates or other methods. We measure values for the initial rates of a reaction at different concentrations of the reactants. From these measurements, we determine the order of the reaction in each reactant. Integrated rate laws are determined by integration of the corresponding differential rate laws. Rate constants for those rate laws are determined from measurements of concentration at various times during a reaction.<\/p>\n<p>The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>integrated rate law for zero-order reactions: [latex]\\left[A\\right]={-}kt+{\\left[A\\right]}_{0},[\/latex] [latex]{t}_{1\\text{\/}2}=\\frac{{\\left[A\\right]}_{0}}{2k}[\/latex]<\/li>\n<li>integrated rate law for first-order reactions: [latex]\\mathrm{ln}\\left[A\\right]={-}kt+{\\left[A\\right]}_{0},\\text{}{t}_{1\\text{\/}2}=\\frac{0.693}{k}[\/latex]<\/li>\n<li>integrated rate law for second-order reactions: [latex]\\frac{1}{\\left[A\\right]}=kt+\\frac{1}{{\\left[A\\right]}_{0}},[\/latex] [latex]{t}_{1\\text{\/}2}=\\frac{1}{{\\left[A\\right]}_{0}k}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of <em>A<\/em> at varying times.<\/li>\n<li>Use the data provided to graphically determine the order and rate constant of the following reaction: [latex]{\\text{SO}}_{2}{\\text{Cl}}_{2}\\rightarrow{\\text{SO}}_{2}+{\\text{Cl}}_{2}[\/latex]<br \/>\n<table id=\"fs-idp120830736\" class=\"medium unnumbered\" summary=\"This table contains two columns and eight rows. The first column is labeled, \u201cTime ( s ),\u201d and the second column is labeled, \u201c[ S O subscript 2 C l subscript 2 ] ( M ).\u201d Under the \u201cTime ( s )\u201d column are the numbers: 0; 5.00 times ten to the third power; 1.00 times ten to the fourth power; 1.50 times ten to the fourth power; 2.50 times ten to the fourth power; 3.00 times ten to the fourth power; and 4.00 times ten to the fourth power. Under the \u201c[ S O subscript 2 C l subscript 2 ] ( M )\u201d column are the numbers: 0.100, 0.0896, 0.0802, 0.0719, 0.0577, 0.0517, and 0.0415.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Time (s)<\/th>\n<th>[SO<sub>2<\/sub>Cl<sub>2<\/sub>] (<em>M<\/em>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>0<\/td>\n<td>0.100<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>5.00 \u00d7 10<sup>3<\/sup><\/td>\n<td>0.0896<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.00 \u00d7 10<sup>4<\/sup><\/td>\n<td>0.0802<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.50 \u00d7 10<sup>4<\/sup><\/td>\n<td>0.0719<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.50 \u00d7 10<sup>4<\/sup><\/td>\n<td>0.0577<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3.00 \u00d7 10<sup>4<\/sup><\/td>\n<td>0.0517<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4.00 \u00d7 10<sup>4<\/sup><\/td>\n<td>0.0415<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Use the data provided in a graphical method to determine the order and rate constant of the following reaction:[latex]2P\\rightarrow Q+W[\/latex]<br \/>\n<table id=\"fs-idm142901952\" class=\"medium unnumbered\" summary=\"This table contains two columns and six rows. The first row is a header row, and it labels each column, \u201cTime ( s ), \u201c and \u201c[ P ] ( M ).\u201d Under the \u201cTime ( s )\u201d column are the numbers: 9.0, 13.0, 18.0, 22.0, and 25.0. Under the \u201c[ P ] ( M )\u201d column are the numbers: 1.077 times ten to the negative 3; 1.068 times ten to the negative 3; 1.055 times ten to the negative 3; 1.046 times ten to the negative 3; and 1.039 times ten to the negative 3.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Time (s)<\/th>\n<th>[P] (<em>M<\/em>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>9.0<\/td>\n<td>1.077 \u00d7 10<sup>-3<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>13.0<\/td>\n<td>1.068 \u00d7 10<sup>-3<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>18.0<\/td>\n<td>1.055 \u00d7 10<sup>-3<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>22.0<\/td>\n<td>1.046 \u00d7 10<sup>-3<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>25.0<\/td>\n<td>1.039 \u00d7 10<sup>-3<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Pure ozone decomposes slowly to oxygen, [latex]{\\text{2O}}_{3}\\left(g\\right)\\rightarrow{\\text{3O}}_{2}\\left(g\\right).[\/latex] Use the data provided in a graphical method and determine the order and rate constant of the reaction.<br \/>\n<table id=\"fs-idp72766816\" class=\"medium unnumbered\" summary=\"This table has two columns and eight rows. The first row is a header row, and it labels each column, \u201cTime ( h ),\u201d and, \u201c[ O subscript 3 ] ( M ).\u201d Under the \u201cTime ( h )\u201d column are the numbers: 0; 2.0 times ten to the third power; 7.6 times ten to the third power; 1.00 times ten to the fourth power; 1.23 times ten to the fourth power; 1.43 times ten to the fourth power; 1.70 times ten to the fourth power. Under the \u201c[ O subscript 3 ] ( M )\u201d column are the numbers: 1.00 times ten to the negative 5; 4.98 times ten to the negative 6; 2.07 times ten to the negative 6; 1.66 times ten to the negative 6; 1.39 times ten to the negative 6; 1.22 times ten to the negative 6; and 1.05 times ten to the negative 6.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Time (h)<\/th>\n<th>[O<sub>3<\/sub>] (<em>M<\/em>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>0<\/td>\n<td>1.00 \u00d7 10<sup>\u22125<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.0 \u00d7 10<sup>\u22123<\/sup><\/td>\n<td>4.98 \u00d7 10<sup>\u22126<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>7.6 \u00d7 10<sup>\u22123<\/sup><\/td>\n<td>2.07 \u00d7 10<sup>\u22126<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.00 \u00d7 10<sup>\u22124<\/sup><\/td>\n<td>1.66 \u00d7 10<sup>\u22126<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.23 \u00d7 10<sup>\u22124<\/sup><\/td>\n<td>1.39 \u00d7 10<sup>\u22126<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.43 \u00d7 10<sup>\u22124<\/sup><\/td>\n<td>1.22 \u00d7 10<sup>\u22126<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.70 \u00d7 10<sup>\u22124<\/sup><\/td>\n<td>1.05 \u00d7 10<sup>\u22126<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>From the given data, use a graphical method to determine the order and rate constant of the following reaction:[latex]2X\\rightarrow Y+Z[\/latex]<br \/>\n<table id=\"fs-idm133654768\" class=\"medium unnumbered\" summary=\"This table contains two columns and nine rows. The first row is a header row, and it labels each column, \u201cTime ( s ),\u201d and \u201c[ X ] ( M ).\u201d Under the \u201cTime ( s )\u201d column are the numbers: 5.0, 10.0, 15.0, 20.0, 25.0, 30.0, 35.0, and 40.0. Under the \u201c[ X ] ( M )\u201d column are the numbers; 0.0990, 0.497, 0.0332, 0.0249, 0.0200, 0.0166, 0.0143, and 0.0125.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Time (s)<\/th>\n<th>[<em>X<\/em>] (<em>M<\/em>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>5.0<\/td>\n<td>0.0990<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>10.0<\/td>\n<td>0.0497<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>15.0<\/td>\n<td>0.0332<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>20.0<\/td>\n<td>0.0249<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>25.0<\/td>\n<td>0.0200<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>30.0<\/td>\n<td>0.0166<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>35.0<\/td>\n<td>0.0143<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>40.0<\/td>\n<td>0.0125<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>What is the half-life for the first-order decay of phosphorus-32 [latex]\\left({}_{15}^{32}\\text{P}\\rightarrow{}_{16}^{32}\\text{S}+{\\text{e}}^{-}\\right)[\/latex] The rate constant for the decay is 4.85 \u00d7 10<sup>\u22122<\/sup> day <sup>\u22121<\/sup>.<\/li>\n<li>What is the half-life for the first-order decay of carbon-14? [latex]\\left({}_{14}^{6}\\text{C}\\rightarrow{}_{14}^{7}\\text{N}+{\\text{e}}^{-}\\right)[\/latex] The rate constant for the decay is 1.21 \u00d7 10<sup>\u22124<\/sup> year<sup>\u22121<\/sup>.<\/li>\n<li>What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 <em>M<\/em>? The rate constant for this second-order reaction is 8.0 \u00d7 10<sup>-8<\/sup> L\/mol\/s.<\/li>\n<li>What is the half-life for the decomposition of O<sub>3<\/sub> when the concentration of O<sub>3<\/sub> is 2.35 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>? The rate constant for this second-order reaction is 50.4 L\/mol\/h.<\/li>\n<li>The reaction of compound <em>A<\/em> to give compounds <em>C<\/em> and <em>D<\/em> was found to be second-order in <em>A<\/em>. The rate constant for the reaction was determined to be 2.42 L\/mol\/s. If the initial concentration is 0.500 mol\/L, what is the value of t<sub>1\/2<\/sub>?<\/li>\n<li>The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol\/L. How long will it take for the concentration to drop to 0.0300 mol\/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?<\/li>\n<li>Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 \u00d7 10<sup>4<\/sup> g\/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 \u00b5g (0.15 \u00d7 10<sup>\u22126<\/sup> g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.<br \/>\n<table id=\"fs-idp79245440\" class=\"medium unnumbered\" summary=\"This table contains two columns and four rows. The first row is a header row, and it labels each column, \u201c[ Penicillin ] ( M ),\u201d and, \u201cRate ( mol \/ L \/ min ).\u201d Under the \u201c[ Penicillin ] ( M )\u201d column are the numbers: 2.0 times ten to the negative six; 3.0 times ten to the negative six; and 4.0 times ten to the negative 6. Under the \u201cRate ( mol \/ L \/ min )\u201d column are the numbers: 1.0 times ten to the negative ten; 1.5 times ten to the negative 10; and 2.0 times ten to the negative 10.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>[Penicillin] (<em>M<\/em>)<\/th>\n<th>Rate (mol\/L\/min)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>2.0 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td>1.0 \u00d7 10<sup>\u221210<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3.0 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td>1.5 \u00d7 10<sup>\u221210<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4.0 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td>2.0 \u00d7 10<sup>\u221210<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?<\/li>\n<li>There are two molecules with the formula C<sub>3<\/sub>H<sub>6<\/sub>. Propene, CH<sub>3<\/sub>CH=CH<sub>2,<\/sub> is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2213\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212304\/CNX_Chem_12_04_Cycloprop_img1.jpg\" alt=\"A structural formula for cyclopropane is shown. Three C H subscript 2 groups are positioned as vertices of an equilateral triangle connected with single bonds represented by line segments.\" width=\"272\" height=\"180\" \/><br \/>\nWhen heated to 499 \u00b0C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of [latex]5.95\\times {10}^{-4}{s}^{-1}[\/latex].\u00a0What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499.5 \u00b0C?<\/li>\n<li>Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is [latex]{}_{9}^{18}\\text{F}\\rightarrow{}_{8}^{18}\\text{O}+{\\text{e}}^{+}.[\/latex] ) Physicians use <sup>18<\/sup>F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>What is the rate constant for the decomposition of fluorine-18?<\/li>\n<li>If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?<\/li>\n<li>How long does it take for 99.99% of the <sup>18<\/sup>F to decay?<\/li>\n<\/ol>\n<\/li>\n<li>Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for [latex]\\frac{1}{64}[\/latex] of the initial dose to remain in the athlete\u2019s body?<\/li>\n<li>Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.<\/li>\n<li>Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 \u00b0C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:<br \/>\n<table id=\"fs-idp121895360\" class=\"medium unnumbered\" summary=\"This table contains three columns and nine rows. The first row is a header row, and it labels each column, \u201cInitial [ C subscript 3 H subscript 5 N subscript 3 O subscript 9 ] ( M ),\u201d \u201ct ( s ),\u201d and \u201c% Decomposed.\u201d Under the \u201cInitial [ C subscript 3 H subscript 5 N subscript 3 O subscript 9 ] ( M )\u201d column are the numbers: 4.88, 3.52, 2.29, 1.81, 5.33, 4.05, 2.95, and 1.72. Under the \u201ct ( s )\u201d column are the numbers: 300, 300, 300, 300, 180, 180, 180, and 180. Under the \u201c% Decomposed\u201d column are the numbers: 52.0, 52.9, 53.2, 53.9, 34.6, 35.9, 36.0, and 35.4.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Initial [C<sub>3<\/sub>H<sub>5<\/sub>N<sub>3<\/sub>O<sub>9<\/sub>] (M)<\/th>\n<th><em>t<\/em> (s)<\/th>\n<th>% Decomposed<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>4.88<\/td>\n<td>300<\/td>\n<td>52.0<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3.52<\/td>\n<td>300<\/td>\n<td>52.9<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.29<\/td>\n<td>300<\/td>\n<td>53.2<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.81<\/td>\n<td>300<\/td>\n<td>53.9<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>5.33<\/td>\n<td>180<\/td>\n<td>34.6<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4.05<\/td>\n<td>180<\/td>\n<td>35.9<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.95<\/td>\n<td>180<\/td>\n<td>36.0<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.72<\/td>\n<td>180<\/td>\n<td>35.4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene (CH<sub>2<\/sub>=CH\u2013CH=CH<sub>2<\/sub>) has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-2214\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212305\/CNX_Chem_12_04_ExSolutio2_img1.jpg\" alt=\"A structural formula for cyclobutene is shown. The figure has two C H subscript 2 groups as the upper two vertices of a square structure. These groups are connected by a single, short line segment. Line segments extend below each of these C H subscript 2 groups to C H groups positioned at the lower two vertices of the square structure. The C H groups are connected with a double line segment indicating a double bond.\" width=\"220\" height=\"156\" \/><br \/>\nThe isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as 2.0 \u00d7 10<sup>\u22124<\/sup> s<sup>\u22121<\/sup> at 150\u00b0C in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150\u00b0C with an initial pressure of 55 torr.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q658620\">Show Selected Answers<\/span><\/p>\n<div id=\"q658620\" class=\"hidden-answer\" style=\"display: none\">\n<p>2.\u00a0Plotting a graph of ln[SO<sub>2<\/sub>Cl<sub>2<\/sub>] versus <em>t<\/em> reveals a linear trend; therefore we know this is a first-order reaction:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-5382\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09033827\/Graph.png\" alt=\"Graph of ln SO2Cl2 M and time. A line connects the following points: coordinates negative 2.303, zero; coordinates negative 2.412, 5 times 10 to the power of three; coordinates negative 2.523, 1 times 10 to the power of 4; coordinates negative 2.632, 1.5 times 10 to the power of 4; coordinates negative 2.852, 2.5 times 10 to the power of 4; coordinates negative 2.962 times 3 times 10 to the power of 4; coordinates negative 3.182, 4 times 10 to the power of 4.\" width=\"471\" height=\"351\" \/><\/p>\n<p>The value of <em>k<\/em> is found from the slope of the line since [latex]\\mathrm{ln}\\left[A\\right]={-}kt+\\mathrm{ln}{\\left[A\\right]}_{0}[\/latex] is in the form of a straight line, y = <em>mx<\/em> + <em>b<\/em>.<\/p>\n<p>[latex]\\frac{\\Delta y}{\\Delta x}=\\frac{\\text{ln}0.0896-\\text{ln}0.100}{5.00\\times {10}^{3}-0\\text{s}}=\\frac{-0.1098}{5.00\\times {10}^{3}\\text{s}}=-2.20\\times {10}^{5}{\\text{s}}^{-1}[\/latex]<\/p>\n<p>4.\u00a0To distinguish a first-order reaction from a second-order reaction, we plot ln[<em>P<\/em>] against <em>t<\/em> and compare that plot with a plot of [latex]\\frac{1}{\\left[P\\right]}[\/latex] versus <em>t<\/em>. The values needed for these plots are abbreviated to include only the data needed for a second-order plot, as the data do not seem to support a first-order reaction:<\/p>\n<table id=\"fs-idm111039952\" class=\"medium unnumbered\" summary=\"This table contains one column and eight rows. The first row is a header row and it labels the column, \u201c1 over &#091; O subscript 3 &#093; ( M superscript negative 1 ).\u201d Under this column are the numbers: 1.00 times ten to the fifth power; 2.01 times ten to the fifth power; 4.83 times ten to the fifth power; 6.02 times ten to the fifth power; 7.19 times ten to the fifth power; 8.20 times ten to the fifth power; 9.52 times ten to the fifth power.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>[latex]\\frac{1}{\\left[{\\text{O}}_{3}\\right]}\\text{(}{M}^{-1}\\text{)}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>1.00 \u00d7 10<sup>5<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.01 \u00d7 10<sup>5<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4.83 \u00d7 10<sup>5<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>6.02 \u00d7 10<sup>5<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>7.19 \u00d7 10<sup>5<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>8.20 \u00d7 10<sup>5<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>9.52 \u00d7 10<sup>5<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The plot is nicely linear, so the reaction is second order.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-5383\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09033910\/Graph2.png\" alt=\"Graph of 1 over O3 M and time in hours. The rise is delta 1 over O3. The run is delta t.\" width=\"316\" height=\"426\" \/><br \/>\n[latex]\\text{slope}=\\frac{9.52\\times {10}^{5}-1.00\\times {10}^{5}}{17\\times {10}^{3}-0}=50.1{\\text{L mol}}^{-1}{\\text{h}}^{-1}[\/latex]<\/p>\n<p>6. \u00a0The half-life is [latex]{t}_{1\\text{\/}2}=\\frac{0.693}{k},[\/latex] where <em>k<\/em> is the rate constant:<\/p>\n<p>[latex]k=\\frac{0.693}{{t}_{1\\text{\/}2}}=4.85\\times {10}^{-2}{\\text{d}}^{-1}[\/latex]<\/p>\n<p>[latex]{t}_{1\\text{\/}2}=\\frac{0.693}{4.85\\times {10}^{-2}{\\text{d}}^{-1}}=\\text{14.3 d}[\/latex]<\/p>\n<p>8.\u00a0In a second-order reaction, the rate is concentration-dependent, [latex]{t}_{1\\text{\/}2}=\\frac{1}{k{\\left[A\\right]}_{0}}.[\/latex]<\/p>\n<p>[latex]{t}_{1\\text{\/}2}=\\frac{1}{k{\\left[A\\right]}_{0}}=\\frac{1}{8.0\\times {10}^{-8}{\\text{L mol}}^{-1}{\\text{s}}^{-1}\\left[0.15M\\right]}=8.3\\times {10}^{7}\\text{s}[\/latex]<\/p>\n<p>10.\u00a0For a second-order reaction, the half-life is concentration-dependent:<\/p>\n<p>[latex]{t}_{1\\text{\/}2}=\\frac{1}{k{\\left[A\\right]}_{0}}=\\frac{1}{2.42{\\text{L mol}}^{-1}{\\text{s}}^{-1}\\times 0.500{\\text{mol L}}^{-1}}=0.826\\text{s}[\/latex]<\/p>\n<p>12.\u00a0The reaction is first order with respect to penicillinase, and the rate doubles as [penicillin] doubles. Thus the rate equation is:<\/p>\n<p>rate = <em>k<\/em>[penicillinase][penicillin]<\/p>\n<p>Using the data in the first row,<\/p>\n<p>[latex]k=\\frac{1.0\\times {10}^{-10}{\\text{mol L}}^{-1}{\\text{min}}^{-1}}{\\left(\\frac{0.15\\times {10}^{-6}{\\text{g L}}^{-1}}{3.0\\times {10}^{4}{\\text{g mol}}^{-1}}\\right)\\left(2.0\\times {10}^{-6}{\\text{mol L}}^{-1}\\right)}=1.0\\times {10}^{7}{\\text{mol}}^{-1}{\\text{min}}^{-1}[\/latex]<\/p>\n<p>14.\u00a0The unit of the provided rate constant indicates the reaction is first order. Using the integrated form of the first-order rate law, we have:<\/p>\n<p>[latex]\\text{ln}\\frac{{\\left[A\\right]}_{0}}{\\left[A\\right]}{-}kt[\/latex]<\/p>\n<p>Let [<em>A<\/em>]<sub>0<\/sub> = 1 , then:<\/p>\n<p>[latex]\\text{ln}\\frac{\\left[1\\right]}{\\left[A\\right]}=5.95\\times {10}^{-4}{\\cancel{\\text{s}}}^{-1}\\times 0.75\\cancel{\\text{h}}\\times 60\\frac{\\cancel{\\text{min}}}{\\cancel{\\text{h}}}\\times 60\\frac{\\cancel{\\text{s}}}{\\cancel{\\text{min}}}=1.606[\/latex]<\/p>\n<p>Convert 1.606, a natural log, to the corresponding number by taking the <em>e<\/em>s of both sides:<\/p>\n<p>[latex]\\frac{1}{\\left[\\text{A}\\right]}=4.98\\text{[A]}=0.20,\\text{so 20% remains}.[\/latex]<\/p>\n<p>16.\u00a0[latex]\\frac{1}{64}=\\frac{1}{{2}^{x}}[\/latex] where <em>x<\/em> represents the number of half-life periods <em>x<\/em> = 6, so (6)(42) = 252 days.<\/p>\n<p>18.\u00a0From the first-order rate law, calculate the value of [<em>A<\/em>], [latex]\\text{ln}\\left(\\frac{{\\left[\\text{A}\\right]}_{0}}{\\left[\\text{A}\\right]}\\right),[\/latex] and <em>k<\/em>. The values are tabulated:<\/p>\n<table id=\"fs-idp20979328\" class=\"medium unnumbered\" summary=\"This table has five columns and nine rows. The first row is a header row, and it labels each column: \u201c&#091; A &#093; subscript 0 ( M ),\u201d \u201c&#091; A &#093; ( M ),\u201d \u201cl n ( &#091; A &#093; subscript 0 over &#091; A &#093; ),\u201d \u201ct ( s ),\u201d and \u201ck times 10 to the third power ( s superscript negative 1 ).\u201d Under the \u201c&#091; A &#093; subscript 0 ( M )\u201d column are the numbers: 4.88, 3.52, 2.29, 1.81, 5.33, 4.05, 2.95, and 1.72. Under the \u201c&#091; A &#093; ( M )\u201d column are the numbers: 2.34, 1.66, 1.07, 0.834, 3.49, 2.61, 1.89, and 1.11. Under the \u201cl n ( &#091; A &#093; subscript 0 over &#091; A &#093; )\u201d column are the numbers: 0.734, 0.752, 0.761, 0.775, 0.423, 0.439, 0.445, and 0.438. Under the \u201ct ( s )\u201d column are the numbers: 300, 300, 300, 300, 180, 180, 180, and 180. Under the \u201ck times 10 to the third power ( s superscript negative 1 )\u201d columns are the numbers: 2.45, 2.51, 2.54, 2.58, 2.35, 2.44, 2.47, and 2.43.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>[A]<sub>0<\/sub> (<em>M<\/em>)<\/th>\n<th>[A] (<em>M<\/em>)<\/th>\n<th>[latex]\\text{ln}\\left(\\frac{{\\left[\\text{A}\\right]}_{0}}{\\left[\\text{A}\\right]}\\right)[\/latex]<\/th>\n<th><em>t<\/em> (s)<\/th>\n<th><em>k<\/em> \u00d7 10<sup>3<\/sup> (s<sup>\u22121<\/sup>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>4.88<\/td>\n<td>2.34<\/td>\n<td>0.734<\/td>\n<td>300<\/td>\n<td>2.45<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3.52<\/td>\n<td>1.66<\/td>\n<td>0.752<\/td>\n<td>300<\/td>\n<td>2.51<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.29<\/td>\n<td>1.07<\/td>\n<td>0.761<\/td>\n<td>300<\/td>\n<td>2.54<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.81<\/td>\n<td>0.834<\/td>\n<td>0.775<\/td>\n<td>300<\/td>\n<td>2.58<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>5.33<\/td>\n<td>3.49<\/td>\n<td>0.423<\/td>\n<td>180<\/td>\n<td>2.35<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4.05<\/td>\n<td>2.61<\/td>\n<td>0.439<\/td>\n<td>180<\/td>\n<td>2.44<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.95<\/td>\n<td>1.89<\/td>\n<td>0.445<\/td>\n<td>180<\/td>\n<td>2.47<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>1.72<\/td>\n<td>1.11<\/td>\n<td>0.438<\/td>\n<td>180<\/td>\n<td>2.43<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>half-life of a reaction (<em>t<\/em><sub>l\/2<\/sub>): <\/strong>time required for half of a given amount of reactant to be consumed<\/p>\n<p><strong>integrated rate law: <\/strong>equation that relates the concentration of a reactant to elapsed time of reaction<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2215\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":74,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2215","chapter","type-chapter","status-publish","hentry"],"part":2992,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2215","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":14,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2215\/revisions"}],"predecessor-version":[{"id":6075,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2215\/revisions\/6075"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/parts\/2992"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2215\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/media?parent=2215"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapter-type?post=2215"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/contributor?post=2215"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/license?post=2215"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}