{"id":2227,"date":"2015-04-22T21:03:47","date_gmt":"2015-04-22T21:03:47","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=2227"},"modified":"2020-12-28T17:19:40","modified_gmt":"2020-12-28T17:19:40","slug":"reaction-mechanisms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/reaction-mechanisms\/","title":{"raw":"Reaction Mechanisms","rendered":"Reaction Mechanisms"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Distinguish net reactions from elementary reactions (steps)<\/li>\r\n \t<li>Identify the molecularity of elementary reactions<\/li>\r\n \t<li>Write a balanced chemical equation for a process given its reaction mechanism<\/li>\r\n \t<li>Derive the rate law consistent with a given reaction mechanism<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp19792192\">Chemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The\u00a0<strong><span id=\"term551\" data-type=\"term\">reaction mechanism<\/span><\/strong>\u00a0(or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs.<\/p>\r\n<p id=\"fs-idm10360272\">The decomposition of ozone, for example, appears to follow a mechanism with two steps:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{O}}_{3}\\left(g\\right)\\rightarrow{\\text{O}}_{2}\\left(g\\right)+\\text{O}\\\\ \\text{O}+{\\text{O}}_{3}\\left(g\\right)\\rightarrow 2{\\text{O}}_{2}\\left(g\\right)\\end{array}[\/latex]<\/p>\r\nWe call each step in a reaction mechanism an <strong>elementary reaction<\/strong>. Elementary reactions occur exactly as they are written and cannot be broken down into simpler steps. Elementary reactions add up to the overall reaction, which, for the decomposition, is:\r\n<p style=\"text-align: center;\">[latex]2{\\text{O}}_{3}\\left(g\\right)\\rightarrow 3{\\text{O}}_{2}\\left(g\\right)[\/latex]<\/p>\r\nNotice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called <strong>intermediates<\/strong>.\r\n<p id=\"fs-idp6340064\">While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react to give three molecules of oxygen, the mechanism of the reaction\u00a0<em data-effect=\"italics\">does not involve the direct collision and reaction of two ozone molecules<\/em>. Instead, one O<sub>3<\/sub>\u00a0decomposes to yield O<sub>2<\/sub>\u00a0and an oxygen atom, and a second O<sub>3<\/sub>\u00a0molecule subsequently reacts with the oxygen atom to yield two additional O<sub>2<\/sub>\u00a0molecules.<\/p>\r\n<p id=\"fs-idm500275472\">Unlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction\u2019s equation undergo only the bond-breaking and\/or making events depicted to yield the product(s). For this reason,\u00a0<em data-effect=\"italics\">the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction<\/em>. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation.<\/p>\r\n\r\n<h2>Unimolecular Elementary Reactions<\/h2>\r\nThe <strong>molecularity<\/strong> of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a <strong>unimolecular reaction<\/strong> involves the rearrangement of a <em>single<\/em> reactant species to produce one or more molecules of product:\r\n<p style=\"text-align: center;\">[latex]A\\rightarrow\\text{products}[\/latex]<\/p>\r\nThe rate equation for a unimolecular reaction is:\r\n<p style=\"text-align: center;\">[latex]\\text{rate}=k[A][\/latex]<\/p>\r\nA unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:\r\n<p style=\"text-align: center;\">[latex]{\\text{O}}_{3}\\rightarrow{\\text{O}}_{2}+\\text{O}[\/latex]<\/p>\r\nillustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism. However, some unimolecular reactions may have only a single reaction in the reaction mechanism. (In other words, an elementary reaction can also be an overall reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, C<sub>4<\/sub>H<sub>8<\/sub>, to ethylene, C<sub>2<\/sub>H<sub>4<\/sub>, is represented by the following chemical equation:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212316\/CNX_Chem_12_06_CyclobD_img1.jpg\" alt=\"In this figure, structural formulas are used to illustrate a chemical reaction. On the left, a structural formula for cyclobutane is shown. This structure is composed of 4 C atoms connected with single bonds in a square shape. Each C atom is bonded to two other C atoms in the structure, leaving two bonds for H atoms pointing outward above, below, left, and right. An arrow points right to two identical ethane molecules with a plus symbol between them. Each of these molecules contains two C atoms connected with a double bond oriented vertically between them. The C atom at the top of these molecules has H atoms bonded above to the right and left. Similarly, the lower C atom has two H atoms bonded below to the right and left.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\nThis equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing first-order behavior:\r\n<p style=\"text-align: center;\">[latex]\\text{rate}=-\\dfrac{\\Delta\\left[{\\text{C}}_{4}{\\text{H}}_{8}\\right]}{\\Delta t}=k\\left[{\\text{C}}_{4}{\\text{H}}_{8}\\right][\/latex]<\/p>\r\nThis agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction.\r\n<h2>Bimolecular Elementary Reactions<\/h2>\r\nA\u00a0<strong><span id=\"term556\" data-type=\"term\">bimolecular reaction<\/span><\/strong>\u00a0involves two reactant species, for example:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}A+B\\rightarrow\\text{products}\\\\ \\text{and}\\\\ 2A\\rightarrow\\text{products}\\end{array}[\/latex]<\/p>\r\nFor the first type, in which the two reactant molecules are different, the rate law is first-order in\u00a0<em data-effect=\"italics\">A<\/em>\u00a0and first order in\u00a0<em data-effect=\"italics\">B<\/em>\u00a0(second-order overall):\r\n<p style=\"text-align: center;\">[latex]\\text{rate} = k[A][B][\/latex]<\/p>\r\nFor the second type, in which two identical molecules collide and react, the rate law is second order in\u00a0<em data-effect=\"italics\">A<\/em>:\r\n<p style=\"text-align: center;\">[latex]\\text{rate} = k[A][B] = k[A]^{2}[\/latex]<\/p>\r\nSome chemical reactions have mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:\r\n<p style=\"text-align: center;\">[latex]{\\text{NO}}_{2}\\left(g\\right)+\\text{CO}\\left(g\\right)\\rightarrow\\text{NO}\\left(g\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"700\"]<img class=\"\" src=\"https:\/\/openstax.org\/resources\/9786aed35d8e9f6d86771aad63ad5543b409678e\" alt=\"This figure provides an illustration of a reaction between two H I molecules using space filling models. H atoms are shown as white spheres, and I atoms are shown as purple spheres. On the left, two H I molecules are shownwith a small white sphere bonded to a much larger purple sphere. The label, \u201cTwo H I molecules,\u201d appears below. An arrow points right to a similar structure in which the two molecules appear pushed together, so that the purple spheres of the two molecules are touching. Below appears the label, \u201cTransition state.\u201d Following another arrow, two white spheres are shown vertically oriented and bonded together with the label, \u201cH subscript 2\u201d above. The H subscript 2 molecule is followed by a plus sign and two purple spheres bonded together with the label, \u201cI subscript 2\u201d above. Below these structures is the label, \u201cHydrogen iodide molecules decompose to produce hydrogen H subscript 2 and iodine I subscript 2.\u201d\" width=\"700\" height=\"93\" \/> Figure 1. The probable mechanism for the reaction between NO2 and CO to yield NO and CO2.[\/caption]\r\n\r\nBimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:\r\n<p style=\"text-align: center;\">[latex]\\text{O}\\left(g\\right)+{\\text{O}}_{3}\\left(g\\right)\\rightarrow{\\text{2O}}_{2}\\left(g\\right)[\/latex]<\/p>\r\n\r\n<h2>Termolecular Elementary Reactions<\/h2>\r\nAn elementary <strong>termolecular reaction<\/strong> involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{2NO}+{\\text{O}}_{2}\\rightarrow 2{\\text{NO}}_{2}\\\\ \\text{rate}=k{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]\\end{array}[\/latex]<\/p>\r\nLikewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{2NO}+{\\text{Cl}}_{2}\\rightarrow 2\\text{NOCl}\\\\ \\text{rate}=k{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{Cl}}_{2}\\right]\\end{array}[\/latex]<\/p>\r\n\r\n<h2>Relating Reaction Mechanisms to Rate Laws<\/h2>\r\n[caption id=\"\" align=\"alignright\" width=\"300\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212320\/CNX_Chem_12_06_Cattle1.jpg\" alt=\"A photo is shown of cattle passing through a narrow chute into a holding pen. A person directs them through the gate with a long white and red pole.\" width=\"300\" height=\"201\" data-media-type=\"image\/jpeg\" \/> Figure\u00a02. A cattle chute is a nonchemical example of a rate-determining step. Cattle can only be moved from one holding pen to another as quickly as one animal can make its way through the chute. (credit: Loren Kerns)[\/caption]\r\n\r\nIt's often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the <strong>rate-limiting step<\/strong> (or rate-determining step) of the reaction.\r\n\r\nAs described above, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, we must determine the overall rate law from experimental data and deduce the mechanism from the rate law (and sometimes from other data). The reaction of NO<sub>2<\/sub> and CO provides an illustrative example:\r\n<p style=\"text-align: center;\">[latex]{\\text{NO}}_{2}\\left(g\\right)+\\text{CO}\\left(g\\right)\\rightarrow{\\text{CO}}_{2}\\left(g\\right)+\\text{NO}\\left(g\\right)[\/latex]<\/p>\r\nFor temperatures above 225 \u00b0C, the rate law has been found to be:\r\n<p style=\"text-align: center;\">[latex]\\text{rate}=k\\left[{\\text{NO}}_{2}\\right]\\left[\\text{CO}\\right][\/latex]<\/p>\r\nThe reaction is first order with respect to NO<sub>2<\/sub> and first-order with respect to CO. This is consistent with a single-step bimolecular mechanism and it is <em>possible<\/em> that this is the mechanism for this reaction at high temperatures.\r\n\r\nAt temperatures below 225 \u00b0C<em>,<\/em> the reaction is described by a rate law that is second order with respect to NO<sub>2<\/sub>:\r\n<p style=\"text-align: center;\">[latex]\\text{rate}=k{\\left[{\\text{NO}}_{2}\\right]}^{2}[\/latex]<\/p>\r\nThis is consistent with a mechanism that involves the following two elementary reactions, the first of which is slower and is therefore the rate-determining step:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{NO}}_{2}\\left(g\\right)+{\\text{NO}}_{2}\\left(g\\right)\\rightarrow{\\text{NO}}_{3}\\left(g\\right)+\\text{NO}\\left(g\\right)\\qquad\\left(\\text{slow}\\right)\\\\ {\\text{NO}}_{3}\\left(g\\right)+\\text{CO}\\left(g\\right)\\rightarrow{\\text{NO}}_{2}\\left(g\\right)+{\\text{CO}}_{2}\\left(g\\right)\\qquad\\left(\\text{fast}\\right)\\end{array}[\/latex]<\/p>\r\nThe rate-determining step gives a rate law showing second-order dependence on the NO<sub>2<\/sub> concentration, and the sum of the two equations gives the net overall reaction.\r\n\r\nIn general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving an <em>equilibrium<\/em> reaction, the rate law for the overall reaction may be more difficult to derive.\r\n\r\na reversible reaction is at\u00a0<em data-effect=\"italics\">equilibrium<\/em>\u00a0when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species N<sub>2<\/sub>O<sub>2<\/sub>. When this reaction is at equilibrium:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{NO}+\\text{NO}\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{2}\\\\ {\\text{rate}}_{\\text{forward}}={\\text{rate}}_{\\text{reverse}}\\\\ {k}_{1}{\\left[\\text{NO}\\right]}^{2}={k}_{-1}\\left[{\\text{N}}_{2}{\\text{O}}_{2}\\right]\\end{array}[\/latex]<\/p>\r\nThis expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:\r\n<p style=\"text-align: center;\">[latex]\\left(\\dfrac{{\\text{k}}_{1}{\\left[\\text{NO}\\right]}^{2}}{{\\text{k}}_{-1}}\\right)=\\left[{\\text{N}}_{2}{\\text{O}}_{2}\\right][\/latex]<\/p>\r\nSince intermediate species concentrations are not used in formulating rate laws for overall reactions, this approach is sometimes necessary, as illustrated in the following example exercise.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Deriving A Rate Law From a Reaction Mechanism<\/h3>\r\nThe two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:\r\n<p style=\"padding-left: 30px;\"><strong>Step 1:<\/strong>\u00a0 [latex]\\text{NO}(\\text{g})+\\text{Cl}_{2}(\\text{g})\\rightleftharpoons{\\text{NOCl}}_{2}(\\text{g})\\qquad\\text{fast}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\"><strong>Step 2:<\/strong>\u00a0 [latex]\\text{NOCl}_{2}(\\text{g})+\\text{NO}(\\text{g})\\longrightarrow{2}\\text{NOCl}(\\text{g})\\qquad\\text{slow}[\/latex]<\/p>\r\n<p id=\"fs-idm352158848\">Use this mechanism to derive the equation and predicted rate law for the overall reaction.<\/p>\r\n[reveal-answer q=\"339226\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"339226\"]\r\n\r\nThe equation for the overall reaction is obtained by adding the two elementary reactions:\r\n<p style=\"text-align: center;\">[latex]2\\text{NO}(\\text{g})+\\text{Cl}_{2}(\\text{g})\\rightleftharpoons{2}\\text{NOCl}(\\text{g})[\/latex]<\/p>\r\nTo derive a rate law from this mechanism, first write rates laws for each of the two steps.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\text{rate}_{1}}&amp; =&amp; {k}_{1}\\left[{\\text{NO}}\\right]\\left[\\text{Cl}_{2}\\right]\\qquad\\text{for the forward reaction of step 1}\\hfill \\\\ \\hfill{\\text{rate}_{-1}}&amp; =&amp; {k}_{-1}\\left[{\\text{NOCl}}_{2}\\right]\\qquad\\text{for the reverse reaction of step 1}\\hfill \\\\ \\hfill{\\text{rate}_{2}}&amp; =&amp; {k}_{2}\\left[\\text{NOCl}_{2}\\right]\\left[\\text{NO}\\right]\\qquad\\text{for step 2}\\hfill \\end{array}[\/latex]<\/p>\r\n<p id=\"fs-idm328804768\">Step 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [NOCl<sub>2<\/sub>]. To remedy this, use the first step\u2019s rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations.<\/p>\r\n<p id=\"fs-idm328530432\">Assuming Step 1 is at equilibrium:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{rate}_{1}=\\text{rate}_{-1}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{k}_{1}\\left[{\\text{NO}}\\right]\\left[\\text{Cl}_{2}\\right]={k}_{-1}\\left[\\text{NOCl}_{2}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[\\text{NOCl}_{2}\\right]=\\left(\\dfrac{{k}_{1}}{{k}_{-1}}\\right)\\left[\\text{NO}\\right]\\left[\\text{Cl}_{2}\\right][\/latex]<\/p>\r\n&nbsp;\r\n\r\nSubstituting this expression into the rate law for Step 2 yields:\r\n<p style=\"text-align: center;\">[latex]\\text{rate}_{2}=\\text{rate}_{\\text{overall}}=\\left[\\text{NOCl}_{2}\\right]=\\left(\\dfrac{{k}_{2}{k}_{1}}{{k}_{-1}}\\right)\\left[\\text{NO}\\right]^{2}\\left[\\text{Cl}_{2}\\right][\/latex]<\/p>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nThe first step of a proposed multistep mechanism is:\r\n<p style=\"text-align: center;\">[latex]\\text{F}_{2}(\\text{g})\\rightleftharpoons{2}\\text{F}(\\text{g})\\qquad\\text{fast}[\/latex]<\/p>\r\n<p id=\"fs-idm373041168\">Derive the equation relating atomic fluorine concentration to molecular fluorine concentration.<\/p>\r\n\r\n<div id=\"fs-idm330964176\" class=\"ui-has-child-title\" data-type=\"note\"><\/div>\r\n[reveal-answer q=\"853909\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"853909\"]\r\n\r\n[latex][F]=\\left(\\dfrac{{\\text{k}}_{1}{\\left[\\text{F}_{2}\\right]}}{{\\text{k}}_{-1}}\\right)^{1\/2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nThe sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The overall rate of a reaction is determined by the rate of the slowest step, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Why are elementary reactions involving three or more reactants very uncommon?<\/li>\r\n \t<li>In general, can we predict the effect of doubling the concentration of <em>A<\/em> on the rate of the overall reaction [latex]A+B\\rightarrow C[\/latex] ? Can we predict the effect if the reaction is known to be an elementary reaction?<\/li>\r\n \t<li>Define these terms:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>unimolecular reaction<\/li>\r\n \t<li>bimolecular reaction<\/li>\r\n \t<li>elementary reaction<\/li>\r\n \t<li>overall reaction<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What is the rate equation for the elementary termolecular reaction [latex]A+2B\\rightarrow\\text{products}[\/latex] ? For [latex]3A\\rightarrow\\text{products}[\/latex] ?<\/li>\r\n \t<li>Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{l}{\\text{Cl}}_{2}+\\text{CO}\\rightarrow{\\text{Cl}}_{2}\\text{CO}\\\\ \\text{rate}=k{\\left[{\\text{Cl}}_{2}\\right]}^{3\\text{\/}2}\\left[\\text{CO}\\right]\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{l}{\\text{PCl}}_{3}+{\\text{Cl}}_{2}\\rightarrow{\\text{PCl}}_{5}\\\\ \\text{rate}=k\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{l}2\\text{NO}+{\\text{H}}_{2}\\rightarrow{\\text{N}}_{2}+{\\text{H}}_{2}\\text{O}\\\\ \\text{rate}=k\\left[\\text{NO}\\right]\\left[{\\text{H}}_{2}\\right]\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\begin{array}{l}2\\text{NO}+{\\text{O}}_{2}\\rightarrow 2{\\text{NO}}_{2}\\\\ \\text{rate}=k{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]\\end{array}[\/latex]<\/li>\r\n \t<li>\u00a0[latex]\\begin{array}{l}\\text{NO}+{\\text{O}}_{3}\\rightarrow{\\text{NO}}_{2}+{\\text{O}}_{2}\\\\ \\text{rate}=k\\left[\\text{NO}\\right]\\left[{\\text{O}}_{3}\\right]\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Write the rate equation for each of the following elementary reactions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{O}}_{3}\\stackrel{\\text{sunlight}}{\\to }{\\text{O}}_{2}+\\text{O}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{O}}_{3}+\\text{Cl}\\rightarrow{\\text{O}}_{2}+\\text{ClO}[\/latex]<\/li>\r\n \t<li>[latex]\\text{ClO}+\\text{O}\\rightarrow\\text{Cl}+{\\text{O}}_{2}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{O}}_{3}+\\text{NO}\\rightarrow{\\text{NO}}_{2}+{\\text{O}}_{2}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{NO}}_{2}+\\text{O}\\rightarrow\\text{NO}+{\\text{O}}_{2}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Nitrogen(II) oxide, NO, reacts with hydrogen, H<sub>2<\/sub>, according to the following equation:\r\n[latex]2\\text{NO}+2{\\text{H}}_{2}\\rightarrow{\\text{N}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]What would the rate law be if the mechanism for this reaction were:\r\n[latex]\\begin{array}{l}2\\text{NO}+{\\text{H}}_{2}\\rightarrow{\\text{N}}_{2}+{\\text{H}}_{2}{\\text{O}}_{2}\\left(\\text{slow}\\right)\\\\ {\\text{H}}_{2}{\\text{O}}_{2}+{\\text{H}}_{2}\\rightarrow 2{\\text{H}}_{2}\\text{O}\\left(\\text{fast}\\right)\\end{array}[\/latex]<\/li>\r\n \t<li>Experiments were conducted to study the rate of the reaction represented by this equation[footnote]This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.[\/footnote].\r\n[latex]\\text{2NO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{N}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]\r\nInitial concentrations and rates of reaction are given here.\r\n<table id=\"fs-idp16941152\" class=\"medium unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row and it labels each column, \u201cExperiment,\u201d \u201cInitial Concentration [ N O ] ( mol \/ L ),\u201d \u201cInitial Concentration, [ H subscript 2 ] ( mol \/ L ),\u201d and \u201cInitial Rate of Formation of N subscript 2 ( mol \/ L min ).\u201d Under the \u201cExperiment\u201d columns are the numbers: 1, 2, 3, and 4. Under the \u201cInitial Concentration [ N O ] ( mol \/ L )\u201d column are the numbers: 0.0060, 0.0060, 0.0010, and 0.0020. Under the \u201cInitial Concentration, [ H subscript 2 ] ( mol \/ L )\u201d column are the numbers: 0.0010, 0.0020, 0.0060, 0.0060. Under the \u201cInitial Rate of Formation of N subscript 2 ( mol \/ L min )\u201d are the numbers: 1.8 times ten to the negative 4; 3.6 times ten to the negative 4; 0.30 times ten to the negative 4; and 1.2 times ten to the negative 4.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Experiment<\/th>\r\n<th>Initial Concentration [NO] (mol\/L)<\/th>\r\n<th>Initial Concentration, [H<sub>2<\/sub>] (mol\/L)<\/th>\r\n<th>Initial Rate of Formation of N<sub>2<\/sub> (mol\/L min)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>1<\/td>\r\n<td>0.0060<\/td>\r\n<td>0.0010<\/td>\r\n<td>1.8 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>0.0060<\/td>\r\n<td>0.0020<\/td>\r\n<td>3.6 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>3<\/td>\r\n<td>0.0010<\/td>\r\n<td>0.0060<\/td>\r\n<td>0.30 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>4<\/td>\r\n<td>0.0020<\/td>\r\n<td>0.0060<\/td>\r\n<td>1.2 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nConsider the following questions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Determine the order for each of the reactants, NO and H<sub>2<\/sub>, from the data given and show your reasoning.<\/li>\r\n \t<li>Write the overall rate law for the reaction.<\/li>\r\n \t<li>Calculate the value of the rate constant, <em>k<\/em>, for the reaction. Include units.<\/li>\r\n \t<li>For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H<sub>2<\/sub> had been consumed.<\/li>\r\n \t<li>The following sequence of elementary steps is a proposed mechanism for the reaction.<\/li>\r\n<\/ol>\r\nStep 1: [latex]\\text{NO}+\\text{NO}\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{2}[\/latex]\r\nStep 2: [latex]{\\text{N}}_{2}{\\text{O}}_{2}+{\\text{H}}_{2}\\rightleftharpoons{\\text{H}}_{2}\\text{O}+{\\text{N}}_{2}\\text{O}[\/latex]\r\nStep 3: [latex]{\\text{N}}_{2}\\text{O}+{\\text{H}}_{2}\\rightleftharpoons{\\text{N}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]\r\n\r\nBased on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.<\/li>\r\n \t<li>The reaction of CO with Cl<sub>2<\/sub> gives phosgene (COCl<sub>2<\/sub>), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:[latex]{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{Cl}\\left(g\\right)[\/latex] (fast, k<sub>1<\/sub> represents the forward rate constant, <em>k<\/em><sub>-1<\/sub> the reverse rate constant)\r\n[latex]\\text{CO}\\left(g\\right)+\\text{Cl}\\left(g\\right)\\rightarrow\\text{COCl}\\left(g\\right)[\/latex] (slow, <em>k<\/em><sub>2<\/sub> the rate constant)\r\n[latex]\\text{COCl}\\left(g\\right)+\\text{Cl}\\left(g\\right)\\rightarrow{\\text{COCl}}_{2}\\left(g\\right)[\/latex] (fast, <em>k<\/em><sub>3<\/sub> the rate constant)\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Write the overall reaction.<\/li>\r\n \t<li>Identify all intermediates.<\/li>\r\n \t<li>Write the rate law for each elementary reaction.<\/li>\r\n \t<li>Write the overall rate law expression.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"780544\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"780544\"]\r\n\r\n2. No. In general, for the overall reaction, we cannot predict the effect of changing the concentration without knowing the rate equation. Yes if the reaction is an elementary reaction, then doubling the concentration of <em>A<\/em> doubles the rate.\r\n\r\n4. In an elementary reaction, the rate constant is multiplied by the concentration of the reactant raised to the power of its stoichiometric coefficient. Rate = <em>k<\/em>[<em>A<\/em>][<em>B<\/em>]<sup>2<\/sup>; Rate = <em>k<\/em>[<em>A<\/em>]<sup>3<\/sup>\r\n\r\n6.\u00a0The rate equation are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Rate<sub>1<\/sub> = <em>k<\/em>[O<sub>3<\/sub>]<\/li>\r\n \t<li>Rate<sub>2<\/sub> = <em>k<\/em>[O<sub>3<\/sub>][Cl]<\/li>\r\n \t<li>Rate<sub>3<\/sub> = <em>k<\/em>[ClO][O]<\/li>\r\n \t<li>Rate<sub>2<\/sub> = <em>k<\/em>[O<sub>3<\/sub>][NO]<\/li>\r\n \t<li>Rate<sub>3<\/sub> = <em>k<\/em>[NO<sub>2<\/sub>][O]<\/li>\r\n<\/ol>\r\n8. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Doubling [H<sub>2<\/sub>] doubles the rate. [H<sub>2<\/sub>] must enter the rate equation to the first power. Doubling [NO] increases the rate by a factor of 4. [NO] must enter the rate law to the second power.<\/li>\r\n \t<li>The rate law is Rate = <em>k<\/em> [NO]<sup>2<\/sup>[H<sub>2<\/sub>].<\/li>\r\n \t<li>1.8 \u00d7 10<sup>\u22124<\/sup> mol\/L\/ min = <em>k<\/em>[0.0060 mol\/L]<sup>2<\/sup>[0.0010 mol\/L], <em>k<\/em> = 5.0 \u00d7 10<sup>3<\/sup> mol<sup>\u22122<\/sup> L<sup>\u22122<\/sup> min<sup>\u22121<\/sup>;<\/li>\r\n \t<li>The reaction has consumed 0.0010 mol\/L of H<sub>2<\/sub>. The amount of NO consumed is the same, 0.0010 mol\/L of NO. Thus 0.0060 \u2212 0.0010 = 0.0050 mol\/L remains.<\/li>\r\n \t<li>Step II is the rate-determining step. If step I gives N<sub>2<\/sub>O<sub>2<\/sub> in adequate amount, steps 1 and 2 combine to give [latex]2\\text{NO}+{\\text{H}}_{2}\\rightarrow{\\text{H}}_{2}\\text{O}+{\\text{N}}_{2}\\text{O}.[\/latex] This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3, which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>bimolecular reaction: <\/strong>elementary reaction involving the collision and combination of two reactant species\r\n\r\n<strong>elementary reaction: <\/strong>reaction that takes place precisely as depicted in its chemical equation\r\n\r\n<strong>intermediate: <\/strong>molecule or ion produced in one step of a reaction mechanism and consumed in another\r\n\r\n<strong>molecularity: <\/strong>number of reactant species (atoms, molecules or ions) involved in an elementary reaction\r\n\r\n<strong>rate-determining step: <\/strong>(also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction\r\n\r\n<strong>reaction mechanism: <\/strong>stepwise sequence of elementary reactions by which a chemical change takes place\r\n\r\n<strong>termolecular reaction: <\/strong>elementary reaction involving the simultaneous collision and combination of three reactant species\r\n\r\n<strong>unimolecular reaction: <\/strong>elementary reaction involving the rearrangement of a single reactant species to produce one or more molecules of product","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Distinguish net reactions from elementary reactions (steps)<\/li>\n<li>Identify the molecularity of elementary reactions<\/li>\n<li>Write a balanced chemical equation for a process given its reaction mechanism<\/li>\n<li>Derive the rate law consistent with a given reaction mechanism<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp19792192\">Chemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The\u00a0<strong><span id=\"term551\" data-type=\"term\">reaction mechanism<\/span><\/strong>\u00a0(or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs.<\/p>\n<p id=\"fs-idm10360272\">The decomposition of ozone, for example, appears to follow a mechanism with two steps:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{O}}_{3}\\left(g\\right)\\rightarrow{\\text{O}}_{2}\\left(g\\right)+\\text{O}\\\\ \\text{O}+{\\text{O}}_{3}\\left(g\\right)\\rightarrow 2{\\text{O}}_{2}\\left(g\\right)\\end{array}[\/latex]<\/p>\n<p>We call each step in a reaction mechanism an <strong>elementary reaction<\/strong>. Elementary reactions occur exactly as they are written and cannot be broken down into simpler steps. Elementary reactions add up to the overall reaction, which, for the decomposition, is:<\/p>\n<p style=\"text-align: center;\">[latex]2{\\text{O}}_{3}\\left(g\\right)\\rightarrow 3{\\text{O}}_{2}\\left(g\\right)[\/latex]<\/p>\n<p>Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called <strong>intermediates<\/strong>.<\/p>\n<p id=\"fs-idp6340064\">While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react to give three molecules of oxygen, the mechanism of the reaction\u00a0<em data-effect=\"italics\">does not involve the direct collision and reaction of two ozone molecules<\/em>. Instead, one O<sub>3<\/sub>\u00a0decomposes to yield O<sub>2<\/sub>\u00a0and an oxygen atom, and a second O<sub>3<\/sub>\u00a0molecule subsequently reacts with the oxygen atom to yield two additional O<sub>2<\/sub>\u00a0molecules.<\/p>\n<p id=\"fs-idm500275472\">Unlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction\u2019s equation undergo only the bond-breaking and\/or making events depicted to yield the product(s). For this reason,\u00a0<em data-effect=\"italics\">the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction<\/em>. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation.<\/p>\n<h2>Unimolecular Elementary Reactions<\/h2>\n<p>The <strong>molecularity<\/strong> of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a <strong>unimolecular reaction<\/strong> involves the rearrangement of a <em>single<\/em> reactant species to produce one or more molecules of product:<\/p>\n<p style=\"text-align: center;\">[latex]A\\rightarrow\\text{products}[\/latex]<\/p>\n<p>The rate equation for a unimolecular reaction is:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rate}=k[A][\/latex]<\/p>\n<p>A unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{O}}_{3}\\rightarrow{\\text{O}}_{2}+\\text{O}[\/latex]<\/p>\n<p>illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism. However, some unimolecular reactions may have only a single reaction in the reaction mechanism. (In other words, an elementary reaction can also be an overall reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, C<sub>4<\/sub>H<sub>8<\/sub>, to ethylene, C<sub>2<\/sub>H<sub>4<\/sub>, is represented by the following chemical equation:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212316\/CNX_Chem_12_06_CyclobD_img1.jpg\" alt=\"In this figure, structural formulas are used to illustrate a chemical reaction. On the left, a structural formula for cyclobutane is shown. This structure is composed of 4 C atoms connected with single bonds in a square shape. Each C atom is bonded to two other C atoms in the structure, leaving two bonds for H atoms pointing outward above, below, left, and right. An arrow points right to two identical ethane molecules with a plus symbol between them. Each of these molecules contains two C atoms connected with a double bond oriented vertically between them. The C atom at the top of these molecules has H atoms bonded above to the right and left. Similarly, the lower C atom has two H atoms bonded below to the right and left.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p>This equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing first-order behavior:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rate}=-\\dfrac{\\Delta\\left[{\\text{C}}_{4}{\\text{H}}_{8}\\right]}{\\Delta t}=k\\left[{\\text{C}}_{4}{\\text{H}}_{8}\\right][\/latex]<\/p>\n<p>This agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction.<\/p>\n<h2>Bimolecular Elementary Reactions<\/h2>\n<p>A\u00a0<strong><span id=\"term556\" data-type=\"term\">bimolecular reaction<\/span><\/strong>\u00a0involves two reactant species, for example:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}A+B\\rightarrow\\text{products}\\\\ \\text{and}\\\\ 2A\\rightarrow\\text{products}\\end{array}[\/latex]<\/p>\n<p>For the first type, in which the two reactant molecules are different, the rate law is first-order in\u00a0<em data-effect=\"italics\">A<\/em>\u00a0and first order in\u00a0<em data-effect=\"italics\">B<\/em>\u00a0(second-order overall):<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rate} = k[A][B][\/latex]<\/p>\n<p>For the second type, in which two identical molecules collide and react, the rate law is second order in\u00a0<em data-effect=\"italics\">A<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rate} = k[A][B] = k[A]^{2}[\/latex]<\/p>\n<p>Some chemical reactions have mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{NO}}_{2}\\left(g\\right)+\\text{CO}\\left(g\\right)\\rightarrow\\text{NO}\\left(g\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<\/p>\n<div style=\"width: 710px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/openstax.org\/resources\/9786aed35d8e9f6d86771aad63ad5543b409678e\" alt=\"This figure provides an illustration of a reaction between two H I molecules using space filling models. H atoms are shown as white spheres, and I atoms are shown as purple spheres. On the left, two H I molecules are shownwith a small white sphere bonded to a much larger purple sphere. The label, \u201cTwo H I molecules,\u201d appears below. An arrow points right to a similar structure in which the two molecules appear pushed together, so that the purple spheres of the two molecules are touching. Below appears the label, \u201cTransition state.\u201d Following another arrow, two white spheres are shown vertically oriented and bonded together with the label, \u201cH subscript 2\u201d above. The H subscript 2 molecule is followed by a plus sign and two purple spheres bonded together with the label, \u201cI subscript 2\u201d above. Below these structures is the label, \u201cHydrogen iodide molecules decompose to produce hydrogen H subscript 2 and iodine I subscript 2.\u201d\" width=\"700\" height=\"93\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The probable mechanism for the reaction between NO2 and CO to yield NO and CO2.<\/p>\n<\/div>\n<p>Bimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{O}\\left(g\\right)+{\\text{O}}_{3}\\left(g\\right)\\rightarrow{\\text{2O}}_{2}\\left(g\\right)[\/latex]<\/p>\n<h2>Termolecular Elementary Reactions<\/h2>\n<p>An elementary <strong>termolecular reaction<\/strong> involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{2NO}+{\\text{O}}_{2}\\rightarrow 2{\\text{NO}}_{2}\\\\ \\text{rate}=k{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]\\end{array}[\/latex]<\/p>\n<p>Likewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{2NO}+{\\text{Cl}}_{2}\\rightarrow 2\\text{NOCl}\\\\ \\text{rate}=k{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{Cl}}_{2}\\right]\\end{array}[\/latex]<\/p>\n<h2>Relating Reaction Mechanisms to Rate Laws<\/h2>\n<div style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212320\/CNX_Chem_12_06_Cattle1.jpg\" alt=\"A photo is shown of cattle passing through a narrow chute into a holding pen. A person directs them through the gate with a long white and red pole.\" width=\"300\" height=\"201\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a02. A cattle chute is a nonchemical example of a rate-determining step. Cattle can only be moved from one holding pen to another as quickly as one animal can make its way through the chute. (credit: Loren Kerns)<\/p>\n<\/div>\n<p>It&#8217;s often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the <strong>rate-limiting step<\/strong> (or rate-determining step) of the reaction.<\/p>\n<p>As described above, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, we must determine the overall rate law from experimental data and deduce the mechanism from the rate law (and sometimes from other data). The reaction of NO<sub>2<\/sub> and CO provides an illustrative example:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{NO}}_{2}\\left(g\\right)+\\text{CO}\\left(g\\right)\\rightarrow{\\text{CO}}_{2}\\left(g\\right)+\\text{NO}\\left(g\\right)[\/latex]<\/p>\n<p>For temperatures above 225 \u00b0C, the rate law has been found to be:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rate}=k\\left[{\\text{NO}}_{2}\\right]\\left[\\text{CO}\\right][\/latex]<\/p>\n<p>The reaction is first order with respect to NO<sub>2<\/sub> and first-order with respect to CO. This is consistent with a single-step bimolecular mechanism and it is <em>possible<\/em> that this is the mechanism for this reaction at high temperatures.<\/p>\n<p>At temperatures below 225 \u00b0C<em>,<\/em> the reaction is described by a rate law that is second order with respect to NO<sub>2<\/sub>:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rate}=k{\\left[{\\text{NO}}_{2}\\right]}^{2}[\/latex]<\/p>\n<p>This is consistent with a mechanism that involves the following two elementary reactions, the first of which is slower and is therefore the rate-determining step:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{NO}}_{2}\\left(g\\right)+{\\text{NO}}_{2}\\left(g\\right)\\rightarrow{\\text{NO}}_{3}\\left(g\\right)+\\text{NO}\\left(g\\right)\\qquad\\left(\\text{slow}\\right)\\\\ {\\text{NO}}_{3}\\left(g\\right)+\\text{CO}\\left(g\\right)\\rightarrow{\\text{NO}}_{2}\\left(g\\right)+{\\text{CO}}_{2}\\left(g\\right)\\qquad\\left(\\text{fast}\\right)\\end{array}[\/latex]<\/p>\n<p>The rate-determining step gives a rate law showing second-order dependence on the NO<sub>2<\/sub> concentration, and the sum of the two equations gives the net overall reaction.<\/p>\n<p>In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving an <em>equilibrium<\/em> reaction, the rate law for the overall reaction may be more difficult to derive.<\/p>\n<p>a reversible reaction is at\u00a0<em data-effect=\"italics\">equilibrium<\/em>\u00a0when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species N<sub>2<\/sub>O<sub>2<\/sub>. When this reaction is at equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{NO}+\\text{NO}\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{2}\\\\ {\\text{rate}}_{\\text{forward}}={\\text{rate}}_{\\text{reverse}}\\\\ {k}_{1}{\\left[\\text{NO}\\right]}^{2}={k}_{-1}\\left[{\\text{N}}_{2}{\\text{O}}_{2}\\right]\\end{array}[\/latex]<\/p>\n<p>This expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(\\dfrac{{\\text{k}}_{1}{\\left[\\text{NO}\\right]}^{2}}{{\\text{k}}_{-1}}\\right)=\\left[{\\text{N}}_{2}{\\text{O}}_{2}\\right][\/latex]<\/p>\n<p>Since intermediate species concentrations are not used in formulating rate laws for overall reactions, this approach is sometimes necessary, as illustrated in the following example exercise.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Deriving A Rate Law From a Reaction Mechanism<\/h3>\n<p>The two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:<\/p>\n<p style=\"padding-left: 30px;\"><strong>Step 1:<\/strong>\u00a0 [latex]\\text{NO}(\\text{g})+\\text{Cl}_{2}(\\text{g})\\rightleftharpoons{\\text{NOCl}}_{2}(\\text{g})\\qquad\\text{fast}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\"><strong>Step 2:<\/strong>\u00a0 [latex]\\text{NOCl}_{2}(\\text{g})+\\text{NO}(\\text{g})\\longrightarrow{2}\\text{NOCl}(\\text{g})\\qquad\\text{slow}[\/latex]<\/p>\n<p id=\"fs-idm352158848\">Use this mechanism to derive the equation and predicted rate law for the overall reaction.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q339226\">Show Solution<\/span><\/p>\n<div id=\"q339226\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation for the overall reaction is obtained by adding the two elementary reactions:<\/p>\n<p style=\"text-align: center;\">[latex]2\\text{NO}(\\text{g})+\\text{Cl}_{2}(\\text{g})\\rightleftharpoons{2}\\text{NOCl}(\\text{g})[\/latex]<\/p>\n<p>To derive a rate law from this mechanism, first write rates laws for each of the two steps.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\text{rate}_{1}}& =& {k}_{1}\\left[{\\text{NO}}\\right]\\left[\\text{Cl}_{2}\\right]\\qquad\\text{for the forward reaction of step 1}\\hfill \\\\ \\hfill{\\text{rate}_{-1}}& =& {k}_{-1}\\left[{\\text{NOCl}}_{2}\\right]\\qquad\\text{for the reverse reaction of step 1}\\hfill \\\\ \\hfill{\\text{rate}_{2}}& =& {k}_{2}\\left[\\text{NOCl}_{2}\\right]\\left[\\text{NO}\\right]\\qquad\\text{for step 2}\\hfill \\end{array}[\/latex]<\/p>\n<p id=\"fs-idm328804768\">Step 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [NOCl<sub>2<\/sub>]. To remedy this, use the first step\u2019s rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations.<\/p>\n<p id=\"fs-idm328530432\">Assuming Step 1 is at equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rate}_{1}=\\text{rate}_{-1}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{k}_{1}\\left[{\\text{NO}}\\right]\\left[\\text{Cl}_{2}\\right]={k}_{-1}\\left[\\text{NOCl}_{2}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\text{NOCl}_{2}\\right]=\\left(\\dfrac{{k}_{1}}{{k}_{-1}}\\right)\\left[\\text{NO}\\right]\\left[\\text{Cl}_{2}\\right][\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Substituting this expression into the rate law for Step 2 yields:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{rate}_{2}=\\text{rate}_{\\text{overall}}=\\left[\\text{NOCl}_{2}\\right]=\\left(\\dfrac{{k}_{2}{k}_{1}}{{k}_{-1}}\\right)\\left[\\text{NO}\\right]^{2}\\left[\\text{Cl}_{2}\\right][\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>The first step of a proposed multistep mechanism is:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{F}_{2}(\\text{g})\\rightleftharpoons{2}\\text{F}(\\text{g})\\qquad\\text{fast}[\/latex]<\/p>\n<p id=\"fs-idm373041168\">Derive the equation relating atomic fluorine concentration to molecular fluorine concentration.<\/p>\n<div id=\"fs-idm330964176\" class=\"ui-has-child-title\" data-type=\"note\"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q853909\">Show Solution<\/span><\/p>\n<div id=\"q853909\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex][F]=\\left(\\dfrac{{\\text{k}}_{1}{\\left[\\text{F}_{2}\\right]}}{{\\text{k}}_{-1}}\\right)^{1\/2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The overall rate of a reaction is determined by the rate of the slowest step, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Why are elementary reactions involving three or more reactants very uncommon?<\/li>\n<li>In general, can we predict the effect of doubling the concentration of <em>A<\/em> on the rate of the overall reaction [latex]A+B\\rightarrow C[\/latex] ? Can we predict the effect if the reaction is known to be an elementary reaction?<\/li>\n<li>Define these terms:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>unimolecular reaction<\/li>\n<li>bimolecular reaction<\/li>\n<li>elementary reaction<\/li>\n<li>overall reaction<\/li>\n<\/ol>\n<\/li>\n<li>What is the rate equation for the elementary termolecular reaction [latex]A+2B\\rightarrow\\text{products}[\/latex] ? For [latex]3A\\rightarrow\\text{products}[\/latex] ?<\/li>\n<li>Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{l}{\\text{Cl}}_{2}+\\text{CO}\\rightarrow{\\text{Cl}}_{2}\\text{CO}\\\\ \\text{rate}=k{\\left[{\\text{Cl}}_{2}\\right]}^{3\\text{\/}2}\\left[\\text{CO}\\right]\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{l}{\\text{PCl}}_{3}+{\\text{Cl}}_{2}\\rightarrow{\\text{PCl}}_{5}\\\\ \\text{rate}=k\\left[{\\text{PCl}}_{3}\\right]\\left[{\\text{Cl}}_{2}\\right]\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{l}2\\text{NO}+{\\text{H}}_{2}\\rightarrow{\\text{N}}_{2}+{\\text{H}}_{2}\\text{O}\\\\ \\text{rate}=k\\left[\\text{NO}\\right]\\left[{\\text{H}}_{2}\\right]\\end{array}[\/latex]<\/li>\n<li>[latex]\\begin{array}{l}2\\text{NO}+{\\text{O}}_{2}\\rightarrow 2{\\text{NO}}_{2}\\\\ \\text{rate}=k{\\left[\\text{NO}\\right]}^{2}\\left[{\\text{O}}_{2}\\right]\\end{array}[\/latex]<\/li>\n<li>\u00a0[latex]\\begin{array}{l}\\text{NO}+{\\text{O}}_{3}\\rightarrow{\\text{NO}}_{2}+{\\text{O}}_{2}\\\\ \\text{rate}=k\\left[\\text{NO}\\right]\\left[{\\text{O}}_{3}\\right]\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Write the rate equation for each of the following elementary reactions:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{O}}_{3}\\stackrel{\\text{sunlight}}{\\to }{\\text{O}}_{2}+\\text{O}[\/latex]<\/li>\n<li>[latex]{\\text{O}}_{3}+\\text{Cl}\\rightarrow{\\text{O}}_{2}+\\text{ClO}[\/latex]<\/li>\n<li>[latex]\\text{ClO}+\\text{O}\\rightarrow\\text{Cl}+{\\text{O}}_{2}[\/latex]<\/li>\n<li>[latex]{\\text{O}}_{3}+\\text{NO}\\rightarrow{\\text{NO}}_{2}+{\\text{O}}_{2}[\/latex]<\/li>\n<li>[latex]{\\text{NO}}_{2}+\\text{O}\\rightarrow\\text{NO}+{\\text{O}}_{2}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Nitrogen(II) oxide, NO, reacts with hydrogen, H<sub>2<\/sub>, according to the following equation:<br \/>\n[latex]2\\text{NO}+2{\\text{H}}_{2}\\rightarrow{\\text{N}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]What would the rate law be if the mechanism for this reaction were:<br \/>\n[latex]\\begin{array}{l}2\\text{NO}+{\\text{H}}_{2}\\rightarrow{\\text{N}}_{2}+{\\text{H}}_{2}{\\text{O}}_{2}\\left(\\text{slow}\\right)\\\\ {\\text{H}}_{2}{\\text{O}}_{2}+{\\text{H}}_{2}\\rightarrow 2{\\text{H}}_{2}\\text{O}\\left(\\text{fast}\\right)\\end{array}[\/latex]<\/li>\n<li>Experiments were conducted to study the rate of the reaction represented by this equation<a class=\"footnote\" title=\"This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.\" id=\"return-footnote-2227-1\" href=\"#footnote-2227-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>.<br \/>\n[latex]\\text{2NO}\\left(g\\right)+2{\\text{H}}_{2}\\left(g\\right)\\rightarrow{\\text{N}}_{2}\\left(g\\right)+2{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<br \/>\nInitial concentrations and rates of reaction are given here.<\/p>\n<table id=\"fs-idp16941152\" class=\"medium unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row and it labels each column, \u201cExperiment,\u201d \u201cInitial Concentration [ N O ] ( mol \/ L ),\u201d \u201cInitial Concentration, [ H subscript 2 ] ( mol \/ L ),\u201d and \u201cInitial Rate of Formation of N subscript 2 ( mol \/ L min ).\u201d Under the \u201cExperiment\u201d columns are the numbers: 1, 2, 3, and 4. Under the \u201cInitial Concentration [ N O ] ( mol \/ L )\u201d column are the numbers: 0.0060, 0.0060, 0.0010, and 0.0020. Under the \u201cInitial Concentration, [ H subscript 2 ] ( mol \/ L )\u201d column are the numbers: 0.0010, 0.0020, 0.0060, 0.0060. Under the \u201cInitial Rate of Formation of N subscript 2 ( mol \/ L min )\u201d are the numbers: 1.8 times ten to the negative 4; 3.6 times ten to the negative 4; 0.30 times ten to the negative 4; and 1.2 times ten to the negative 4.\" data-label=\"\">\n<thead>\n<tr valign=\"top\">\n<th>Experiment<\/th>\n<th>Initial Concentration [NO] (mol\/L)<\/th>\n<th>Initial Concentration, [H<sub>2<\/sub>] (mol\/L)<\/th>\n<th>Initial Rate of Formation of N<sub>2<\/sub> (mol\/L min)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>1<\/td>\n<td>0.0060<\/td>\n<td>0.0010<\/td>\n<td>1.8 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>0.0060<\/td>\n<td>0.0020<\/td>\n<td>3.6 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>3<\/td>\n<td>0.0010<\/td>\n<td>0.0060<\/td>\n<td>0.30 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>4<\/td>\n<td>0.0020<\/td>\n<td>0.0060<\/td>\n<td>1.2 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Consider the following questions:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Determine the order for each of the reactants, NO and H<sub>2<\/sub>, from the data given and show your reasoning.<\/li>\n<li>Write the overall rate law for the reaction.<\/li>\n<li>Calculate the value of the rate constant, <em>k<\/em>, for the reaction. Include units.<\/li>\n<li>For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H<sub>2<\/sub> had been consumed.<\/li>\n<li>The following sequence of elementary steps is a proposed mechanism for the reaction.<\/li>\n<\/ol>\n<p>Step 1: [latex]\\text{NO}+\\text{NO}\\rightleftharpoons{\\text{N}}_{2}{\\text{O}}_{2}[\/latex]<br \/>\nStep 2: [latex]{\\text{N}}_{2}{\\text{O}}_{2}+{\\text{H}}_{2}\\rightleftharpoons{\\text{H}}_{2}\\text{O}+{\\text{N}}_{2}\\text{O}[\/latex]<br \/>\nStep 3: [latex]{\\text{N}}_{2}\\text{O}+{\\text{H}}_{2}\\rightleftharpoons{\\text{N}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\n<p>Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.<\/li>\n<li>The reaction of CO with Cl<sub>2<\/sub> gives phosgene (COCl<sub>2<\/sub>), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:[latex]{\\text{Cl}}_{2}\\left(g\\right)\\rightleftharpoons2\\text{Cl}\\left(g\\right)[\/latex] (fast, k<sub>1<\/sub> represents the forward rate constant, <em>k<\/em><sub>-1<\/sub> the reverse rate constant)<br \/>\n[latex]\\text{CO}\\left(g\\right)+\\text{Cl}\\left(g\\right)\\rightarrow\\text{COCl}\\left(g\\right)[\/latex] (slow, <em>k<\/em><sub>2<\/sub> the rate constant)<br \/>\n[latex]\\text{COCl}\\left(g\\right)+\\text{Cl}\\left(g\\right)\\rightarrow{\\text{COCl}}_{2}\\left(g\\right)[\/latex] (fast, <em>k<\/em><sub>3<\/sub> the rate constant)<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Write the overall reaction.<\/li>\n<li>Identify all intermediates.<\/li>\n<li>Write the rate law for each elementary reaction.<\/li>\n<li>Write the overall rate law expression.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q780544\">Show Selected Solutions<\/span><\/p>\n<div id=\"q780544\" class=\"hidden-answer\" style=\"display: none\">\n<p>2. No. In general, for the overall reaction, we cannot predict the effect of changing the concentration without knowing the rate equation. Yes if the reaction is an elementary reaction, then doubling the concentration of <em>A<\/em> doubles the rate.<\/p>\n<p>4. In an elementary reaction, the rate constant is multiplied by the concentration of the reactant raised to the power of its stoichiometric coefficient. Rate = <em>k<\/em>[<em>A<\/em>][<em>B<\/em>]<sup>2<\/sup>; Rate = <em>k<\/em>[<em>A<\/em>]<sup>3<\/sup><\/p>\n<p>6.\u00a0The rate equation are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Rate<sub>1<\/sub> = <em>k<\/em>[O<sub>3<\/sub>]<\/li>\n<li>Rate<sub>2<\/sub> = <em>k<\/em>[O<sub>3<\/sub>][Cl]<\/li>\n<li>Rate<sub>3<\/sub> = <em>k<\/em>[ClO][O]<\/li>\n<li>Rate<sub>2<\/sub> = <em>k<\/em>[O<sub>3<\/sub>][NO]<\/li>\n<li>Rate<sub>3<\/sub> = <em>k<\/em>[NO<sub>2<\/sub>][O]<\/li>\n<\/ol>\n<p>8. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Doubling [H<sub>2<\/sub>] doubles the rate. [H<sub>2<\/sub>] must enter the rate equation to the first power. Doubling [NO] increases the rate by a factor of 4. [NO] must enter the rate law to the second power.<\/li>\n<li>The rate law is Rate = <em>k<\/em> [NO]<sup>2<\/sup>[H<sub>2<\/sub>].<\/li>\n<li>1.8 \u00d7 10<sup>\u22124<\/sup> mol\/L\/ min = <em>k<\/em>[0.0060 mol\/L]<sup>2<\/sup>[0.0010 mol\/L], <em>k<\/em> = 5.0 \u00d7 10<sup>3<\/sup> mol<sup>\u22122<\/sup> L<sup>\u22122<\/sup> min<sup>\u22121<\/sup>;<\/li>\n<li>The reaction has consumed 0.0010 mol\/L of H<sub>2<\/sub>. The amount of NO consumed is the same, 0.0010 mol\/L of NO. Thus 0.0060 \u2212 0.0010 = 0.0050 mol\/L remains.<\/li>\n<li>Step II is the rate-determining step. If step I gives N<sub>2<\/sub>O<sub>2<\/sub> in adequate amount, steps 1 and 2 combine to give [latex]2\\text{NO}+{\\text{H}}_{2}\\rightarrow{\\text{H}}_{2}\\text{O}+{\\text{N}}_{2}\\text{O}.[\/latex] This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3, which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>bimolecular reaction: <\/strong>elementary reaction involving the collision and combination of two reactant species<\/p>\n<p><strong>elementary reaction: <\/strong>reaction that takes place precisely as depicted in its chemical equation<\/p>\n<p><strong>intermediate: <\/strong>molecule or ion produced in one step of a reaction mechanism and consumed in another<\/p>\n<p><strong>molecularity: <\/strong>number of reactant species (atoms, molecules or ions) involved in an elementary reaction<\/p>\n<p><strong>rate-determining step: <\/strong>(also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction<\/p>\n<p><strong>reaction mechanism: <\/strong>stepwise sequence of elementary reactions by which a chemical change takes place<\/p>\n<p><strong>termolecular reaction: <\/strong>elementary reaction involving the simultaneous collision and combination of three reactant species<\/p>\n<p><strong>unimolecular reaction: <\/strong>elementary reaction involving the rearrangement of a single reactant species to produce one or more molecules of product<\/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-2227\">\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 2e. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/\">https:\/\/openstax.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>: Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-2227-1\">This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. <a href=\"#return-footnote-2227-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":17,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry 2e\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2227","chapter","type-chapter","status-publish","hentry"],"part":2992,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/2227","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":20,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/2227\/revisions"}],"predecessor-version":[{"id":7658,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/2227\/revisions\/7658"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/parts\/2992"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/2227\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/media?parent=2227"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapter-type?post=2227"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/contributor?post=2227"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/license?post=2227"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}