{"id":2741,"date":"2016-08-24T17:54:58","date_gmt":"2016-08-24T17:54:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2741"},"modified":"2016-08-24T21:21:51","modified_gmt":"2016-08-24T21:21:51","slug":"calculations-with-equilibrium-constants","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/calculations-with-equilibrium-constants\/","title":{"raw":"Calculations with Equilibrium Constants","rendered":"Calculations with Equilibrium Constants"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Perform equilibrium constant calculations.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Iron-poor blood?<\/h3>\r\n<p id=\"x-ck12-NTE5OTkwNGEyNWY0NTI5YTgyYTY1ODMzMDhjNTI0Zjg.-fmb\"><span class=\"x-ck12-img-inline\"><img class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/20140811155756732047.png\" alt=\"Ferrozine is used to assess the amount of iron in human blood because it tightly binds iron\" width=\"300\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NThjYTY4MDRiNWU0YzJlZDljNTAzZjVmOTNlNzhiZDg.-dgj\">Iron is an important component of red cells. Patients who have low iron will usually be anemic and have a lower than normal number of red blood cells. One way to assess serum iron concentration is with the use of Ferrozine, a complex organic molecule. Ferrozine forms a product with Fe <sup> 3+ <\/sup> , producing a pink color. In order to determine factors affecting the reaction, we need to measure the equilibrium constant. If the equilibrium does not lie far in the direction of products, precautions need to be taken when using this material to measure iron in serum.<\/p>\r\n\r\n<\/div>\r\n<h2 id=\"x-ck12-NDM4MTgxMWM0OTFiNjc4MTUzMDU2ZmVkNzEyOWI1N2E.-ksz_3-uxe\"><strong> Calculations with Equilibrium Constants <\/strong><\/h2>\r\n<p id=\"x-ck12-OWZmNjM1NWIxMTJiMmRhOWVkMzBmODY4OTUxMDM3MGU.-b58\">The general value of the equilibrium constant gives us information about whether the reactants or the products are favored at equilibrium. Since the product concentrations are in the numerator of the equilibrium expression, a <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212940\/24dff1268489ad0d5ec416270930b17a.png\" alt=\"K_{eq} &gt; 1\" width=\"60\" height=\"18\" \/> means that the products are favored over the reactants. A\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212940\/560a084f8eb953d9ffd0361ce00615a3.png\" alt=\"K_{eq}&lt; 1\" width=\"60\" height=\"18\" \/> means that the reactants are favored over the products.<\/p>\r\n<p id=\"x-ck12-NjJmZmUyNjUxYWE4MWU3N2QzMThlODljNjNlNmFjZDc.-sol\">Though it would often seem that the\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value would have various units depending on the values of the exponents in the expression, the general rule is that any units are dropped. All\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> values will be reported as having no units.<\/p>\r\n\r\n<h3>Sample Problem: Calculating an Equilibrium Constant<\/h3>\r\n<p id=\"x-ck12-MjI1YTY1MmI1ZTMxM2FlMGFjNjFjMjRkOGFlZjc1M2Q.-vxn\">Equilibrium occurs when nitrogen monoxide gas reacts with oxygen gas to form nitrogen dioxide gas.<\/p>\r\n<p id=\"x-ck12-lub\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212940\/94249b258c9573ab707f0ee4233722bc.png\" alt=\"2text{NO}(g)+text{O}_2(g) rightleftarrows 2text{NO}_2(g)\" width=\"220\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-ZDc3OWUyODRlYjlkMTAyYjZlYzY2ZTRmNjYwYzdlZmU.-nfn\">At equilibrium at 230\u00b0C, the concentrations are measured to be [NO] = 0.0542 M, [O <sub> 2 <\/sub> ] = 0.127 M, and [NO <sub> 2 <\/sub> ] = 15.5 M. Calculate the equilibrium constant at this temperature.<\/p>\r\n<p id=\"x-ck12-ODFjNTAwNDljM2RhMjljYTgyMTUxYWY2Nzk5ZWFhNGI.-ld8\"><em> Step 1: List the known values and plan the problem <\/em> .<\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-fnh\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NDkyMDcxOTUyMzUxOTU3ZDJkNmFiOWMzMGFkNzUxNWQ.-vi2\">\r\n \t<li>[NO] = 0.0542 M<\/li>\r\n \t<li>[O <sub> 2 <\/sub> ] = 0.127 M<\/li>\r\n \t<li>[NO <sub> 2 <\/sub> ] = 15.5 M<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-vef\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MjA2M2MxNjA4ZDZlMGJhZjgwMjQ5YzQyZTJiZTU4MDQ.-bbe\">\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-OTE2NjQwNzNkMjE0ZTE0NjMyZTE2ZjYwYTkwYzExMjM.-fuq\">The equilibrium expression is first written according to the general form in the text. The equilibrium values are substituted into the expression and the value calculated.<\/p>\r\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-t5b\"><em> Step 2: Solve <\/em> .<\/p>\r\n<p id=\"x-ck12-dyg\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212941\/f9a24caa9722758de9ed904c42858f7f.png\" alt=\"K_{eq}=frac{left [text{NO}_2 right ]^2}{left [text{NO} right ]^2 left [text{O}_2 right ]}\" width=\"133\" height=\"50\" \/><\/p>\r\n<p id=\"x-ck12-MWE2NGVkMjllMTc1YWE1OTYxMzEwMWM0MGU1YjI1Njc.-muc\">Substituting in the concentrations at equilibrium:<\/p>\r\n<p id=\"x-ck12-sfv\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212942\/1a0a28abdf432ee6d512ef41548ff1b1.png\" alt=\"K_{eq}=frac{left ( 15.5right )^2}{left (0.0542 right )^2 left (0.127right )}=6.44 times 10^5\" width=\"287\" height=\"49\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-ywv\"><em> Step 3: Think about your result <\/em> .<\/p>\r\n<p id=\"x-ck12-MjdhZWY2NmQxZjFiZTdkMjRhMmEwYTY2ZTczYmRkZmM.-gon\">The equilibrium concentration of the product NO <sub> 2 <\/sub> is significantly higher than the concentrations of the reactants NO and O <sub> 2 <\/sub> . As a result, the\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value is much larger than 1, an indication that the product is favored at equilibrium.<\/p>\r\n<p id=\"x-ck12-NmYxNjAyNWViYjMxY2YwMjM3NTgzNzllNzUwOWM3YmE.-q93\">The equilibrium expression only shows those substances whose concentrations are variable during the reaction. A pure solid or a pure liquid does not have a concentration that will vary during a reaction. Therefore, an equilibrium expression omits pure solids and liquids and only shows the concentrations of gases and aqueous solutions. The decomposition of mercury(II) oxide can be shown by the following equation, followed by its equilibrium expression.<\/p>\r\n<p id=\"x-ck12-t7v\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212943\/c5ecde134c0043a5f225a0af8a263015.png\" alt=\"2text{HgO}(s) rightleftarrows 2text{Hg}(l)+text{O}_2(g) qquad quad K_{eq}=[text{O}_2]\" width=\"348\" height=\"20\" \/><\/p>\r\n<p id=\"x-ck12-ZGNjMzE1N2EyNGYyMDIwYjAyNTk2N2E3ZWNjYjAyNzk.-bby\">The stoichiometry of an equation can also be used in a calculation of an equilibrium constant. At 40\u00b0C, solid ammonium carbamate decomposes to ammonia and carbon dioxide gases.<\/p>\r\n<p id=\"x-ck12-g1l\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212944\/f75fd6969bb9b0dd1313a09063298fcb.png\" alt=\"text{NH}_4text{CO}_2text{NH}_2(s) rightleftarrows 2text{NH}_3(g)+text{CO}_2(g)\" width=\"297\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-OGM3MDUwZGRjYTE5OGY2ZjFjYzk1YmIzMWE5ZWJhODA.-unq\">At equilibrium, the [CO <sub> 2 <\/sub> ] is found to be 4.71\u00a0\u00d7\u00a010 <sup> -3 <\/sup> M. Can the\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value be calculated from just that information? Because the ammonium carbamate is a solid, it is not present in the equilibrium expression.<\/p>\r\n<p id=\"x-ck12-cft\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212945\/c21923b1dd61b6b231a2587e0fce4a4b.png\" alt=\"K_{eq}=[text{NH}_3]^2 [text{CO}_2]\" width=\"146\" height=\"23\" \/><\/p>\r\n<p id=\"x-ck12-ZmQwNGJlYzJiZjU1ZWQ3MDY3ODMyYmQxNGNiYWM1MjY.-2ck\">The stoichiometry of the chemical equation indicates that as the ammonium carbamate decomposes, 2 mol of ammonia gas is produced for every 1 mol of carbon dioxide. Therefore, at equilibrium, the concentration of the ammonia will be twice the concentration of carbon dioxide. So [NH <sub> 3 <\/sub> ] = 2\u00a0\u00d7 (4.71\u00a0\u00d7 10 <sup> -3 <\/sup> ) = 9.42\u00a0\u00d7 10 <sup> -3 <\/sup> M. Substituting these values into the\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> expression:<\/p>\r\n<p id=\"x-ck12-nxo\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212946\/4b20af4979fd8f206b49cea89289d492.png\" alt=\"K_{eq}=(9.42 times 10^{-3})^2 (4.71 times 10^{-3})=4.18 times 10^{-7}\" width=\"379\" height=\"23\" \/><\/p>\r\n\r\n<h3>Using Equilibrium Constants<\/h3>\r\n<p id=\"x-ck12-MGZkZjgxMmI1NTkxMzI3ZTVhNWI5MWE1ZDBjNjNkMTE.-o82\">The equilibrium constants are known for a great many reactions. Hydrogen and bromine gases combine to form hydrogen bromide gas. At 730\u00b0C, the equation and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> are given below.<\/p>\r\n<p id=\"x-ck12-8j9\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212947\/ee71af302adf63ffeeb64f50edb46371.png\" alt=\"text{H}_2(g)+text{Br}_2(g) rightleftarrows 2text{HBr}(g) qquad K_{eq}=2.18 times 10^6\" width=\"387\" height=\"23\" \/><\/p>\r\n<p id=\"x-ck12-MzdkMDQyMzA3M2VkMGExNWY2Y2UyODJmYjkwOGM3Mzg.-sqt\">A certain reaction is begun with only HBr. When the reaction mixture reaches equilibrium at 730\u00b0C, the concentration of bromine gas is measured to be 0.00243 M. What is the concentration of the H <sub> 2 <\/sub> and the HBr at equilibrium?<\/p>\r\n<p id=\"x-ck12-YWY1MmM5Yzk4ZTc4ZDExNzAzN2RiMTg4M2Y2ODIxNDI.-asi\">Since the reaction begins with only HBr and the mole ratio of H <sub> 2 <\/sub> to Br <sub> 2 <\/sub> is 1:1, the concentration of H <sub> 2 <\/sub> at equilibrium is also 0.00243 M. The equilibrium expression can be rearranged to solve for the concentration of HBr at equilibrium.<\/p>\r\n<p id=\"x-ck12-nzy\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212949\/e0a93cf755ac5cfe9f62297a91e59373.png\" alt=\"K_{eq} &amp; = frac{[text{HBr}]^2}{[text{H}_2][text{Br}_2]}\\left [ text{HBr} right ] &amp; = sqrt{K_{eq}[text{H}_2][text{Br}_2]} \\&amp; = sqrt{2.18 times 10^6 (0.00243)(0.00243)} = 3.59 text{ M}\" width=\"386\" height=\"114\" \/><\/p>\r\n<p id=\"x-ck12-NzY4MDdmMTk4OWJhNzYwZjBhYTJmYTZhMWVmOGQyN2I.-yfs\">Since the value of the equilibrium constant is very high, the concentration of HBr is much greater than that of H <sub> 2 <\/sub> and Br <sub> 2 <\/sub> at equilibrium.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-NWE5ZTVmMTY5Mjc1N2ZiYzBjM2FiZjljY2I2MjQxNzQ.-ezb\">\r\n \t<li>Calculation of an equilibrium constant is described.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Practice<\/h3>\r\n<p id=\"x-ck12-YzFlZGZkYjhiMDQyNWIxYmUyNzNkZDllYjgwOGI1ZDA.-vvr\"><a href=\"http:\/\/www.chemtopics.com\/unit10\/ppu10.pdf\" target=\"_blank\">Work problems 1-4 at\u00a0ChemTopics.com.<\/a><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ZGZlM2ZjZWQ2ZDFiZDQzNTRiZjAyNmYzZmI0OTdiY2I.-gj5\">\r\n \t<li>What are the units for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> ?<\/li>\r\n \t<li>Why is the temperature specified in equilibrium problems?<\/li>\r\n \t<li>Why don\u2019t we include solids or liquids in equilibrium calculations?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"836080\"]Show References[\/reveal-answer]\r\n[hidden-answer a=\"836080\"]\r\n<h2>References<\/h2>\r\n<ol>\r\n \t<li>User:Yikrazuul\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Ferrozine.svg\">http:\/\/commons.wikimedia.org\/wiki\/File:Ferrozine.svg <\/a>.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Perform equilibrium constant calculations.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Iron-poor blood?<\/h3>\n<p id=\"x-ck12-NTE5OTkwNGEyNWY0NTI5YTgyYTY1ODMzMDhjNTI0Zjg.-fmb\"><span class=\"x-ck12-img-inline\"><img decoding=\"async\" class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/20140811155756732047.png\" alt=\"Ferrozine is used to assess the amount of iron in human blood because it tightly binds iron\" width=\"300\" \/><\/span><\/p>\n<p id=\"x-ck12-NThjYTY4MDRiNWU0YzJlZDljNTAzZjVmOTNlNzhiZDg.-dgj\">Iron is an important component of red cells. Patients who have low iron will usually be anemic and have a lower than normal number of red blood cells. One way to assess serum iron concentration is with the use of Ferrozine, a complex organic molecule. Ferrozine forms a product with Fe <sup> 3+ <\/sup> , producing a pink color. In order to determine factors affecting the reaction, we need to measure the equilibrium constant. If the equilibrium does not lie far in the direction of products, precautions need to be taken when using this material to measure iron in serum.<\/p>\n<\/div>\n<h2 id=\"x-ck12-NDM4MTgxMWM0OTFiNjc4MTUzMDU2ZmVkNzEyOWI1N2E.-ksz_3-uxe\"><strong> Calculations with Equilibrium Constants <\/strong><\/h2>\n<p id=\"x-ck12-OWZmNjM1NWIxMTJiMmRhOWVkMzBmODY4OTUxMDM3MGU.-b58\">The general value of the equilibrium constant gives us information about whether the reactants or the products are favored at equilibrium. Since the product concentrations are in the numerator of the equilibrium expression, a <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212940\/24dff1268489ad0d5ec416270930b17a.png\" alt=\"K_{eq} &gt; 1\" width=\"60\" height=\"18\" \/> means that the products are favored over the reactants. A\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212940\/560a084f8eb953d9ffd0361ce00615a3.png\" alt=\"K_{eq}&lt; 1\" width=\"60\" height=\"18\" \/> means that the reactants are favored over the products.<\/p>\n<p id=\"x-ck12-NjJmZmUyNjUxYWE4MWU3N2QzMThlODljNjNlNmFjZDc.-sol\">Though it would often seem that the\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value would have various units depending on the values of the exponents in the expression, the general rule is that any units are dropped. All\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> values will be reported as having no units.<\/p>\n<h3>Sample Problem: Calculating an Equilibrium Constant<\/h3>\n<p id=\"x-ck12-MjI1YTY1MmI1ZTMxM2FlMGFjNjFjMjRkOGFlZjc1M2Q.-vxn\">Equilibrium occurs when nitrogen monoxide gas reacts with oxygen gas to form nitrogen dioxide gas.<\/p>\n<p id=\"x-ck12-lub\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212940\/94249b258c9573ab707f0ee4233722bc.png\" alt=\"2text{NO}(g)+text{O}_2(g) rightleftarrows 2text{NO}_2(g)\" width=\"220\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-ZDc3OWUyODRlYjlkMTAyYjZlYzY2ZTRmNjYwYzdlZmU.-nfn\">At equilibrium at 230\u00b0C, the concentrations are measured to be [NO] = 0.0542 M, [O <sub> 2 <\/sub> ] = 0.127 M, and [NO <sub> 2 <\/sub> ] = 15.5 M. Calculate the equilibrium constant at this temperature.<\/p>\n<p id=\"x-ck12-ODFjNTAwNDljM2RhMjljYTgyMTUxYWY2Nzk5ZWFhNGI.-ld8\"><em> Step 1: List the known values and plan the problem <\/em> .<\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-fnh\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-NDkyMDcxOTUyMzUxOTU3ZDJkNmFiOWMzMGFkNzUxNWQ.-vi2\">\n<li>[NO] = 0.0542 M<\/li>\n<li>[O <sub> 2 <\/sub> ] = 0.127 M<\/li>\n<li>[NO <sub> 2 <\/sub> ] = 15.5 M<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-vef\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-MjA2M2MxNjA4ZDZlMGJhZjgwMjQ5YzQyZTJiZTU4MDQ.-bbe\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value<\/li>\n<\/ul>\n<p id=\"x-ck12-OTE2NjQwNzNkMjE0ZTE0NjMyZTE2ZjYwYTkwYzExMjM.-fuq\">The equilibrium expression is first written according to the general form in the text. The equilibrium values are substituted into the expression and the value calculated.<\/p>\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-t5b\"><em> Step 2: Solve <\/em> .<\/p>\n<p id=\"x-ck12-dyg\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212941\/f9a24caa9722758de9ed904c42858f7f.png\" alt=\"K_{eq}=frac{left [text{NO}_2 right ]^2}{left [text{NO} right ]^2 left [text{O}_2 right ]}\" width=\"133\" height=\"50\" \/><\/p>\n<p id=\"x-ck12-MWE2NGVkMjllMTc1YWE1OTYxMzEwMWM0MGU1YjI1Njc.-muc\">Substituting in the concentrations at equilibrium:<\/p>\n<p id=\"x-ck12-sfv\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212942\/1a0a28abdf432ee6d512ef41548ff1b1.png\" alt=\"K_{eq}=frac{left ( 15.5right )^2}{left (0.0542 right )^2 left (0.127right )}=6.44 times 10^5\" width=\"287\" height=\"49\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-ywv\"><em> Step 3: Think about your result <\/em> .<\/p>\n<p id=\"x-ck12-MjdhZWY2NmQxZjFiZTdkMjRhMmEwYTY2ZTczYmRkZmM.-gon\">The equilibrium concentration of the product NO <sub> 2 <\/sub> is significantly higher than the concentrations of the reactants NO and O <sub> 2 <\/sub> . As a result, the\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value is much larger than 1, an indication that the product is favored at equilibrium.<\/p>\n<p id=\"x-ck12-NmYxNjAyNWViYjMxY2YwMjM3NTgzNzllNzUwOWM3YmE.-q93\">The equilibrium expression only shows those substances whose concentrations are variable during the reaction. A pure solid or a pure liquid does not have a concentration that will vary during a reaction. Therefore, an equilibrium expression omits pure solids and liquids and only shows the concentrations of gases and aqueous solutions. The decomposition of mercury(II) oxide can be shown by the following equation, followed by its equilibrium expression.<\/p>\n<p id=\"x-ck12-t7v\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212943\/c5ecde134c0043a5f225a0af8a263015.png\" alt=\"2text{HgO}(s) rightleftarrows 2text{Hg}(l)+text{O}_2(g) qquad quad K_{eq}=[text{O}_2]\" width=\"348\" height=\"20\" \/><\/p>\n<p id=\"x-ck12-ZGNjMzE1N2EyNGYyMDIwYjAyNTk2N2E3ZWNjYjAyNzk.-bby\">The stoichiometry of an equation can also be used in a calculation of an equilibrium constant. At 40\u00b0C, solid ammonium carbamate decomposes to ammonia and carbon dioxide gases.<\/p>\n<p id=\"x-ck12-g1l\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212944\/f75fd6969bb9b0dd1313a09063298fcb.png\" alt=\"text{NH}_4text{CO}_2text{NH}_2(s) rightleftarrows 2text{NH}_3(g)+text{CO}_2(g)\" width=\"297\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-OGM3MDUwZGRjYTE5OGY2ZjFjYzk1YmIzMWE5ZWJhODA.-unq\">At equilibrium, the [CO <sub> 2 <\/sub> ] is found to be 4.71\u00a0\u00d7\u00a010 <sup> -3 <\/sup> M. Can the\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> value be calculated from just that information? Because the ammonium carbamate is a solid, it is not present in the equilibrium expression.<\/p>\n<p id=\"x-ck12-cft\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212945\/c21923b1dd61b6b231a2587e0fce4a4b.png\" alt=\"K_{eq}=[text{NH}_3]^2 [text{CO}_2]\" width=\"146\" height=\"23\" \/><\/p>\n<p id=\"x-ck12-ZmQwNGJlYzJiZjU1ZWQ3MDY3ODMyYmQxNGNiYWM1MjY.-2ck\">The stoichiometry of the chemical equation indicates that as the ammonium carbamate decomposes, 2 mol of ammonia gas is produced for every 1 mol of carbon dioxide. Therefore, at equilibrium, the concentration of the ammonia will be twice the concentration of carbon dioxide. So [NH <sub> 3 <\/sub> ] = 2\u00a0\u00d7 (4.71\u00a0\u00d7 10 <sup> -3 <\/sup> ) = 9.42\u00a0\u00d7 10 <sup> -3 <\/sup> M. Substituting these values into the\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> expression:<\/p>\n<p id=\"x-ck12-nxo\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212946\/4b20af4979fd8f206b49cea89289d492.png\" alt=\"K_{eq}=(9.42 times 10^{-3})^2 (4.71 times 10^{-3})=4.18 times 10^{-7}\" width=\"379\" height=\"23\" \/><\/p>\n<h3>Using Equilibrium Constants<\/h3>\n<p id=\"x-ck12-MGZkZjgxMmI1NTkxMzI3ZTVhNWI5MWE1ZDBjNjNkMTE.-o82\">The equilibrium constants are known for a great many reactions. Hydrogen and bromine gases combine to form hydrogen bromide gas. At 730\u00b0C, the equation and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> are given below.<\/p>\n<p id=\"x-ck12-8j9\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212947\/ee71af302adf63ffeeb64f50edb46371.png\" alt=\"text{H}_2(g)+text{Br}_2(g) rightleftarrows 2text{HBr}(g) qquad K_{eq}=2.18 times 10^6\" width=\"387\" height=\"23\" \/><\/p>\n<p id=\"x-ck12-MzdkMDQyMzA3M2VkMGExNWY2Y2UyODJmYjkwOGM3Mzg.-sqt\">A certain reaction is begun with only HBr. When the reaction mixture reaches equilibrium at 730\u00b0C, the concentration of bromine gas is measured to be 0.00243 M. What is the concentration of the H <sub> 2 <\/sub> and the HBr at equilibrium?<\/p>\n<p id=\"x-ck12-YWY1MmM5Yzk4ZTc4ZDExNzAzN2RiMTg4M2Y2ODIxNDI.-asi\">Since the reaction begins with only HBr and the mole ratio of H <sub> 2 <\/sub> to Br <sub> 2 <\/sub> is 1:1, the concentration of H <sub> 2 <\/sub> at equilibrium is also 0.00243 M. The equilibrium expression can be rearranged to solve for the concentration of HBr at equilibrium.<\/p>\n<p id=\"x-ck12-nzy\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212949\/e0a93cf755ac5cfe9f62297a91e59373.png\" alt=\"K_{eq} &amp; = frac{[text{HBr}]^2}{[text{H}_2][text{Br}_2]}\\left [ text{HBr} right ] &amp; = sqrt{K_{eq}[text{H}_2][text{Br}_2]} \\&amp; = sqrt{2.18 times 10^6 (0.00243)(0.00243)} = 3.59 text{ M}\" width=\"386\" height=\"114\" \/><\/p>\n<p id=\"x-ck12-NzY4MDdmMTk4OWJhNzYwZjBhYTJmYTZhMWVmOGQyN2I.-yfs\">Since the value of the equilibrium constant is very high, the concentration of HBr is much greater than that of H <sub> 2 <\/sub> and Br <sub> 2 <\/sub> at equilibrium.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-NWE5ZTVmMTY5Mjc1N2ZiYzBjM2FiZjljY2I2MjQxNzQ.-ezb\">\n<li>Calculation of an equilibrium constant is described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-YzFlZGZkYjhiMDQyNWIxYmUyNzNkZDllYjgwOGI1ZDA.-vvr\"><a href=\"http:\/\/www.chemtopics.com\/unit10\/ppu10.pdf\" target=\"_blank\">Work problems 1-4 at\u00a0ChemTopics.com.<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZGZlM2ZjZWQ2ZDFiZDQzNTRiZjAyNmYzZmI0OTdiY2I.-gj5\">\n<li>What are the units for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212938\/c64985e4b3631a26ef1d3fb0b88af84c.png\" alt=\"K_{eq}\" width=\"27\" height=\"18\" \/> ?<\/li>\n<li>Why is the temperature specified in equilibrium problems?<\/li>\n<li>Why don\u2019t we include solids or liquids in equilibrium calculations?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836080\">Show References<\/span><\/p>\n<div id=\"q836080\" class=\"hidden-answer\" style=\"display: none\">\n<h2>References<\/h2>\n<ol>\n<li>User:Yikrazuul\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Ferrozine.svg\">http:\/\/commons.wikimedia.org\/wiki\/File:Ferrozine.svg <\/a>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":17,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2741","chapter","type-chapter","status-publish","hentry"],"part":2340,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2741","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2741\/revisions"}],"predecessor-version":[{"id":2999,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2741\/revisions\/2999"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2340"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2741\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2741"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2741"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2741"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2741"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}