{"id":360,"date":"2016-01-11T20:00:08","date_gmt":"2016-01-11T20:00:08","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/hesss-law-2\/"},"modified":"2020-07-27T16:05:38","modified_gmt":"2020-07-27T16:05:38","slug":"hesss-law","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/hesss-law\/","title":{"raw":"Hess\u2019s Law","rendered":"Hess\u2019s Law"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Learn how to combine chemical equations and their enthalpy changes.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nNow that we understand that chemical reactions occur with a simultaneous change in energy, we can apply the concept more broadly. To start, remember that some chemical reactions are rather difficult to perform. For example, consider the combustion of carbon to make carbon monoxide:\r\n\r\n\\[\\ce{2C(s)}+\\ce{O2(g)}\\rightarrow \\ce{2CO(g)}\\quad \\Delta H=?\\]\r\n\r\nIn reality, this is extremely difficult to do; given the opportunity, carbon will react to make another compound, carbon dioxide:\r\n\r\n\\[\\ce{2C(s)}+\\ce{O2(g)}\\rightarrow \\ce{2CO2(g)}\\quad \\Delta H=-393.5\\text{ kJ}\\]\r\n\r\nIs there a way around this? Yes. It comes from the understanding that chemical equations can be treated like algebraic equations, with the arrow acting like the equals sign. Like algebraic equations, chemical equations can be combined, and if the same substance appears on both sides of the arrow, it can be cancelled out (much like a spectator ion in ionic equations). For example, consider these two reactions:\r\n\r\n\\[\\begin{array}{rrl}\r\n\\ce{2C(s)}+\\ce{2O2(g)}&\\rightarrow&\\ce{2CO2(g)} \\\\ \\\\\r\n\\ce{2CO2(g)}&\\rightarrow&\\ce{2CO(g)}+\\ce{O2(g)}\r\n\\end{array}\\]\r\n\r\nIf we added these two equations by combining all the reactants together and all the products together, we would get:\r\n\r\n\\[\\ce{2C(s)}+\\ce{2O2(g)}+\\ce{2CO2(g)}\\rightarrow \\ce{2CO2(g)}+\\ce{2CO(g)}+\\ce{O2(g)}\\]\r\n\r\nWe note that 2CO2<\/sub>(g) appears on both sides of the arrow, so they cancel:\r\n\r\n\\[\\ce{2C(s)}+\\ce{2O2(g)}+\\cancel{\\ce{2CO2(g)}}\\rightarrow \\cancel{\\ce{2CO2(g)}}+\\ce{2CO(g)}+\\ce{O2(g)}\\]\r\n\r\nWe also note that there are 2 mol of O2<\/sub> on the reactant side, and 1 mol of O2<\/sub> on the product side. We can cancel 1 mol of O2<\/sub> from both sides:\r\n\r\n\\[\\ce{2C(s)}+\\ce{\\cancel{2}O2(g)}\\rightarrow \\ce{2CO(g)}+\\cancel{\\ce{O2(g)}}\\]\r\n\r\nWhat do we have left?\r\n\r\n\\[\\ce{2C(s)}+\\ce{O2(g)}\\rightarrow \\ce{2CO(g)}\\]\r\n\r\nThis is the reaction we are looking for! So by algebraically combining chemical equations, we can generate new chemical equations that may not be feasible to perform.\r\n\r\nWhat about the enthalpy changes? [pb_glossary id=\"1712\"]Hess\u2019s law[\/pb_glossary]\u00a0states that when chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. Two corollaries immediately present themselves:\r\n
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    1. If a chemical reaction is reversed, the sign on \u0394H<\/i> is changed.<\/li>\r\n \t
    2. If a multiple of a chemical reaction is taken, the same multiple of the \u0394H<\/i> is taken as well.<\/li>\r\n<\/ol>\r\nWhat are the equations being combined? The first chemical equation is the combustion of C, which produces CO2<\/sub>:\r\n\r\n\\[\\ce{2C(s)}+\\ce{2O2(g)}\\rightarrow \\ce{2CO2(g)}\\]\r\n\r\nThis reaction is two times the reaction to make CO2<\/sub> from C(s) and O2<\/sub>(g), whose enthalpy change is known:\r\n\r\n\\[\\ce{C(s)}+\\ce{O2(g)}\\rightarrow \\ce{CO2(g)}\\quad \\Delta H=-393.5\\text{ kJ}\\]\r\n\r\nAccording to the first corollary, the first reaction has an energy change of two times \u2212393.5 kJ, or \u2212787.0 kJ:\r\n\r\n\\[\\ce{2C(s)}+\\ce{2O2(g)}\\rightarrow \\ce{2CO2(g)}\\quad \\Delta H=-787.0\\text{ kJ}\\]\r\n\r\nThe second reaction in the combination is related to the combustion of CO(g):\r\n\r\n\\[\\ce{2CO(g)}+\\ce{O2(g)}\\rightarrow \\ce{2CO2(g)}\\quad \\Delta H=-566.0\\text{ kJ}\\]\r\n\r\nThe second reaction in our combination is the reverse<\/i> of the combustion of CO. When we reverse the reaction, we change the sign on the \u0394H<\/i>:\r\n\r\n\\[\\ce{2CO2(g)}\\rightarrow \\ce{2CO(g)}+\\ce{O2(g)}\\quad \\Delta H=+566.0\\text{ kJ}\\]\r\n\r\nNow that we have identified the enthalpy changes of the two component chemical equations, we can combine the \u0394H<\/i> values and add them:\r\n

      \\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\ce{2C(s)}+\\ce{\\cancel{2}O2(g)}&\\rightarrow&\\cancel{\\ce{2CO2(g)}} \\\\\r\n\\cancel{\\ce{2CO2(g)}}&\\rightarrow&\\ce{2CO(g)}+\\cancel{\\ce{O2(g)}} \\\\\r\n\\midrule\r\n\\ce{2C(s)}+\\ce{O2(g)}&\\rightarrow&\\ce{2CO(g)}\r\n\\end{array}\r\n&\r\n\\begin{array}{rrl}\r\n\\Delta H&=&-787.0\\text{ kJ} \\\\\r\n\\Delta H&=&+566.0\\text{ kJ} \\\\\r\n\\Delta H&=&-787.0+566.0\\text{ kJ}=-221.0\\text{ kJ}\r\n\\end{array}\r\n\\end{array}\\)<\/p>\r\nHess\u2019s law is very powerful. It allows us to combine equations to generate new chemical reactions whose enthalpy changes can be calculated, rather than directly measured.\r\n

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      Example 7.14<\/p>\r\n\r\n<\/header>\r\n

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      Problem<\/h1>\r\nDetermine the enthalpy change of:\r\n

      \\(\\ce{C2H4}+\\ce{3O2}\\rightarrow \\ce{2CO2}+\\ce{2H2O}\\quad \\Delta H=?\\)<\/p>\r\nfrom these reactions:\r\n\r\n\\[\\begin{array}{rrll}\r\n\\ce{C2H2}+\\ce{H2}&\\rightarrow &\\ce{C2H4}&\\Delta H = -174.5\\text{ kJ} \\\\ \\\\\r\n\\ce{2C2H2}+\\ce{5O2}&\\rightarrow&\\ce{4CO2}+\\ce{2H2O}&\\Delta H =-1,692.2\\text{ kJ} \\\\ \\\\\r\n\\ce{2CO2}+\\ce{H2}&\\rightarrow&\\ce{2O2}+\\ce{C2H2}&\\Delta H =-167.5\\text{ kJ}\r\n\\end{array}\\]\r\n

      Solution<\/h2>\r\nWe will start by writing chemical reactions that put the correct number of moles of the correct substance on the proper side. For example, our desired reaction has C2<\/sub>H4<\/sub> as a reactant, and only one reaction from our data has C2<\/sub>H4<\/sub>. However, it has C2<\/sub>H4<\/sub> as a product. To make it a reactant, we need to reverse the reaction, changing the sign on the \u0394H<\/i>:\r\n\r\n\\[\\ce{C2H4}\\rightarrow \\ce{C2H2}+\\ce{H2}\\quad \\Delta H=+174.5\\text{ kJ}\\]\r\n\r\nWe need CO2<\/sub> and H2<\/sub>O as products. The second reaction has them on the proper side, so let us include one of these reactions (with the hope that the coefficients will work out when all our reactions are added):\r\n\r\n\\[\\ce{2C2H2}+\\ce{5O2}\\rightarrow \\ce{4CO2}+\\ce{2H2O}\\quad \\Delta H=-1,692.2\\text{ kJ}\\]\r\n\r\nWe note that we now have 4 mol of CO2<\/sub> as products; we need to get rid of 2 mol of CO2<\/sub>. The last reaction has 2CO2<\/sub> as a reactant. Let us use it as written:\r\n\r\n\\[\\ce{2CO2}+\\ce{H2}\\rightarrow \\ce{2O2}+\\ce{C2H2}\\quad \\Delta H=-167.5\\text{ kJ}\\]\r\n

      We combine these three reactions, modified as stated:<\/p>\r\n

      \\(\\begin{array}{ll}\r\n\\begin{array}{rrl}\r\n\\\\ \\\\\r\n\\ce{C2H4}&\\rightarrow&\\ce{C2H2}+\\ce{H2} \\\\\r\n\\ce{2C2H2}+\\ce{5O2}&\\rightarrow&\\ce{4CO2}+\\ce{2H2O} \\\\\r\n\\ce{2CO2}+\\ce{H2}&\\rightarrow&\\ce{2O2}+\\ce{C2H2} \\\\\r\n\\midrule\r\n\\ce{C2H4}+\\ce{2C2H2}+\\ce{5O2}+\\ce{2CO2}+\\ce{H2}&\\rightarrow&\\ce{C2H2}+\\ce{H2}+\\ce{4CO2}+\\ce{2H2O}+\\ce{2O2}+\\ce{C2H2}\r\n\\end{array}\r\n&\r\n\\begin{array}{l}\r\n\\Delta H=+174.5\\text{ kJ} \\\\\r\n\\Delta H=-1,692.2\\text{ kJ} \\\\\r\n\\Delta H=-167.5\\text{ kJ}\r\n\\end{array}\r\n\\end{array}\\)<\/p>\r\nWhat cancels? 2C2<\/sub>H2<\/sub>, H2<\/sub>, 2O2<\/sub>, and 2CO2<\/sub>. What is left is:\r\n\r\n\\[\\ce{C2H4}+\\ce{3O2}\\rightarrow \\ce{2CO2}+\\ce{2H2O}\\]\r\n\r\nWhich is the reaction we are looking for. The \u0394H<\/i> of this reaction is the sum of the three \u0394H<\/i> values:\r\n\r\n\\[\\Delta H =+174.5-1,692.2-167.5=-1,685.2\\text{ kJ}\\]\r\n

      Test Yourself<\/h1>\r\nGiven the following thermochemical equations:\r\n

      \\(\\begin{array}{rrll}\r\n\\ce{Pb}+\\ce{Cl2}&\\rightarrow&\\ce{PbCl2}&\\Delta H=-223\\text{ kJ} \\\\\r\n\\ce{PbCl2}+\\ce{Cl2}&\\rightarrow&\\ce{PbCl4}&\\Delta H=-87\\text{ kJ}\r\n\\end{array}\\)<\/p>\r\nDetermine \u0394H<\/i> for:\r\n\r\n\\[\\ce{2PbCl2}\\rightarrow \\ce{Pb}+\\ce{PbCl4}\\]\r\n

      Answer<\/h2>\r\n+136 kJ\r\n\r\n<\/div>\r\n<\/div>\r\n
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      Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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