{"id":2433,"date":"2019-04-22T18:09:14","date_gmt":"2019-04-22T18:09:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-introductorychemistry\/chapter\/hesss-law-2\/"},"modified":"2019-04-29T12:33:32","modified_gmt":"2019-04-29T12:33:32","slug":"hesss-law-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductorychemistry\/chapter\/hesss-law-2\/","title":{"raw":"Hess\u2019s Law","rendered":"Hess\u2019s Law"},"content":{"raw":"
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Learning Objective<\/h3>\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\n

    Now 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:<\/p>\r\n2 C(s) +\u00a0O2<\/sub>(g) \u2192\u00a02 CO(g) \u0394H<\/em>\u00a0=\u00a0?<\/span><\/span>\r\n

    In reality, this is extremely difficult to do; given the opportunity, carbon will react to make another compound, carbon dioxide:<\/p>\r\n2 C(s) +\u00a0O2<\/sub>(g) \u2192\u00a02 CO2<\/sub>(g) \u0394H<\/em>\u00a0=\u00a0\u2212393.5 kJ<\/span><\/span>\r\n

    Is 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 canceled out (much like a spectator ion in ionic equations). For example, consider these two reactions:<\/p>\r\n2 C(s) +\u00a02 O2<\/sub>(g) \u2192\u00a02 CO2<\/sub>(g)<\/span><\/span>\r\n2 CO2<\/sub>(g) \u2192\u00a02 CO(g) +\u00a0O2<\/sub>(g)<\/span><\/span>\r\n

    If we added these two equations by combining all the reactants together and all the products together, we would get<\/p>\r\n2 C(s) +\u00a02 O2<\/sub>(g) +\u00a02 CO2<\/sub>(g) \u2192\u00a02 CO2<\/sub>(g) +\u00a02 CO(g) +\u00a0O2<\/sub>(g)<\/span><\/span>\r\n

    We note that 2 CO2<\/sub>(g) appears on both sides of the arrow, so they cancel:<\/p>\r\n\"Screen<\/a>\r\n

    We 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:<\/p>\r\n\"Screen<\/a>\r\n

    What do we have left?<\/p>\r\n2 C(s) +\u00a0O2<\/sub>(g) \u2192\u00a02 CO(g)<\/span><\/span>\r\n

    This 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.<\/p>\r\n

    What about the enthalpy changes? Hess\u2019s law\u00a0<\/a><\/span>states that when chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. Two corollaries immediately present themselves:<\/p>\r\n\r\n

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    1. If a chemical reaction is reversed, the sign on \u0394H<\/em> is changed.<\/li>\r\n \t
    2. If a multiple of a chemical reaction is taken, the same multiple of the \u0394H<\/em> is taken as well.<\/li>\r\n<\/ol>\r\n

      What are the equations being combined? The first chemical equation is the combustion of C, which produces CO2<\/sub>:<\/p>\r\n2 C(s) +\u00a02 O2<\/sub>(g) \u2192\u00a02 CO2<\/sub>(g)<\/span><\/span>\r\n

      This reaction is two times the reaction to make CO2<\/sub> from C(s) and O2<\/sub>(g), whose enthalpy change is known:<\/p>\r\nC(s) +\u00a0O2<\/sub>(g) \u2192\u00a0CO2<\/sub>(g) \u0394H<\/em>\u00a0=\u00a0\u2212393.5 kJ<\/span><\/span>\r\n

      According to the first corollary, the first reaction has an energy change of two times \u2212393.5 kJ, or \u2212787.0 kJ:<\/p>\r\n2 C(s) +\u00a02 O2<\/sub>(g) \u2192\u00a02 CO2<\/sub>(g) \u0394H<\/em>\u00a0=\u00a0\u2212787.0 kJ<\/span><\/span>\r\n

      The second reaction in the combination is related to the combustion of CO(g):<\/p>\r\n2 CO(g) +\u00a0O2<\/sub>(g) \u2192\u00a02 CO2<\/sub>(g) \u0394H<\/em>\u00a0=\u00a0\u2212566.0 kJ<\/span><\/span>\r\n

      The second reaction in our combination is the reverse<\/em> of the combustion of CO. When we reverse the reaction, we change the sign on the \u0394H<\/em>:<\/p>\r\n2 CO2<\/sub>(g) \u2192\u00a02 CO(g) +\u00a0O2<\/sub>(g) \u0394H<\/em>\u00a0=\u00a0+566.0 kJ<\/span><\/span>\r\n

      Now that we have identified the enthalpy changes of the two component chemical equations, we can combine the \u0394H<\/em> values and add them:<\/p>\r\n

      \"Enthalpy<\/a><\/p>\r\n

      Hess\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.<\/p>\r\n\r\n

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      Example 10<\/h3>\r\n

      Determine the enthalpy change of<\/p>\r\nC2<\/sub>H4<\/sub> +\u00a03 O2<\/sub> \u2192\u00a02 CO2<\/sub> +\u00a02 H2<\/sub>O \u0394H<\/em>\u00a0=\u00a0?<\/span><\/span>\r\n

      from these reactions:<\/p>\r\nC2<\/sub>H2<\/sub> +\u00a0H2<\/sub> \u2192\u00a0C2<\/sub>H4<\/sub> \u0394H<\/em>\u00a0=\u00a0\u2212174.5 kJ<\/span><\/span>\r\n2 C2<\/sub>H2<\/sub> +\u00a05 O2<\/sub> \u2192\u00a04 CO2<\/sub> +\u00a02 H2<\/sub>O \u0394H<\/em>\u00a0=\u00a0\u22121,692.2 kJ<\/span><\/span>\r\n2 CO2<\/sub> +\u00a0H2<\/sub> \u2192\u00a02 O2<\/sub> +\u00a0C2<\/sub>H2<\/sub> \u0394H<\/em>\u00a0=\u00a0\u2212167.5 kJ<\/span><\/span>\r\n

      Solution<\/p>\r\n

      We 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<\/em>:<\/p>\r\nC2<\/sub>H4<\/sub> \u2192\u00a0C2<\/sub>H2<\/sub> +\u00a0H2<\/sub> \u0394H<\/em>\u00a0=\u00a0+174.5 kJ<\/span><\/span>\r\n

      We 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):<\/p>\r\n2 C2<\/sub>H2<\/sub> +\u00a05 O2<\/sub> \u2192\u00a04 CO2<\/sub> +\u00a02 H2<\/sub>O \u0394H<\/em>\u00a0=\u00a0\u22121,692.2 kJ<\/span><\/span>\r\n

      We 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:<\/p>\r\n2 CO2<\/sub> +\u00a0H2<\/sub> \u2192\u00a02 O2<\/sub> +\u00a0C2<\/sub>H2<\/sub> \u0394H<\/em>\u00a0=\u00a0\u2212167.5 kJ<\/span><\/span>\r\n

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

      \"Enthalpy<\/a><\/p>\r\n\r\n

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      What cancels? 2 C2<\/sub>H2<\/sub>, H2<\/sub>, 2 O2<\/sub>, and 2 CO2<\/sub>. What is left is<\/p>\r\nC2<\/sub>H4<\/sub> +\u00a03 O2<\/sub> \u2192\u00a02 CO2<\/sub> +\u00a02H2<\/sub>O<\/span><\/span>\r\n

      which is the reaction we are looking for. The \u0394H<\/em> of this reaction is the sum of the three \u0394H<\/em> values:<\/p>\r\n\u0394H<\/em>\u00a0=\u00a0+174.5 \u2212 1,692.2 \u2212 167.5 = \u22121,685.2 kJ<\/span><\/span>\r\n

      Test Yourself<\/em><\/p>\r\n

      Given the thermochemical equations<\/p>\r\nPb +\u00a0Cl2<\/sub> \u2192\u00a0PbCl2<\/sub> \u0394H<\/em>\u00a0=\u00a0\u2212223 kJ<\/span><\/span>\r\nPbCl2<\/sub> +\u00a0Cl2<\/sub> \u2192\u00a0PbCl4<\/sub> \u0394H<\/em>\u00a0=\u00a0\u221287 kJ<\/span><\/span>\r\n

      determine \u0394H<\/em> for<\/p>\r\n2 bCl2<\/sub> \u2192\u00a0Pb +\u00a0PbCl4<\/sub><\/span><\/span>\r\n

      Answer<\/em><\/p>\r\n

      +136 kJ<\/p>\r\n\r\n<\/div>\r\n

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      Key Takeaway<\/h3>\r\n