{"id":7379,"date":"2021-06-08T21:56:00","date_gmt":"2021-06-08T21:56:00","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/stoichiometry-calculations-using-enthalpy\/"},"modified":"2021-09-24T20:43:17","modified_gmt":"2021-09-24T20:43:17","slug":"stoichiometry-calculations-using-enthalpy","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/stoichiometry-calculations-using-enthalpy\/","title":{"raw":"Stoichiometry Calculations Using Enthalpy","rendered":"Stoichiometry Calculations Using Enthalpy"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Perform stoichiometry calculations using energy changes from thermochemical equations.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nIn Chapter 5 \"Stoichiometry and the Mole\"<\/a>, we related quantities of one substance to another in a chemical equation by performing calculations that used the balanced chemical equation; the balanced chemical equation provided equivalences that we used to construct conversion factors. For example, observe the following balanced chemical equation:\r\n

    [latex]\\ce{2H2(g)}+\\ce{O2(g)}\\rightarrow \\ce{2H2O(\\ell)}[\/latex]<\/p>\r\nIn this equation, we recognize the following equivalences:\r\n

    [latex]2\\text{ mol }\\ce{H2}\\Leftrightarrow 1\\text{ mol }\\ce{O2}\\Leftrightarrow 2\\text{ mol }\\ce{H2O}[\/latex]<\/p>\r\nWhere \u21d4 is the mathematical symbol for \u201cis equivalent to.\u201d In our thermochemical equation, however, we have another quantity \u2014 energy change:\r\n

    [latex]\\ce{2H2(g)}+\\ce{O2(g)}\\rightarrow \\ce{2H2O(\\ell)} \\hspace{10mm}\\Delta H =-570\\text{ kJ}[\/latex]<\/p>\r\nThis new quantity allows us to add another equivalence to our list:\r\n

    [latex]2\\text{ mol }\\ce{H2}\\Leftrightarrow 1\\text{ mol }\\ce{O2}\\Leftrightarrow 2\\text{ mol }\\ce{H2O} \\Leftrightarrow -570\\text{ kJ}[\/latex]<\/p>\r\nThat is, we can now add an energy amount to the equivalences \u2014 the enthalpy change of a balanced chemical reaction. This equivalence can also be used to construct conversion factors so that we can relate enthalpy change to amounts of substances reacted or produced.\r\n\r\nNote that these equivalences address a concern. When an amount of energy is listed for a balanced chemical reaction, what amount(s) of reactants or products does it refer to? The answer is that relates to the number of moles of the substance as indicated by its coefficient in the balanced chemical reaction. Thus, 2 mol of H2<\/sub> are related to \u2212570 kJ, while 1 mol of O2<\/sub> is related to \u2212570 kJ. This is why the unit on the energy change is kJ, not kJ\/mol.\r\n\r\nFor example, consider the following thermochemical equation:\r\n

    [latex]\\ce{H2(g)}+\\ce{Cl2(g)}\\rightarrow \\ce{2HCl(g)}\\hspace{10mm}\\Delta H = -184.6\\text{ kJ}[\/latex]<\/p>\r\n

    The equivalences for this thermochemical equation are:<\/p>\r\n

    [latex]1\\text{ mol }\\ce{H2}\\Leftrightarrow 1\\text{ mol }\\ce{Cl2} \\Leftrightarrow 2\\text{ mol }\\ce{HCl}\\Leftrightarrow -184.6\\text{ kJ}[\/latex]<\/p>\r\nSuppose we asked how much energy is given off when 8.22 mol of H2<\/sub> react. We would construct a conversion factor between the number of moles of H2<\/sub> and the energy given off, \u2212184.6 kJ:\r\n

    [latex]8.22\\text{ }\\cancel{\\text{mol }\\ce{H2}}\\times \\dfrac{-184.6\\text{ kJ}}{1\\text{ }\\cancel{\\text{mol }\\ce{H2}}}=-1,520\\text{ kJ}[\/latex]<\/p>\r\nThe negative sign means that this much energy is given off.\r\n

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    Example 7.6<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nDetermine how much energy is given off when 222.4 g of N2<\/sub> reacts in the following thermochemical equation:\r\n

    [latex]\\ce{N2(g)}+\\ce{3H2(g)}\\rightarrow \\ce{2NH3(g)}\\hspace{10 mm}\\Delta H=-91.8\\text{ kJ}[\/latex]<\/p>\r\n\r\n

    Solution<\/h2>\r\nThe balanced thermochemical equation relates the energy change to moles, not grams, so we first convert the amount of N2<\/sub> to moles and then use the thermochemical equation to determine the energy change:\r\n

    [latex]222.4\\text{ }\\cancel{\\text{g }\\ce{N2}}\\times\\dfrac{1\\text{ }\\cancel{\\text{mol }\\ce{N2}}}{28.00\\text{ }\\cancel{\\text{g }\\ce{N2}}}\\times\\dfrac{-91.8\\text{ kJ}}{1\\text{ }\\cancel{\\text{mol }\\ce{N2}}}=-729\\text{ kJ}[\/latex]<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nDetermine how much heat is given off when 1.00 g of H2<\/sub> reacts in the following thermochemical equation:\r\n

    [latex]\\ce{N2(g)}+\\ce{3H2(g)}\\rightarrow \\ce{2NH3(g)}\\hspace{10 mm}\\Delta H=-91.8\\text{ kJ}[\/latex]<\/p>\r\n\r\n

    Answer<\/h2>\r\n\u221215.1 kJ\r\n\r\n<\/div>\r\n<\/div>\r\nLike any stoichiometric quantity, we can start with energy and determine an amount, rather than the other way around.\r\n
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    Example 7.7<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nDetermine what mass of NO can be made if 558 kJ of energy are supplied, given the following thermochemical equation:\r\n

    [latex]\\ce{N2(g)}+\\ce{O2(g)}\\rightarrow \\ce{2NO(g)}\\hspace{10 mm}\\Delta H=180.6\\text{ kJ}[\/latex]<\/p>\r\n\r\n

    Solution<\/h2>\r\nThis time, we start with an amount of energy:\r\n

    [latex]558\\text{ \\cancel{kJ}}\\times \\dfrac{2 \\text{ \\cancel{mol NO}}}{180.6\\text{ \\cancel{kJ}}}\\times \\dfrac{30.0\\text{ g NO}}{1\\text{ \\cancel{mol NO}}}=185\\text{ g NO}[\/latex]<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nGiven the following equation, how many grams of N2<\/sub> will react if 100.0 kJ of energy are supplied?\r\n

    [latex]\\ce{N2(g)}+\\ce{O2(g)}\\rightarrow \\ce{2NO(g)}\\hspace{10mm}\\Delta H=180.6\\text{ kJ}[\/latex]<\/p>\r\n\r\n

    Answer<\/h2>\r\n15.5 g\r\n\r\n<\/div>\r\n<\/div>\r\n
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    Chemistry Is Everywhere: Welding with Chemical Reactions<\/h1>\r\nOne very energetic reaction is called the thermite reaction<\/i>. Its classic reactants are aluminum metal and iron(III) oxide; the reaction produces iron metal and aluminum oxide:\r\n

    [latex]\\ce{2Al(s)}+\\ce{Fe2O3(s)}\\rightarrow\\ce{Al2O3(s)}+\\ce{2Fe(s)}\\hspace{10 mm}\\Delta H=-850.2\\text{ kJ}[\/latex]<\/p>\r\nWhen properly done, the reaction gives off so much energy that the iron product comes off as a liquid<\/i>. (Iron normally melts at 1,536\u00b0C.) If carefully directed, the liquid iron can fill spaces between two or more metal parts and, after it quickly cools, can weld the metal parts together.\r\n\r\nThermite reactions are used for this purpose even today. For civilian purposes, they are used to re-weld broken locomotive axles that cannot be easily removed for repair. They are used to weld railroad tracks together. Thermite reactions can also be used to separate thin pieces of metal if, for whatever reason, a torch doesn\u2019t work.\r\n\r\n[caption id=\"attachment_338\" align=\"aligncenter\" width=\"400\"]\"A Figure 7.2 \"Thermite Reaction.\" A small clay pot contains a thermite mixture. It is reacting at high temperature in the photo and will eventually produce molten metal to join the railroad tracks below it.[\/caption]\r\n\r\nThermite reactions are also used for military purposes. Thermite mixtures are frequently used with additional components as incendiary devices \u2014 devices that start fires. Thermite reactions are also useful in disabling enemy weapons: a piece of artillery doesn\u2019t work so well when it has a hole melted into its barrel because of a thermite reaction!\r\n\r\n<\/div>\r\n

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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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