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5.4 Figure 1 Some reactions are too slow to be studied experimentally. Hess’s Law of Additivity of Reaction Enthalpies Calorimetry is an accurate technique for determining enthalpy changes, but how do chemists deal with chemical systems that cannot be analyzed using this technique? For example, the rusting of iron (Figure 1) is extremely slow and, therefore, the resulting temperature change would be too small to be measured using a conventional calorimeter. Similarly, the reaction to produce carbon monoxide from its elements is impossible to measure with a calorimeter because the combustion of carbon produces both carbon dioxide and carbon monoxide simultaneously. Chemists have devised a number of methods to deal with this problem. These methods are based on the principle that net (or overall) changes in some properties of a system are independent of the way the system changes from the initial state to the final state. An analogy for this concept is shown in Figure 2. The net vertical distance that the bricks rise is the same whether they go up in one stage or in two stages. The same principle applies to enthalpy changes: If a set of reactions occurs in different steps but the initial reactants and final products are the same, the overall enthalpy change is the same (Figure 3). Figure 2 In this illustration, bricks on a construction site are being moved from the ground up to the second floor, but there are two different paths that the bricks could follow. In one path, they could move from the ground up to the third floor and then be carried down to the second floor. In the other path, they would be carried up to the second floor in a single step. In both cases, the overall change in position — one floor up — is the same. 322 Chapter 5 NEL Section 5.4 Potential Energy Diagram Showing Additive Enthalpy Changes NO –56 kJ Ep +90 kJ NO2 +34 kJ N2 and O2 Reaction Progress Figure 3 In this potential energy diagram, nitrogen gas and oxygen gas combine to form nitrogen dioxide, but there are two different paths to reach the products. In one path, nitrogen (N2) and oxygen (O2) gases react to form nitrogen monoxide (NO), a reaction for which H +90 kJ. Then, nitrogen monoxide and more oxygen react to form nitrogen dioxide (NO2) gas, a reaction for which H 56 kJ. In the other path, nitrogen (N2) and oxygen (O2) gases react directly to form nitrogen dioxide (NO2) gas. In both cases, the overall enthalpy change, H 34 kJ, is the same. Predicting H Using Hess’s Law Based on experimental measurements of enthalpy changes, the Swiss chemist G. H. Hess suggested that there is a mathematical relationship among a series of reactions leading from a set of reactants to a set of products. This generalization has been tested in many experiments and is now accepted as the law of additivity of reaction enthalpies, also known as Hess’s Law The value of the H for any reaction that can be written in steps equals the sum of the values of H for each of the individual steps. Another way to state Hess’s law is: If two or more equations with known enthalpy changes can be added together to form a new “target” equation, then their enthalpy changes may be similarly added together to yield the enthalpy change of the target equation. Hess’s law can also be written as an equation using the uppercase Greek letter (pronounced “sigma”) to mean “the sum of.” Htarget H1 H2 H3 … or Htarget Hknown Hess’s discovery allowed chemists to determine the enthalpy change of a reaction without direct calorimetry, using two familiar rules for chemical equations and enthalpy changes: • If a chemical equation is reversed, then the sign of ∆H changes. • If the coefficients of a chemical equation are altered by multiplying or dividing by a constant factor, then the ∆H is altered in the same way. NEL Thermochemistry 323 SAMPLE problem Using Hess’s Law to Find H 1. What is the enthalpy change for the formation of two moles of nitrogen monoxide from its elements? N2(g) O2(g) → 2 NO(g) H° ? This reaction, which may be called the target equation to distinguish it clearly from the other equations, is difficult to study calorimetrically since the combustion of nitrogen produces nitrogen dioxide as well as nitrogen monoxide. However, we can measure the enthalpy of complete combustion in excess oxygen (to nitrogen dioxide) for both nitrogen and nitrogen monoxide by calorimetry. Consider the following two known reference equations: LEARNING TIP Generally, a subscript on a H value indicates a molar enthalpy value, expressed in kJ/mol. The “known” equations in Hess’s law problems are exceptions. The subscript is used as a convenience in distinguishing equations from each other and the units of the Hn values are kilojoules. 1 2 (1) N2(g) O2(g) → NO2(g) (2) NO(g) 1 2 O2(g) → NO2(g) H°1 +34 kJ H°2 –56 kJ If we work with these two equations, which may be called known equations, and then add them together, we obtain the chemical equation for the formation of nitrogen monoxide. The first term in the target equation for the formation of nitrogen monoxide is one mole of nitrogen gas. We therefore need to double equation (1) so that N2(g) will appear on the reactant side when we add the equations. However, from equation (2) we want 2 mol of NO(g) to appear as a product, so we must reverse equation (2) and double each of its terms (including the enthalpy change). Effectively, we have multiplied known equation (1) by +2, and multiplied known equation (2) by –2. 2 (1): N2(g) 2 O2(g) → 2 NO2(g) H° 2(+34) kJ –2 (2): 2 NO2(g) → 2 NO(g) O2(g) H° –2(–56) kJ Note that the sign of the enthalpy change in equation (2) will change, since the equation has been reversed. Now add the reactants, products, and enthalpy changes to obtain a net reaction equation. Note that 2 NO2(g) can be cancelled because it appears on both sides of the net equation. Similarly, O2(g) can be cancelled from each side of the equation, yielding the target equation: N2(g) 2 O2(g) + 2 NO 2(g) → 2 NO2(g) 2 NO(g) + O2(g) or N2(g) O2(g) → 2 NO(g) Now we can apply Hess’s law: If the known equations can be added together to form the target equation, then their enthalpy changes can be added together. H ° (2 34) kJ + (–2 (–56)) kJ 68 kJ 112 kJ H ° +180 kJ The enthalpy change for the formation of two moles of nitrogen monoxide from its elements is 180 kJ. When manipulating the known equations, you should check the target equation and plan ahead to ensure that the substances end up on the correct sides and in the correct amounts. 324 Chapter 5 NEL Section 5.4 2. What is the enthalpy change for the formation of one mole of butane ( C4H10 ) gas from its elements? The reaction is: 4 C(s) + 5 H2(g) → C4H10(g) H° ? The following known equations, determined by calorimetry, are provided: (1) C4H10(g) + 13 2 O2(g) → 4 CO2(g) + 5 H2O(g) H°1 2657.4 kJ (2) C(s) + O2(g) → CO2(g) H°2 393.5 kJ (3) 2 H2(g) + O2(g) → 2 H2O(g) H°3 483.6 kJ Reversing known equation (1), which will require multiplying its H by 1, will make C4H10(g) a product; multiplying known equation (2) by 4 will provide the required amount of C(s) reactant; and multiplying known equation (3) by 5/2 will provide the required amount of H2(g) reactant. Cancellation when the equations are added will determine whether the required amount of O2(g) remains. → C4H10(g) 1 (1): 4 CO2(g) + 5 H2O(g) 13 2 O2(g) H° 1(657.4) kJ 4 (2): 4 C(s) 4 O2(g) → 4 CO2(g) H° 4(393.5) kJ 5 2 → 5 H2O(g) H° (3): 5 H2(g) 5 2 O2(g) 4 CO2(g) 5 H2O(g) + 4 C(s) 13 2 2(g) 5 H2(g) → C4H10(g) O 13 2 5 (483.6) 2 kJ 2(g) O +4 CO2(g) 5 H2O(g) or 4 C(s) 5 H2(g) → C4H10(g) If the known equations can be added together to form the target equation, then their enthalpy changes can be added together. In this case, H °total (+2657.4) + ( 1574.0) + ( 1209.0) kJ H °total 125.6 kJ The enthalpy change for the formation of one mole of butane is 125.6 kJ. Example Determine the enthalpy change involved in the formation of two moles of liquid propanol. 6 C(s) 8 H2(g) O2(g) → 2 C3H7OH(l) The standard enthalpies of combustion of propanol, carbon, and hydrogen gas at SATP are 2008, 394, and 286 kJ/mol, respectively. Solution The known equations are (1) C3H7OH(l) 9 2 (2) C(s) O2(g) (3) H2(g) + NEL 1 2 O2(g) O2(g) → 3 CO2(g) 4 H2O(l) H°1 –2008 kJ → CO2(g) H°2 –394 kJ → H2O(l) H°3 –286 kJ Thermochemistry 325 2 (1): 6 CO2(g) 8 H2O(l) → 2 C3H7OH(l) 9 O2(g) H° 2(2008 kJ) 6 (2): 6 C(s) 6 O2(g) → 6 CO2(g) H° 6(394 kJ) 8 (3): 8 H2(g) 4 O2(g) → 8 H2O(l) H° 8(286 kJ) → 2 C3H7OH(l) H° –636 kJ 6 C(s) 8 H2(g) O2(g) The enthalpy change involved in the formation of propanol is 636 kJ. Practice Understanding Concepts Answers 1. The enthalpy changes for the formation of aluminum oxide and iron(III) oxide from 1. 851.5 kJ their elements are: 2. 131.3 kJ 3 2 H°1 –1675.7 kJ 3 2 H°2 –824.2 kJ (1) 2 Al(s) O2(g) → Al2O3(s) 3. 524.8 kJ (2) 2 Fe(s) O2(g) → Fe2O3(s) Calculate the enthalpy change for the following target reaction. Fe2O3(s) 2 Al(s) Figure 4 Electric power generating stations that use coal as a fuel are only 30% to 40% efficient. Coal gasification and combustion of the coal gas provide one alternative to burning coal. Efficiency is improved by using both a combustion turbine and a steam turbine to produce electricity. → Al2O3(s) 2 Fe(s) H2O(g) C(s) → CO(g) H2(g) H° ? Calculate the standard enthalpy change for this reaction from the following chemical equations and enthalpy changes. (1) 2 C(s) O2(g) → 2 CO(g) H°1 221.0 kJ (2) 2 H2(g) O2(g) → 2 H2O(g) H°2 483.6 kJ coal gas heat recovery steam generator gasifier coal gas combustion turbine coal slurry pulverized coal 326 Chapter 5 ? 2. Coal gasification converts coal into a combustible mixture of carbon monoxide and hydrogen, called coal gas (Figure 4), in a gasifier. oxygen mixed with water H° mineral slag ash and sulfur removal to exhaust stack steam generator generator power out power out boiler water NEL Section 5.4 3. The coal gas described in the previous question can be used as a fuel, for example, in a combustion turbine. CO(g) H2(g) O2(g) → CO2(g) H2O(g) H° ? Predict the change in enthalpy for this combustion reaction from the following information. (1) 2 C(s) O2(g) → 2 CO(g) H°1 –221.0 kJ (2) C(s) O2(g) → CO2(g) H°2 –393.5 kJ (3) 2 H2(g) O2(g) → 2 H2O(g) H°3 –483.6 kJ INVESTIGATION 5.4.1 Hess’s Law (p. 351) Use calorimetry to determine your own “known” equations and use them to calculate the molar enthalpy of combustion of magnesium. Multistep Energy Calculations In practice, energy calculations rarely involve only a single-step calculation of heat or enthalpy change. Several energy calculations might be required, involving a combination of energy change definitions such as • heat flows, q mc∆T • enthalpy changes, ∆H n∆Hr • Hess’s law, ∆Htarget ∆Hknown In these multi-step problems, ∆H is often found by using standard molar enthalpies or Hess’s law and then equated to the transfer of heat, q. As shown in the following sample problem, if we know the enthalpy change of a reaction and the quantity of reactant or product, we can predict how much energy will be absorbed or released. Solving Multistep Enthalpy Problems SAMPLE problem In the Solvay process for the production of sodium carbonate (or washing soda), one step is the endothermic decomposition of sodium hydrogen carbonate: 2 NaHCO3(s) 129.2 kJ → Na2CO3(s) CO2(g) H2O(g) What quantity of chemical energy, H, is required to decompose 100.0 kg of sodium hydrogen carbonate? First, calculate the energy absorbed per mole of NaHCO3, that is, the molar enthalpy of reaction with respect to sodium hydrogen carbonate. H nHr H Hr n 129.2 kJ 2 mol Hr 64.6 kJ/mol This means that 64.6 kJ of energy is required for every mole of NaHCO3 decomposed. Converting 100.0 kg to an amount in moles and multiplying by the molar enthalpy will give us the required enthalpy change, H, for the equation. NEL Thermochemistry 327 1 mol nNaHCO3 100.0 kg 84.01 g 1.190 kmol H nHr 64 .6 kJ 1.190 kmol 1 mol H 76.9 MJ The decomposition of 100 kg of sodium hydrogen carbonate requires 76.9 MJ of energy. Example How much energy can be obtained from the roasting of 50.0 kg of zinc sulfide ore? ZnS(s) 3 2 O2(g) → ZnO(s) SO2(g) You are given the following thermochemical equations. 1 2 (1) ZnO(s) → Zn(s) (2) S(s) O2(g) → SO2(g) H°2 296.8 kJ (3) ZnS(s) → Zn(s) S(s) H°3 206.0 kJ O2(g) H°1 350.5 kJ Solution MZnS 97.44 g/mol 1 (1): Zn(s) 1 2 O2(g) → ZnO(s) H° 1(350.5 kJ) 1 (2): S(s) O2(g) → SO2(g) H° 1(296.8 kJ) 1 (3): ZnS(s) → Zn(s) S(s) H° 1(206.0 kJ) _____________________________________________________________________ ZnS(s) 3 2 O2(g) → ZnO(s) SO2(g) H° 441.3 kJ According to Hess’s law, 441.3 kJ of energy can be obtained from the roasting of 1 mol of ZnS for which reaction H°r 441.3 kJ/mol. 1 mol nZnS 50.0 kg 97.44 g 513 mol H ° nZnSH°r (441.3 kJ) 513 mol 1 mol H ° 2.26 105 kJ or 226 MJ According to Hess’s law, 226 MJ of energy can be obtained from the roasting of 50.0 kg of zinc sulfide ore. 328 Chapter 5 NEL Section 5.4 Practice Understanding Concepts 4. Ethyne gas may be reduced by reaction with hydrogen gas to form ethane gas in Answers the following reduction reaction: 4. 2.39 MJ C2H2(g) 2 H2(g) → C2H6(g) 5. 2.20 MJ Predict the enthalpy change for the reduction of 200 g of ethyne, using the following information. 5 (1) C2H2(g) O2(g) 2 1 (2) H2 O2(g) 2 7 (3) C2H6(g) + O2(g) 2 → 2 CO2(g) H2O(l) H°1 1299 kJ → H2O(l) H°2 286 kJ → 2 CO2(g) 3 H2O(l) H°3 1560 kJ 5. As an alternative to combustion of coal gas described earlier in this section, coal gas can undergo a process called methanation. 3 H2(g) CO(g) → CH4(g) H2O(g) H ? Determine the enthalpy change involved in the reaction of 300 g of carbon monoxide in this methanation reaction, using the following reference equations and enthalpy changes. (1) 2 H2(g) O2(g) → 2 H2O(g) H°1 483.6 kJ (2) 2 C(s) + O2(g) → 2 CO(g) H°2 221.0 kJ (3) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) (4) C(s) + O2(g) NEL → CO2(g) H°3 802.7 kJ H°4 393.5 kJ Thermochemistry 329 Section 5.4 Questions Understanding Concepts Applying Inquiry Skills 1. (a) Write three balanced thermochemical equations to represent the combustions of one mole each of octane, hydrogen, and carbon, given that their molar enthalpies of combustion are, respectively, –5.47 MJ, –285.8 kJ, and –393.5 kJ/mol. (b) Use Hess’s law to predict the enthalpy change for the formation of octane from its elements. 8 C(s) 9 H2(g) → C8H18(l) using the following information: (2) NO NO2 + Na2O → 2 NaNO2 Question Can Hess’s law be verified experimentally by combining enthalpies of reaction? Three calorimetry experiments are performed, with the choice of chemical systems such that the enthalpy changes of two of the reactions should equal the enthalpy change of the third. The three thermochemical reactions are: HCl(g) + NaNO2(s) → HNO2 (g) + NaCl(s) → 2 HCl Na2O Hess’s law. Complete the Experimental Design, Prediction, and Analysis sections of the investigation report. Experimental Design H ? 2. Predict the enthalpy change for the reaction (1) 2 NaCl H2O 4. A series of calorimetric experiments is perfomed to test H °1 507 kJ H °2 –427 kJ (3) NO NO2 → N2O + O2 H °3 –43 kJ (4) 2 HNO2 → N2O + O2 + H2O H °4 34 kJ 3. Bacteria sour wines and beers by converting ethanol (C2H5OH) into acetic acid (CH3COOH). The reaction is C2H5OH O2 → CH3COOH H2O (1) HBr(aq) KOH(aq) → H2O(l) + KBr(aq) (2) KOH(s) → KOH(aq) (3) KOH(s) HBr(aq) → H2O(l) KBr(aq) H1 ? kJ H2 ? kJ H3 ? kJ (a) Use Hess’s law to show how two of the thermochemical equations can be added together to yield the third thermochemical equation. Evidence See Table 1. The molar enthalpies of combustion of ethanol and acetic acid are, respectively, 1367 kJ/mol and 875 kJ/mol. Write thermochemical equations for the combustions, and use Hess’s law to determine the enthalpy change for the conversion of ethanol to acetic acid. Analysis (b) Use the experimental values to calculate the enthalpy change in each system. Evaluation (c) Calculate a percentage error in the experiment, given that the H for one equation should equal exactly the sum of the other two. Table 1 Observations for Hess’s Law Investigation 330 Chapter 5 Observation Experiment 1 Experiment 2 Experiment 3 quantity of reactant 1 100.0 mL of 1.00 mol/L KOH(aq) 5.61 g KOH(s) 5.61 g KOH(s) quantity of reactant 2 100.0 mL of 1.00 mol/L HBr(aq) N/A 200.0 mL of 0.50 mol/L HBr(aq) initial temperature 20.0°C 20.0°C 20.0°C final temperature 22.5°C 24.1°C 26.7°C NEL