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Thermochemistry: Energy Changes in Chemical Reactions

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Thermochemistry: Energy Changes in Chemical Reactions CHAPTER 9 | Thermochemistry: Energy Changes in Chemical Reactions 9.1. Collect and Organize From the depiction of a diesel engine piston (Figure P9.1), we are to describe how the internal energy of the trapped gases changes when the piston moves up. Analyze Here, the system we are considering is the gases. As the piston moves up, the gases in the cylinder are compressed. Solve Upon compression the molecules are squeezed together and the change in volume on the system is negative. Therefore, work is done on the system and w = –PΔV, where ΔV is negative, so w is positive and the internal energy change E = q + w is positive. Here, the internal energy of the gases in the cylinder increases. Think About It If work is done by the system, in this situation where the gases expand, the internal energy of the gases in the cylinder decreases. 9.2. Collect and Organize From the depiction of a diesel engine piston in which the fuel has been ignited and the cylinder is pushed down (Figure P9.2), we are to describe how the internal energy of the trapped gases changes and assign signs to the quantities E, q, and w. Analyze Here, the system we are considering is the gases in the cylinder. As the gases in the cylinder expand, the piston moves down. Upon ignition, the system releases heat to the surroundings. Solve Upon expansion, the change in volume of the system is positive. Therefore, work is done by the system on the surroundings. In this expansion, ΔV is positive, so w is negative because w = –PΔV. Because heat is transferred from the ignited gases to the surroundings, q is negative. The internal energy change, E = q + w, is therefore negative and the internal energy of the gases in the cylinder decreases. Think About It When the piston moves up to compress the gases in the cylinder, work is done on the system by the surroundings and the surroundings heat up the sample, so w and q are both positive for that situation. 9.3. Collect and Organize All the molecules shown in Figure P9.3 are hydrocarbons with either five or six carbon atoms. On the basis of differences in their structures, we are to predict which has the highest and which has the lowest fuel value. Analyze For hydrocarbons, as the hydrogen-to-carbon ratio decreases the fuel value also decreases. Therefore, we will distinguish between the hydrocarbon in this group that has the most and the least fuel value on the basis of the hydrogen-to-carbon ratio. Solve The hydrogen-to-carbon ratios of those hydrocarbons are as follows: (a) C6H14 H:C = 2.33:1 (b) C6H12 H:C = 2:1 (c) C5H12 H:C = 2.4:1 (d) C6H14 H:C = 2.33:1 From those ratios, (c) C5H12 has the highest fuel value and (b) C6H12 has the lowest fuel value. 1 2 | Chapter 9 Think About It Although structurally different, (a) and (d) have the same H:C ratio and therefore the same fuel value. 9.4. Collect and Organize The diagram in Figure P9.4 shows the reaction between N2 and H2 in a closed vessel that is maintained by the piston at constant pressure while being held at constant temperature. The diagram shows that as the reaction proceeds, the volume of the reaction decreases. We are to determine whether work is done by or on the system, calculate the enthalpy of the reaction by using standard enthalpy data, and predict the direction of heat flow in the reaction. Analyze The balanced reaction to prepare 1 mol of ammonia from nitrogen and hydrogen is 1 3 /2 N2(g) + /2 H2(g) → NH3(g) The fact that the volume of the reaction vessel is smaller after the reaction will help us determine the sign of work for the reaction (the system). Once we calculate the enthalpy of the reaction in part b, we will be able to determine the direction of heat flow for the reaction. Solve (a) Because the volume of the vessel is less after the reaction (ΔV is negative), the reaction was compressed, and therefore work was done on the system by the surroundings. (b) The enthalpy of the reaction under standard conditions is ° ° 1 ° 3 ° ΔH rxn = ⎣⎡1 mol (ΔH f )NH3(g)⎦⎤ – ⎣⎡ 2 mol(ΔH f )N2 (g) + 2 mol(ΔH f )H2 (g)⎦⎤ 1 3 = [1 mole × –46.1 kJ/mol]−[ 2 mol × 0 kJ/mol + 2 mol × 0 kJ/mol)] = −46.1 kJ (c) The negative value of the enthalpy for that reaction means that the reaction is exothermic; heat flows out from the reaction to the surroundings. Think About It The reverse reaction would be endothermic and would have an increase in volume of the reaction vessel. 9.5. Collect and Organize For the reaction depicted in Figure P9.4, we are to assign signs to the values ΔE, q, and w and then determine the percent yield of the reaction if each molecule represents 1 mol. Analyze From our answer to Problem 9.4, we know that the reaction is exothermic and is compressed in going from reactants to products. From the diagram we have 3 mol of N2 and 9 mol of H2 as reactants that form 4 mol of NH3, with 3 mol of H2 and 1 mol of N2 remaining at the end of the reaction. Solve Because the reaction volume is less at the end of the reaction, the sign of w is positive, showing that work was done on the system. Because the reaction is exothermic, the sign of q is negative, showing flow of heat from the system to the surroundings. The change in E for the reaction is negative if the heat released by the reaction is greater than the work done on the system by the surroundings by decrease of volume. For the percent yield we consider the system in comparison with the balanced equation N2(g) + 3 H2(g) → 2 NH3(g) The 3 mol of N2 present in the reaction mixture would be expected to use 9 mol of hydrogen (which we have exactly) and yield 6 mol of ammonia. Only 4 mol of ammonia is produced, so the percent yield of the reaction is experimental yield 4 mol NH ×100 = 3 ×100 = 67% theoretical yield 6 mol NH3 Thermochemistry | 3 Think About It A 100% yield for this reaction would have used up all the nitrogen and the hydrogen initially present in the reaction mixture. 9.6. Collect and Organize ° The enthalpy diagram in Figure P9.6 shows the relative enthalpies of formation (∆Hf, kilojoules per mole) for the elements C, H2, and O2 along with that of C2H2, CO, H2O, and CO2. From the diagram we are to explain why all the elements have the same standard enthalpy of formation, explain why C2H2 is an “endothermic compound,” and predict which reaction has the more negative enthalpy change: acetylene reacting with oxygen to form carbon monoxide, or acetylene reacting with oxygen to form carbon dioxide. Analyze ° ° All the elements have ∆Hf = 0. The ∆Hf is positive for C2H2 and negative for CO, H2O, and CO2. For reactions, the enthalpy is calculated by subtracting the enthalpy of the reactants from the enthalpy of the products. Solve ° (a) The elements are all on the same horizontal line (∆Hf = 0) because the enthalpy of formation of an element in its standard state is defined as zero. (b) C2H2 is an endothermic compound because the enthalpy of its formation is positive. Therefore, energy would have to be added to the elements for C2H2 to form. (c) The only differences in the two reactions of acetylene with oxygen, 2 C22 H (gg )+ 3 O 2 ( )→ 4 CO( g )+ 2 H 2 O( g ) 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O(g) are the amounts of oxygen required for the reaction and the product formed (CO or CO2). The enthalpy values for the reactions are calculated by subtracting the sum of the enthalpies of formation of the reactants (multiplied by the number of moles in the balanced equation for each product) from the sum of enthalpies of formation of the products (again multiplied by their molar amounts from the balanced equation). For the first reaction: ΔH = ⎡4 mol (ΔH )CO(g) + 2 mol (ΔH )H O(g)⎤ – ⎡2 mol (ΔH )C H (g) + 3 mol (ΔH )O (g)⎤ rxn ⎣ f f 2 ⎦ ⎣ f 2 2 f 2 ⎦ For the second reaction: ΔH = ⎡4 mol (ΔH )CO (g) + 2 mol (ΔH )H O(g)⎤ – ⎡2 mol (ΔH )C H (g) + 5 mol (ΔH )O (g)⎤ rxn ⎣ f 2 f 2 ⎦ ⎣ f 2 2 f 2 ⎦ Because the enthalpy of formation of oxygen is zero, the amount of oxygen used in the second reaction does not contribute to any enthalpy of reaction difference between those two reactions. However, the enthalpy of formation of CO2 is more negative than that of CO. That makes the enthalpy of the second reaction, in which the combustion of acetylene forms carbon dioxide, more negative. Think About It If you consider the magnitudes of the enthalpies of the two combustion reactions from Figure P9.6 you can estimate the enthalpy of the combustion reaction to form CO as approximately –1400 kJ and the reaction to form CO2 as approximately –2480 kJ. 9.7. Collect and Organize The diagram in Figure P9.7 shows that the volume of the cylinder on the product side is greater than that on the reactant side.
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