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Unformatted text preview: Chapter 21: The Chemistry of the Main Group Elements 932 End-of-Chapter Solutions for Chapter 21 Summary Problem Answer: (a) V = 8 × 10 4 L (b) (i) Δ H° = –99.89 kJ (ii) 1. Increase 2. Increase 3. No change 4. Increase 5. Decrease (iii) K c = 50.0 (c) Δ G° = –70.87, K p = 0.89, K p = 0.15 Strategy and Explanation: (a) To determine the amount of sulfur dioxide gas produced, the amount of sulfur in the barrels of crude oil must first be determined. 26 barrels oil " 42 gal oil 1 barrel oil " 4 quarts 1 gal " 1 L 1.056710 quarts " 1000 mL 1 L " 0.83 g oil 1 mL oil " 3 g S 100 g oil " 1 mol S 32.065 g S " 1 mol SO 2 1 mol S = 3209.9 mol SO 2 # 3 " 10 3 mol SO 2 (round to 1 sig fig) The moles of sulfur dioxide produced are then inserted into the ideal gas law, PV=nRT; recall that the temperature must be converted to Kelvin and to use the appropriate value of the gas constant, R. V = 3209.9 mol SO 2 ( ) 0.08206 L " atm mol " K # $ % & ’ ( 25 + 273.15 ( ) K 1 atm = 78533.4 L ) 8 * 10 4 L (round to 1 sig fig) (b) (i) The enthalpy change is the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants. Enthalpies of formation are found in Appendix J. Δ H° = (moles of product) " # H f o (product) $ % & ’ ( ) * + (moles of reactant) " # H f o (reactant) $ % & ’ ( ) * SO 2 (g) + 1 2 O 2 (g) SO 3 (g) Δ H° = Δ H f o (SO 3 (g)) – [ Δ H f o (SO 2 (g)) + 1 2 Δ H f o (O 2 (g))] Δ H° = (1 mol) × (–395.72 kJ/mol) – [(1 mol) × (–296.820 kJ/mol) + ( 1 2 mol) × (0 kJ/mol)] = –99.89 kJ (ii) Use the methods described in Section 14.6 to answer this question. Changes in concentrations or pressures of reactants and products or changing the available energy in the system may take the system away from equilibrium. The response to that change will be to partially counteract the change. Changing the amounts of solids or liquids will not affect the equilibrium position. Only temperature can affect the value of the equilibrium constant. Because this reaction is exothermic, energy is evolved. SO 2 (g) + 1 2 O 2 (g) SO 3 (g) Δ H° = –99.89 kJ 1. Increase of pressure in the system can be counteracted by reducing the number of moles of gas. Because the reactants have 1.5 mol of gas and the products have only 1 mol of gas, we will predict that more SO 3 will form, to establish a new equilibrium with greater [SO 3 ]. 2. Decreasing the temperature can be counteracted by evolving energy. Because the reaction is exothermic, we will predict that more SO 3 will form, to establish a new equilibrium with greater [SO 3 ]. 3. Adding a catalyst speeds the reaction rate, but will not change the position of equilibrium, so the [SO 3 ] will not change....
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- Fall '08