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zumdahl_chemprin_6e_csm_ch06

zumdahl_chemprin_6e_csm_ch06 - CHAPTER 6 CHEMICAL...

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149 CHAPTER 6 CHEMICAL EQUILIBRIUM Characteristics of Chemical Equilibrium 10. a. The rates of the forward and reverse reactions are equal. b. There is no net change in the composition (as long as temperature is constant). 11. 2 NOCl(g) 2 NO(g) + Cl 2 (g) K = 1.6 × 10 5 mol/L The expression for K is the product concentrations divided by the reactant concentrations. When K has a value much less than one, the product concentrations are relatively small, and the reactant concentrations are relatively large. 2 NO(g) N 2 (g) + O 2 (g) K = 1 × 10 31 When K has a value much greater than one, the product concentrations are relatively large, and the reactant concentrations are relatively small. In both cases, however, the rate of the forward reaction equals the rate of the reverse reaction at equilibrium (this is a definition of equilibrium). 12. No, equilibrium is a dynamic process. Both reactions: H 2 O + CO H 2 + CO 2 and H 2 + CO 2 H 2 O + CO are occurring at equal rates. Thus 14 C atoms will be distributed between CO and CO 2 . 13. No, it doesn't matter which direction the equilibrium position is reached. Both experiments will give the same equilibrium position because both experiments started with stoichiometric amounts of reactants or products. 14. H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) K = ] CO ][ O H [ ] CO ][ H [ 2 2 2 = 2.0 K is a unitless number because there is an equal number of moles of product gases as moles of reactant gases in the balanced equation. Therefore, we can use units of molecules per liter instead of moles per liter to determine K.
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150 CHAPTER 6 CHEMICAL EQUILIBRIUM We need to start somewhere, so let’s assume 3 molecules of CO react. If 3 molecules of CO react, then 3 molecules of H 2 O must react, and 3 molecules each of H 2 and CO 2 are formed. We would have 6 ! 3 = 3 molecules of CO, 8 ! 3 = 5 molecules of H 2 O, 0 + 3 = 3 molecules of H 2 , and 0 + 3 = 3 molecules of CO 2 present. This will be an equilibrium mixture if K = 2.0: K = 5 3 L CO molecules 3 L O H molecules 5 L CO molecules 3 L H molecules 3 2 2 2 = Because this mixture does not give a value of K = 2.0, this is not an equilibrium mixture. Let’s try 4 molecules of CO reacting to reach equilibrium. Molecules CO remaining = 6 ! 4 = 2 molecules of CO Molecules H 2 O remaining = 8 ! 4 = 4 molecules of H 2 O Molecules H 2 present = 0 + 4 = 4 molecules of H 2 Molecules CO 2 present = 0 + 4 = 4 molecules of CO 2 K = 0 . 2 L CO molecules 2 L O H molecules 4 L CO molecules 4 L H molecules 4 2 2 2 = Because K = 2.0 for this reaction mixture, we are at equilibrium. 15. When equilibrium is reached, there is no net change in the amount of reactants and products present because the rates of the forward and reverse reactions are equal to each other. The first diagram has 4 A 2 B molecules, 2 A 2 molecules, and 1 B 2 molecule present. The second diagram has 2 A 2 B molecules, 4 A 2 molecules, and 2 B 2 molecules. Therefore, the first
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