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Chapter 15 Chemical Equilibrium 305 Assume that x is small compared to 1.0. = 0.00316 x = 0.00316 Check assumption: 1.0 X 100% = 0.32%; assumption is valid. [B] = = x = 0.00316 = 0.0032 [A] = 1.0 - 0.00316 = 0.9968 = 1.0 M Check: Plug the values into the equilibrium expression: Kc = (0.0032)(0.0032) (1.0) = 1.024 10⁻⁵ = 1.0 10⁻⁵, which is the equilibrium constant; so the values are correct Le Châtelier's Principle 15.63 Given: + at equilibrium Find: What is the effect of each of the following? (a) COCl₂ is added to the reaction mixture. Adding COCl₂ increases the concentration of COCl₂ and causes the reaction to shift to the left. (b) Cl₂ is added to the reaction mixture. Adding Cl₂ increases the concentration of Cl₂ and causes the reaction to shift to the right. (c) COCl₂ is removed from the reaction mixture. Removing the COCl₂ decreases the concentration of COCl₂ and causes the reaction to shift to the right. 15.65 Given: 2 2 KCl(s) + at equilibrium Find: What is the effect of each of the following? (a) O₂ is removed from the reaction mixture. Removing the O₂ decreases the concentration of O₂ and causes the reaction to shift to the right. (b) is added to the reaction mixture. Adding does not cause any change in the reaction. KCI is a solid, and the concentration remains constant; so the addition of more solid does not change the equilibrium concentration. (c) KCIO₃ is added to the reaction mixture. Adding KCIO₃ does not cause any change in the reaction. is a solid, and the concentration remains constant; the addition of more solid does not change the equilibrium concentration. (d) O₂ is added to the reaction mixture. Adding O₂ increases the concentration of O₂ and causes the reaction to shift to the left. 15.67 (a) Given: at equilibrium Find: the effect of increasing the volume The chemical equation has 2 moles of gas on the right and 1 mole of gas on the left. Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (toward the side with more moles of gas particles). (b) Given: 2H₂S(g) 2H₂(g) + Find: the effect of decreasing the volume The chemical equation has 3 moles of gas on the right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the left (toward the side with fewer moles of gas particles). (c) Given: + 1L Find: the effect of decreasing the volume The chemical equation has 2 moles of gas on the right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure but causes no shift in the reaction because the moles are equal on both sides. 15.69 Given: C(s) + is endothermic. Find: the effect of increasing the temperature Because the reaction is endothermic, we can think of the heat as a reactant. Increasing the temperature is equivalent to adding a reactant, causing the reaction to shift to the right. This will cause an increase in the concentration of products and a decrease in the concentration of reactants; therefore, the value of K will increase. Find: the effect of decreasing the temperature Because the reaction is endothermic, we can think of the heat as a reactant. Decreasing the temperature is equivalent to removing a reactant, causing the reaction to shift to the left. This will cause a decrease in the concentration of products and an increase in the concentration of reactants; therefore, the value of K will decrease. Copyright © 2017 Pearson Education, Inc.