Consider the following system at equilibrium where Delta H^0=108 kJ, and } K_{c}=1.29 * 10^{-2} at 600K: COCl_{2}(g) --->CO(g)+Cl2(g). (1) If the temperature on the equilibrium system is suddenly decreased:,The value of Kc: (a)increases (b)decreases (c)remains the same (2)The value of Qc is (a)less than Kc (b) greater than Kc (c) equal to Kc(3)The reaction must (a) run in the forward direction to reestablish equilibrium (b)run in the reverse direction to reestablish equilibrium (c)remain in the current position, since it is already at equilibrium. (4) The concentration of Cl2 will (a)increase (b)decrease (c)remain the same
Question
Consider the following system at equilibrium where Delta H^0=108 kJ, and } K_{c}=1.29 * 10^{-2} at 600K: COCl_{2}(g) --->CO(g)+Cl2(g). (1) If the temperature on the equilibrium system is suddenly decreased:,The value of Kc: (a)increases (b)decreases (c)remains the same (2)The value of Qc is (a)less than Kc (b) greater than Kc (c) equal to Kc(3)The reaction must (a) run in the forward direction to reestablish equilibrium (b)run in the reverse direction to reestablish equilibrium (c)remain in the current position, since it is already at equilibrium. (4) The concentration of Cl2 will (a)increase (b)decrease (c)remain the same
Solution
(1) The reaction is endothermic (Delta H^0 > 0), so according to Le Chatelier's principle, decreasing the temperature will shift the equilibrium to the left to compensate for the change. This means the value of Kc will (b) decrease.
(2) When the system is at equilibrium, Qc = Kc. However, since the system is no longer at equilibrium after the temperature change, Qc is not equal to the new Kc. Since the reaction shifts to the left, there will be more reactants and fewer products, so Qc will be (a) less than Kc.
(3) Since the reaction shifts to the left after the temperature decrease, the reaction must (b) run in the reverse direction to reestablish equilibrium.
(4) Since the reaction shifts to the left, the concentration of the product Cl2 will (b) decrease.
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