Consider the following equilibrium for which H = 114.44 kJ: 2 Cl2(g) + 2 H2O(g)

Question

Consider the following equilibrium for which H = 114.44 kJ: 2 Cl2(g) + 2 H2O(g) 4 HCl(g) + O2(g) If a mixture of the gases involved in this reaction is allowed to come to equilibrium, and the volume of the reaction mixture is doubled, what will the reaction do in response to this change, and how will the value of K (the equilibrium constant) be affected? a) The reaction will shift toward reactants with an increase in K. b) The reaction will shift toward products with no change in K. c) The reaction will shift toward reactants with a decrease in K. d) The reaction will shift toward reactants with no change in K. e) The reaction will not shift and the value of K will not change.

Explanation / Answer

equilibrium is given by keq = ([HCL]eq^4 * [O2]eq )/ ([Cl2]eq^2 * [H20]eq^2). Now if the concentration of reactants is doubled,

Q = (1/4)*Keq, which means Q < Keq. So the reaction will proceed towards products side and Keq is a constant. So C> option is the answer.

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