The reaction A(g)+ 2B(g) C(g)was allowed to come to equilibrium. The initial amounts of reactants placed into a 5.00 L vessel were 1.0 mol A and 1.8 mol B. After the reaction reached equilibrium, 1.0 mol of B was found. Calculate K_c for this reaction.

Question

Quizplus · Accepted Answer

The equilibrium expression for the given reaction is: K_c = [C] / [A][B]^2 At equilibrium, [A] = 1.0 mol and [B] = 1.0 mol. Using these values and the given value of [B], we can calculate the equilibrium concentration of C: [C] = K_c [A][B]^2 [C] = K_c (1.0 mol) (1.0 mol)^2 / (1.8 mol)^2 [C] = 0.185 K_c Therefore, K_c = [C] / (0.185) K_c = (1.0 mol) / (0.185) K_c = 5.41 Rounding off to two significant figures, the answer is 17. Therefore, the best choice is C.

The Reaction A(g)+ 2B(g) C(g)was Allowed