For the reaction 2X(g) + Y(g) 2Z(g), K_c = 1.00 ×10³ at 500 K. If at equilibrium the concentration of X is 0.20 M and the concentration of Y is 0.50 M, what is the equilibrium concentration of Z?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is: K_c = [Z]^2 / ([X]^2 [Y]) Substituting the given equilibrium concentrations: 1.00 ×10³ = [Z]^2 / (0.20)^2 (0.50) Solving for [Z]: [Z]^2 = 1.00 ×10³ × (0.20)^2 (0.50) [Z]^2 = 0.001 [Z] = 0.0316 M Since there are 2 moles of Z formed for every 2 moles of X and 1 mole of Y consumed, the equilibrium concentration of Z is: [Z] = 0.0316 M x 2 = 0.0632 M Rounding to two significant figures, the equilibrium concentration of Z is 4.5 M, which is answer choice D.

For the Reaction 2X(g) + Y(g) 2Z(g), Kc =

For the Reaction 2X(g) + Y(g) 2Z(g), K_c =