Question

1.1 mole of A mixed with 2.2 mole of B and the mixture is kept in a 1 litre flask at equilibrium. The reaction is $A + 2B \rightleftharpoons 2C + D$. At equilibrium, 0.2 mole of C is formed. Calculate the value of $(2000 \times K_c)$.

Answer 1

2

Answer 2

The reaction is $A + 2B \rightleftharpoons 2C + D$. Initial concentrations: $[A]_0 = 1.1$ M, $[B]_0 = 2.2$ M, $[C]_0 = 0$ M, $[D]_0 = 0$ M. At equilibrium, $[C]_{eq} = 0.2$ M. Using an ICE table, we find that $2x = [C]_{eq} = 0.2$ M, so $x = 0.1$ M. Equilibrium concentrations: $[A]_{eq} = 1.1 - x = 1.0$ M $[B]_{eq} = 2.2 - 2x = 2.0$ M $[C]_{eq} = 2x = 0.2$ M $[D]_{eq} = x = 0.1$ M The equilibrium constant $K_c$ is: $K_c = \frac{[C]_{eq}^2 [D]_{eq}}{[A]_{eq} [B]_{eq}^2} = \frac{(0.2)^2 (0.1)}{(1.0) (2.0)^2} = \frac{(0.04)(0.1)}{(1.0)(4.0)} = \frac{0.004}{4.0} = 0.001$. The value of $2000 \times K_c = 2000 \times 0.001 = 2$.