Question

A 500 ml vessel contains 1.5 M each of A, B, C and D at equilibrium. If 0.5 M each of C and D are taken out, the value of $K_c$ for A + B $\rightleftharpoons$ C + D will be :

• A. $\frac{1}{9}$
• B. $\frac{8}{9}$
• C. $\frac{4}{9}$
• D. 1.0

Answer 1

Answer: (D)

1.0

Answer 2

The problem asks for the value of the equilibrium constant, $K_c$, for the reaction A + B $\rightleftharpoons$ C + D.

1. Initial Equilibrium: At equilibrium, the concentrations are given as: [A] = 1.5 M [B] = 1.5 M [C] = 1.5 M [D] = 1.5 M

2. Calculate $K_c$ from Initial Equilibrium: The equilibrium constant expression for the reaction A + B $\rightleftharpoons$ C + D is: $K_c = \frac{[C][D]}{[A][B]}$ Substitute the initial equilibrium concentrations: $K_c = \frac{(1.5)(1.5)}{(1.5)(1.5)} = \frac{2.25}{2.25} = 1.0$

3. Effect of Disturbing Equilibrium on $K_c$: The value of the equilibrium constant ($K_c$) depends only on temperature for a given reaction. It does not change with changes in concentration of reactants or products, volume, or pressure. When concentrations are changed (e.g., by removing C and D), the system will shift to a new equilibrium state, but the ratio of products to reactants at this new equilibrium (i.e., $K_c$) will remain the same, provided the temperature is constant. The problem does not mention any change in temperature.

Therefore, even if 0.5 M each of C and D are taken out, the value of $K_c$ for the reaction will remain the same.

The value of $K_c$ is 1.0.