Question

The equilibrium constant at 298 K for the reaction \( A + B \rightleftharpoons C + D \) is 100. If the initial concentrations of all the four species were 1 M each, then equilibrium concentration of D (in mol/L) will be

• A. 1.818
• B. 1.182
• C. 0.818
• D. 0.182

Hint:

For equilibrium calculations, use the ICE table (Initial, Change, Equilibrium) to track the changes in concentration and apply the equilibrium constant expression to solve for unknown concentrations.

Answer 1

The Correct Option is A

The equilibrium expression for the reaction is: \[ K_c = \frac{[C][D]}{[A][B]} = 100 \] At equilibrium, if the initial concentrations of all species are 1 M, we assume the change in concentration is \(x\). Therefore, the concentrations at equilibrium are: - [A] = 1 - x - [B] = 1 - x - [C] = 1 + x - [D] = 1 + x Substitute these values into the equilibrium expression: \[ 100 = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} \] Solving for \(x\), we get: \[ 100 = \frac{(1 + x)^2}{(1 - x)^2} \] Simplifying: \[ (1 + x)^2 = 100(1 - x)^2 \] Expanding both sides and solving for \(x\), we find \(x = 0.818\). Thus, the equilibrium concentration of D is: \[ 1 + x = 1 + 0.818 = 1.818 \, \text{M} \]