Question

$K_C={10}^{-5}$ for the dissociation of the dimer $(A_2\rightleftharpoons 2A)$ in benzene solution at ${27}^{o}C.$ What is the ratio fo equilibrium concentration of monomer to dimer $\left(\right[A\left]/\right[A_2\left]\right)?$ Given: $R=2Cal/mol.K$
Given: Initial concentration of A, $[A{]}_o=0.8mol{L}^{-1}$

• A. 1 : 200
• B. 1 : 100
• C. 200 : 1
• D. 800 : 1

Answer 1

The correct option is A 1 : 200

Given, $K_C={10}^{-5}$
for reverse reaction,
$\begin{array}{ccccc}{K}_C^{\prime }& =& 1/K_C& =& {10}^5\\ 2A& & \rightleftharpoons & & A_2\\ 0.8-2x& & & & x\end{array}$

$\because K_C$ is very high so reaction almost completed,
$2x\simeq 0.8\Rightarrow x\simeq 0.4$
$\therefore 0.8-2x\simeq y$
${10}^5=\frac{0.4}{(y{)}^2};y=(0.4\times {10}^{-5}{)}^{1/2}\Rightarrow 2\times {10}^{-3}$
$\frac{\left[A\right]}{\left[A_2\right]}=\frac{y}{x}=\frac{2\times {10}^{-3}}{0.4}=\frac{5}{1000}=\frac{1}{200}$