Question

At ${1200}^{\circ }C$, the following equilibrium is established between chlorine atoms and molecule:

$C{l}_2\left(g\right)\rightleftharpoons 2Cl\left(g\right)$

The composition of the equilibrium mixture may be determined by measuring the rate of effusion of the mixture through a pin hole. It is found that at ${1200}^{\circ }C$ and $1$ atm pressure the mixture effuses $1.16$ times as fast as krypton effuses under the same condition. The equilibrium constant $K_c$ is:

• A. $6.71\times {10}^{-4}$
• B. $6.92\times {10}^{-4}$
• C. $7.71\times {10}^{-4}$
• D. $7.34\times {10}^{-4}$

Answer 1

Answer: (A)

$6.71\times {10}^{-4}$

The rate of effusion is inversely proportional to the molecular weight.
The molecular weight of krypton is 83.8.
$\frac{1.16}{1}=\sqrt{\frac{83.8}{M}}$
$M=62.27$
Thus, the average molecular weight of the mixture of $C{l}_2$ and Cl is 62.27.
Let x be the mole fraction of $C{l}_2$. The mole fraction of Cl will be $1-x$.
Hence, $71x+35.5(1-x)=62.27$
$x=0.754$
The mole fractions of $C{l}_2$ and $Cl$ are 0.754 and 0.246 respectively. They are also equal to the partial pressures as total pressure is 1 atm.
$K_p=\frac{P_C{l}^2}{P_{C{l}_2}}=\frac{{0.246}^2}{0.754}=0.08$
As $\Delta n=1$, $K_P=K_c\left(RT\right)$
Thus, $0.08=K_c(0.082\times 1473)$
$K_c=6.71\times {10}^{-4}$