Question

In a closed cylinder of capacity $24.6L$, the following reaction occurs at ${27}^{\circ }C$.

$A_2\left(s\right)\rightleftharpoons B_2\left(s\right)+2C\left(g\right)$

At equilibrium, $1g$ of $B_2\left(s\right)$ (molar mass $=50gmo{l}^{-1})$ is present. The equilibrium constant $K_p$ for the equilibrium in $at{m}^2$ unit is:

$[R=0.082Latm{K}^{-1}mo{l}^{-1}]$

• A. $1.6\times {10}^{-2}$
• B. $1.6\times {10}^{-5}$
• C. $1.6\times {10}^{-3}$
• D. $1.6\times {10}^{-4}$
• E. $1.6\times {10}^{-1}$

Answer 1

The correct option is C $1.6\times {10}^{-3}$

Equilibrium concentration of $B_2$.

$[B_2{]}_{equilibrium}=\frac{mass}{molarmass\times volume}$

$=\frac{1}{50\times 24.0}=8.13\times {10}^{-4}$

For the reaction,

$\underset{\underset{\text{a - x}}{\text{a}}}{A_2\left(s\right)}\rightleftharpoons \underset{\underset{x}{0}}{B_2\left(s\right)}+\underset{\underset{2x}{0}}{2C\left(g\right)}$

$=8.13\times {10}^{-3}$

$\therefore \left[C\right]=2\times 8.13\times {10}^{-4}=1.63\times {10}^{-3}$

$K_C=[C{]}^2=[1.63\times {10}^{-3}{]}^2=2.66\times {10}^{-6}$

Again from, $K_P=K_C(RT{)}^{\Delta n_g}$

Here, $\Delta n_g=2-0=2$

$K_P=2.66\times {10}^{-6}\times (0.082\times 300{)}^2$

$\approx 1.6\times {10}^{-3}$.

Hence, the correct option is $\text{C}$