Question

$PC{l}_5,PC{l}_{3}andC{l}_2$ are at equilibrium at $500$K in a closed container and their concentrations are $0.8\times {10}^{-3}mol{L}^{-1},1.2\times {10}^{-3}mol{L}^{-1}$ and $1.2\times {10}^{-3}mol{L}^{-1}$ respectively.
The value of $K_c$ for the reaction:
$PC{l}_{5\left(g\right)}\rightleftharpoons PC{l}_{3\left(g\right)}+C{l}_{2\left(g\right)}$ will be:

• A. $1.8\times {10}^{3}mol{L}^{-1}$
• B. $1.8\times {10}^{-3}mol{L}^{-1}$
• C. $1.8\times {10}^{-3}Lmo{l}^{-1}$
• D. $0.55\times {10}^{4}Lmo{l}^{-1}$

Answer 1

The correct option is B $1.8\times {10}^{-3}mol{L}^{-1}$

The given reaction is :-

${PCl}_5\left(g\right)\rightleftharpoons {PCl}_3\left(g\right)+{Cl}_2\left(g\right)$

$K_C=\frac{\left[{PCl}_3\right]\left[{Cl}_2\right]}{\left[{PCl}_5\right]}$

$=\frac{1.2\times {10}^{-3}\times 1.2\times {10}^{-3}}{0.8\times {10}^{-3}}=1.8\times {10}^{-3}{molL}^{-1}$