Question

$PCl_5$, $PCl_3$ and $Cl_2$ 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 :
$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} +Cl_{2(g)}$ will be

• A. $1.8\times10^{3}\,mol\,L^{-1}$
• B. $1.8\times10^{-3}\,L\,mol^{-1}$
• C. $1.8\times10^{3}\,L\,mol^{-1}$
• D. $0.55 \times10^4\,L\,mol^{-1}$

Answer 1

Answer: (B)

For the reaction :
$PCl_{5(g)} rightleftharpoons PCl_{3(g)} +Cl_{2(g)}$
$K_{c}=\frac{\left[PCl_{3}\right]\left[Cl_{2}\right]}{\left[PCl_{5}\right]}$
$=\frac{\left(1.2\times10^{-3}\right)^{2}}{0.8\times10^{-3}}$
$=1.8\times10^{-3}$