Question

At a given temperature, the equilibrium constant for reaction

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ is $2.4\times {10}^{-3}$. At the same temperature, the

equilibrium constant for reaction $PC{l}_3\left(g\right)+C{l}_2(g\rightleftharpoons (PC{l}_5\left(g\right)$ is

• A. 2.4 x 10^-3
• B. -2.4 x 10^-3
• C. 4.2 x 10²
• D. 4.8 x 10^-2

Answer 1

The correct option is C

4.2 x 10²

Reaction is reversed. Hence

$K=\frac{1}{2.4\times {10}^{-3}}=4.2\times {10}^2$