Question

At a given temperature, the equilibrium constant for the reaction $PC{l}_5\left(g\right)\leftrightharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ is $2.4\times {10}^{-3}$. At the same temperature, the equilibrium constant for the reaction $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\leftrightharpoons PC{l}_5\left(g\right)$ is:

• A. $2.4\times {10}^{-3}$
• B. $-2.4\times {10}^{-3}$
• C. $4.2\times {10}^2$
• D. $4.8\times {10}^{-2}$

Answer 1

The correct option is C $4.2\times {10}^2$

When the reaction $PC{l}_5\left(g\right)\leftrightharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ is reversed, the reaction $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\leftrightharpoons PC{l}_5\left(g\right)$ is obtained.

Hence, the value of the equilibrium constant of the second reaction is the reciprocal of the value of the equilibrium constant of the first reaction,

i.e., $K_2=\frac{1}{K_1}$

But $K_1=2.4\times {10}^{-3}$

Substitute value in the above expression.

$K_2=\frac{1}{2.4\times {10}^{-3}}\approx 4.2\times {10}^2$.