Question

For the reaction $PC{l}_5\left(g\right)\iff PC{l}_3\left(g\right)+C{l}_2\left(g\right),K_c$ is:

• A. $\frac{n_{PC{l}_3}\times n_{C{l}_2}}{n_{PC{l}_5}}$
• B. $\frac{n_{PC{l}_3}\times n_{C{l}_2}}{n_{PC{l}_5}}\times \frac{1}{V}$
• C. $\frac{n_{PC{l}_5}}{n_{PC{l}_3}}\times n_{C{l}_2}\times \frac{1}{V}$
• D. $\frac{n_{PC{l}_3}\times n_{C{l}_2}}{n_{PC{l}_5}}\times V$

Answer 1

The correct option is B $\frac{n_{PC{l}_3}\times n_{C{l}_2}}{n_{PC{l}_5}}\times \frac{1}{V}$

The reaction is $PC{l}_5\left(g\right)\iff PC{l}_3\left(g\right)+C{l}_2\left(g\right)$.

The expression for the equilibrium constant is, $Kc=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$.

But $\left[PC{l}_3\right]=\frac{n_{PC{l}_3}}{V}$, $\left[C{l}_2\right]=\frac{n_{C{l}_2}}{V}$, $\left[PC{l}_5\right]=\frac{n_{PC{l}_5}}{V}.$

Substituting these values in the expression for the equilibrium constant expression, we get

$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}=\frac{\frac{n_{PC{l}_3}}{V}\times \frac{n_{C{l}_2}}{V}}{\frac{n_{PC{l}_5}}{V}}=\frac{n_{PC{l}_3}\times n_{C{l}_2}}{n_{PC{l}_5}}\times \frac{1}{V}$.