Question

The equilibrium constant $K_c$ for the gaseous phase reaction at $523\text{K}$: $PC{l}_3+C{l}_2\rightleftharpoons PC{l}_5$ is $23.10{\text{L mol}}^{-1}$. Calculate $K_p$ at $503\text{K}$

• A. $0.05{\text{atm}}^{-1}$
• B. $0.65{\text{atm}}^{-1}$
• C. $0.56{\text{atm}}^{-1}$
• D. $1.00{\text{atm}}^{-1}$

Answer 1

The correct option is C $0.56{\text{atm}}^{-1}$

$PC{l}_3+C{l}_2\rightleftharpoons PC{l}_5$
$K_p=K_c(RT{)}^{\Delta n}$
$\Delta n=1-2=-1$
$K_p=K_c(RT{)}^{-1}$
$\Rightarrow K_p=23.10\times (0.0821\times 503{)}^{-1}$
$\Rightarrow K_p=\frac{23.10}{0.0821\times 503}=0.5593{\text{atm}}^{-1}$