Question

In the reaction $C\left(s\right)+{CO}_2\rightleftharpoons 2CO\left(g\right)$, the equilibrium pressure is 12 atm. If 50% ${CO}_2$ reacts, calculate $K_P$ :

• A. $K_P=0.25atm$
• B. $K_P=4atm$
• C. $K_P=16atm$
• D. $K_P=8atm$

Answer 1

The correct option is C $K_P=16atm$

$C\left(s\right)+{CO}_2\rightleftharpoons 2CO\left(g\right)$

__ 1 0 (Gaseous moles before dissociation)

__ $(1-\frac{50}{100})$ $\frac{2\times 100}{50}$ (Gaseous moles after dissociation)

Total moles = 1.5 and $△n=1$
Total pressure given at equilibrium = 12 atm
$K_P=\frac{{{(n}_{CO})}^2}{n_{{CO}_2}}\times {\left[\frac{P}{\sum n}\right]}^{\sum n}=\frac{{\left(1\right)}^2}{0.5}\times {\left(\frac{12}{1.5}\right)}^1$
$K_P=\frac{12}{1.5\times 0.5}=16atm$