Question

${CH}_3-CO-{CH}_3\left(g\right)\rightleftharpoons {CH}_3-{CH}_3\left(g\right)+CO\left(g\right)$
Initial pressure of ${CH}_{3}CO{CH}_3$ is $100mm$. When equilibrium set up and mol fraction of $CO\left(g\right)$ is $1/3$ hence $K$ is:

• A. $100mm$
• B. $50mm$
• C. $25mm$
• D. $0.6mm$

Answer 1

Answer: (B)

$50mm$

${C{H}_{3}COC{H}_3}_{\left(9\right)}\rightleftharpoons {C{H}_3-C{H}_3}_{\left(9\right)}+C{O}_{\left(9\right)}$

At $t=0$ $100mm$ $0$ $0$

At $t=equilibrium$ $(100-p)$ $p$ $p$

But $\frac{p}{100-p+p+p}=\frac{1}{3}$

$\therefore \frac{p}{100+p}=\frac{1}{3}$

$\therefore 3P=100+P$

$\therefore P=50mm$

$\therefore K=\frac{P\times P}{(100-P)}=\frac{(50{)}^2}{(100-50)}=50mm$