Question

$C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3\left(g\right)\rightleftharpoons C{H}_3-C{H}_3\left(g\right)+CO\left(g\right)$

The initial pressure of ${CH}_{3}CO{CH}_3$ is $100mmofHg$. When equilibrium is achieved, the mole fraction of $CO\left(g\right)$ is $1/3$ hence, $K_p$ is:

• A. $100mm$
• B. $50mm$
• C. $25mm$
• D. $150mm$

Answer 1

The correct option is B $50mm$

$Explanation:$

Equation $C{H}_{3}COC{H}_3\left(g\right)\rightleftharpoons C{H}_{3}C{H}_3\left(g\right)+CO\left(g\right)$

Pressure

at $t=0$ $100$ $0$ $0$

at $t=t$ $100-x$ $x$ $x$

$P_T=$Total pressure$=100-x+x+x$

According to $Raoul{t}^{\prime }sLaw$

$P_A=\left(P_T\right)\left(X_A\right)$

$P_{CO}=\left(P_T\right)\left(X_{CO}\right)$

where,

$P\left(CO\right)=$Partial pressure of $CO$

$P_T=$Total pressure

$X_{CO}=$Mole fraction of $CO$

But give that $X_{CO}=\frac{1}{3}$

$x=P_{T0}\times \frac{1}{3}$

$x=\frac{100+x}{3}\Rightarrow 2x=100\Rightarrow x=50$

$K_P=\frac{P_{CO}{P}_{C{H}_{3}C{H}_3}}{P_{C{H}_{3}COC{H}_3}}$$\Rightarrow$ $K_P=\frac{50\times 50}{50}=50mm$

$\therefore K_P=50mm$

Hence the correct answer is option $B$