Chemical And Ionic Equilibrium Result Question 48
48. For the reaction, $\mathrm{CO}(g)+2 \mathrm{H}_2(g) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(g)$ hydrogen gas is introduced into a five litre flask at $327^{\circ} \mathrm{C}$, containing $0.2$ mole of $\mathrm{CO}(g)$ and a catalyst, until the pressure is $4.92$ atm . At this point $0.1$ mole of $\mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$ is formed. Calculate the equilibrium constant, $K_p$ and $K_c$.
(1990, 5M)
Show Answer
Solution:
$\begin{aligned} & \quad \quad \quad \quad \mathrm{CO}(g)\quad + \quad 2 \mathrm{H}_2(g) \quad \rightleftharpoons \quad \mathrm{CH}_3 \mathrm{OH}(g) \\ & \text { Mole : } 0.2-0.10 \quad\quad x-0.20 \quad\quad \quad 0.10 \quad \Rightarrow \text { Total moles }=x \\ & \Rightarrow \quad x=\frac{4.92 \times 5}{0.082 \times 600}=0.5 \\ & \Rightarrow \quad \text { moles of } \mathrm{H}_2 \text { at equilibrium }=x-0.2=0.3 \\ & \text { Partial pressures : } \quad \mathrm{CO}=\frac{0.1}{0.5} p, \mathrm{H}_2=\frac{0.3}{0.5} p \\ & \qquad \mathrm{CH}_3 \mathrm{OH}=\frac{0.1}{0.5} \end{aligned}$
$\begin{aligned} & K_p=\frac{\frac{p}{5}}{\left(\frac{p}{5}\right)\left(\frac{3}{5} p\right)^2}=\frac{25}{9 p^2}=\frac{25}{9(4.92)^2}=0.11 \mathrm{~atm}^{-2} \\ & \text { Concentrations : }[\mathrm{CO}]=\frac{0.1}{5} \mathrm{M}, \quad\left[\mathrm{H}_2\right]=\frac{0.3}{5} \mathrm{M}, \\ & {\left[\mathrm{CH}_3 \mathrm{OH}\right]=\frac{0.1}{5} \mathrm{M} \Rightarrow K_c=\frac{(0.1 / 5)}{(0.1 / 5)(0.3 / 5)^2}=277.77 \mathrm{M}^{-2} .} \end{aligned}$
BETA
