Chemical and Ionic Equilibrium 1 Question 46
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 $4.92 \mathrm{~atm}$. At this point 0.1 mole of $\mathrm{CH}{3} \mathrm{OH}(g)$ is formed. Calculate the equilibrium constant, $K{p}$ and $K_{c}$.
$(1990,5 M)$
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Solution:
- $\mathrm{CO}(g)+2 \mathrm{H}{2}(g) \rightleftharpoons \mathrm{CH}{3} \mathrm{OH}(g)$
Mole : $0.2-0.10 \quad x-0.20 \quad 0.10 \Rightarrow$ Total moles $=x$
$\Rightarrow \quad x=\frac{4.92 \times 5}{0.082 \times 600}=0.5$
$\Rightarrow$ moles of $\mathrm{H}_{2}$ at equilibrium $=x-0.2=0.3$
Partial pressures : $\quad \mathrm{CO}=\frac{0.1}{0.5} p, \mathrm{H}_{2}=\frac{0.3}{0.5} p$,
$$ \mathrm{CH}_{3} \mathrm{OH}=\frac{0.1}{0.5} p $$
$$ K_{p}=\frac{\frac{p}{5}}{\left(\frac{p}{5}\right)\left(\frac{3}{5} p\right)^{2}}=\frac{25}{9 p^{2}}=\frac{25}{9(4.92)^{2}}=0.11 \mathrm{~atm}^{-2} $$
Concentrations : $[\mathrm{CO}]=\frac{0.1}{5} \mathrm{M},\left[\mathrm{H}_{2}\right]=\frac{0.3}{5} \mathrm{M}$,
$$ \left[\mathrm{CH}{3} \mathrm{OH}\right]=\frac{0.1}{5} \mathrm{M} \Rightarrow K{c}=\frac{(0.1 / 5)}{(0.1 / 5)(0.3 / 5)^{2}}=277.77 \mathrm{M}^{-2} $$
