Chemical and Ionic Equilibrium - Result Question 48
####48. For the reaction, $CO(g)+2 H _2(g) \rightleftharpoons CH _3 OH(g)$
hydrogen gas is introduced into a five litre flask at $327^{\circ} C$, containing 0.2 mole of $CO(g)$ and a catalyst, until the pressure is $4.92 atm$. At this point 0.1 mole of $CH _3 OH(g)$ is formed. Calculate the equilibrium constant, $K _p$ and $K _c$.
$(1990,5 M)$
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Solution:
- $CO(g)+2 H _2(g) \rightleftharpoons CH _3 OH(g)$
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 $H _2$ at equilibrium $=x-0.2=0.3$
Partial pressures : $\quad CO=\frac{0.1}{0.5} p, H _2=\frac{0.3}{0.5} p$,
$$ CH _3 OH=\frac{0.1}{0.5} p $$
$$ K _p=\frac{\frac{p}{5}}{\left(\frac{p}{5}\right)\left(\frac{3}{5} p\right)^{2}}=\frac{25}{9 p^{2}}=\frac{25}{9(4.92)^{2}}=0.11 atm^{-2} $$
Concentrations : $[CO]=\frac{0.1}{5} M,\left[H _2\right]=\frac{0.3}{5} M$,
$$ \left[CH _3 OH\right]=\frac{0.1}{5} M \Rightarrow K _c=\frac{(0.1 / 5)}{(0.1 / 5)(0.3 / 5)^{2}}=277.77 M^{-2} $$
