Chemical and Ionic Equilibrium - Result Question 47
47. $0.15$ mole of $CO$ taken in a $2.5 L$ flask is maintained at $750 K$ along with a catalyst so that the following reaction can take place:
$ CO(g)+2 H _2(g) \rightleftharpoons CH _3 OH(g) $
Hydrogen is introduced until the total pressure of the system is $8.5$ $ atm$ at equilibrium and $0.08$ mole of methanol is formed.
Calculate (i) $K _p$ and $K _c$ and (ii) the final pressure if the same amount of $CO$ and $H _2$ as before are used, but with no catalyst so that the reaction does not take place.
$(1993,5 M)$
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Solution:
- $\underset{0.15-0.08}{\mathrm{CO}(g)}+\underset{ x-0.16}{2 \mathrm{H}_2(g)} \rightleftharpoons \underset{0.08}{ \mathrm{CH}_3 \mathrm{OH}(g)}$
Total moles at equilibrium $=x-0.01$
$ x-0.01=\dfrac{8.5 \times 2.5}{0.082 \times 750}=0.34 \Rightarrow x=0.35 $
(i) Partial pressures :
$ \begin{aligned} \mathrm{CO} & =\dfrac{0.07}{0.34} \times 8.5 \\ \mathrm{H}_2 & =\dfrac{0.18}{0.34} \times 8.5 \\ \mathrm{CH}_3 \mathrm{OH} & =\dfrac{0.08}{0.34} \times 8.5 \\ K_p & =\dfrac{0.08}{(0.07)(0.18)^2} \times\left(\dfrac{0.34}{8.5}\right)^2=0.056 \end{aligned} $
(ii) Concentrations :
$ \begin{aligned} {\left[\mathrm{CH}_3 \mathrm{OH}\right] } & =\dfrac{0.08}{2.5}=0.032 \mathrm{M} \\ {\left[\mathrm{H}_2\right] } & =\dfrac{0.18}{2.5}=0.072 \mathrm{M} \\ {[\mathrm{CO}] } & =\dfrac{0.07}{2.5}=0.028 \mathrm{M} \\ K_c & =\dfrac{0.032}{(0.028)(0.072)^2}=213.33 \end{aligned} $
BETA
