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First Order Reaction

The gas phase...

Question

The gas phase decomposition of dimethyl ether follows first order kinetics.
$C H_{3} O C H_{3} (g) ⟶ C H_{4} (g) + H_{2} (g) + C O (g)$
The reaction is carried out in a constant volume container at $500^{o} C$ and has a half life of $14.5$ minutes. Initially only dimethyl ether is present at a pressure of $0.40$ atm. What is the total pressure after $12$ minutes? Assume ideal gas behaviour.

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Solution

$k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{14.5} = 0.047793 {m i n}^{- 1}$
Let the pressure of dimethyl ether after $12$ minutes be $p$ atm.
Applying first order equation,
$k = \frac{2.303}{t} {log}_{10} \frac{p_{0}}{p}$
${log}_{10} \frac{0.4}{p} = \frac{0.047793 \times 12}{2.303} = 0.2490$
or $\frac{0.4}{p} = 1.7743$
or $p = \frac{0.4}{1.7743}$
$= 0.2254$ atm
Decrease in pressur, $x = 0.4 - 0.2254 = 0.1746$ atm
$\begin{matrix} C H_{3} O C H_{3} (g) p_{0} - x \end{matrix} \begin{matrix} ⟶ \end{matrix} \begin{matrix} C H_{4} (g) x \end{matrix} \begin{matrix} + \end{matrix} \begin{matrix} H_{2} (g) x \end{matrix} \begin{matrix} + \end{matrix} \begin{matrix} C O (g) x \end{matrix}$
Total pressure $= p_{0} + 2 x$
$= 0.4 + 2 \times 0.1746$
$= 0.7492$ atm

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