Question

The gas phase decomposition of dimethyl ether follows first order kinetics.
$C{H}_{3}OC{H}_3\left(g\right)\to C{H}_4\left(g\right)+H_2\left(g\right)+CO\left(g\right)$
The reaction is carried out in a constant volume container at ${500}^{\circ }\text{C}$ and has a half life of $30.1$ minutes. Initially, only dimethyl ether is present at a presure of $0.396\text{atm}$. What is the total pressure (in $\text{atm}$ after $12$ minutes? Assume ideal gas behavior. (Give your answer upto two decimal)
(Given: ${10}^{0.12}=1.32$)

Answer 1

The rate constant of the first order reaction is given by
$k=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{30.10}=0.02303{\text{min}}^{-1}$
Let the pressure of dimethyl ether after $12$ minutes be $P\text{atm}$.
Applying first order equation,
$k=\frac{2.303}{t}{\log }_{10}\frac{P_0}{P}\Rightarrow {\log }_{10}\frac{0.396}{P}=\frac{0.02303\times 12}{2.303}=0.12\Rightarrow \frac{0.396}{P}=1.32\Rightarrow P=\frac{0.396}{1.32}=0.3\text{atm}$
Decrease in pressure, $x=0.396-0.3=0.096\text{atm}$
$$\begin{array}{cccc}C{H}_{3}OC{H}_3\left(g\right)& \to C{H}_4\left(g\right)+& H_2\left(g\right)+& CO\left(g\right)\\ (P_0-x)& x& x& x\end{array}$$
$\therefore \text{Total pressure}=P_0+2x=0.396+2\times 0.096=0.588\text{atm}$