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 ${500}^{o}C$ and has a half life of 14 minutes. Initially only dimethyl ether is present at a pressure of 0.40 atm. What is the total pressure after 12 minutes in $atm$? Assume ideal gas behaviour. Take: Antilog (0.26) = 1.82

Answer 1

Let the pressure of dimethyl ether after 12 minutes be $Patm$.
Apply first order equation,

$k=\frac{2.303}{t}log\frac{P_0}{P}$



$log\frac{0.4}{P}=\frac{k\times 12}{2.303}$

We know,
$t_{1/2}=\frac{0.693}{k}$

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{14}\approx 0.05$

$\therefore$
$log\frac{0.4}{P}=\frac{0.05\times 12}{2.303}$
$log\frac{0.4}{P}=0.26$
$\frac{0.4}{P}=1.82$
$P=\frac{0.4}{1.82}\approx 0.220atm$

let x is is pressure of reactant which is already converted to product

Decrease in pressure,
$x=P_0-P=0.4-0.22=0.18atm$

$C{H}_3\underset{P_0-x}{OC{H}_3\left(g\right)\to }\underset{x}{C{H}_4\left(g\right)}+\underset{x}{H_2\left(g\right)}+\underset{x}{CO\left(g\right)}$
Total pressure $=P_0+2x$
$=0.4+2\times 0.18$
$=0.76atm$