question_answer

The gas phase decomposition of dimethyl ether follows first order kinetics. \[C{{H}_{3}}-O-C{{H}_{3}}(g)\to C{{H}_{4}}(g)+{{H}_{2}}(g)+CO(g)\]The reaction is carried out in a constant volume container at \[500 {}^\circ C\] and has a half-life of 14.5 minutes. Initially, only dimethyl ether is present at a pressure of 0.40 atmosphere. What is the total pressure of the system after 12 minutes? Assume ideal gas behaviour.

A) 0.75 atm

B) 0.55 atm

C) 0.68 atm

D) 0.85 atm

Correct Answer: A

Solution :

Calculation of initial concentration \[(a)\,\] of\[C{{H}_{3}}OC{{H}_{3}}\]

\[PV=nRT\]or \[\frac{n}{V}=\frac{P}{RT}\]

\[=\frac{P}{RT}=\frac{0.4}{0.082\times 773}=6.31\times {{10}^{-3}}mol\,{{L}^{-1}}\]

calculation of k

\[k=\frac{0.693}{{{t}_{{}^{1}/{}_{2}}}}=\frac{0.693}{14.5}=4.78\times {{10}^{-2}}{{\min }^{-1}}\]

Substituting the values in the first order equation.

\[k=\frac{2.303}{t}\log \frac{a}{a-x}\]

\[4.78\times {{10}^{-2}}=\frac{2.303}{12}\log \frac{a}{a-x}\]

\[\frac{a}{a-x}=1.77446\]

\[a-x=\frac{6.31\times 1{{0}^{-3}}}{1.77446}mol\,{{L}^{-1}}\]

\[=3.556\times 1{{0}^{-3}}mol\,{{L}^{-1}}\]

\[\therefore \]\[x=(6.310-3.556)\times {{10}^{-3}}mol\,{{L}^{-1}}\]

\[=2.754\times {{10}^{-3}}mol\,{{L}^{-1}}\]

\[\therefore \] After 12 min. Total no. of \[mol\,{{L}^{-1}}\]

\[=a+2x=6.31\times {{10}^{-3}}+2\times 2.754\times {{10}^{-3}}\]

\[=11.818\times {{10}^{-3}}\]

\[\therefore \]\[P=\frac{n}{V}RT=11.818\times {{10}^{-3}}\times 0.082\times 773\]

\[=0.75atm\]