Question

The decomposition of $AB\left(g\right)\to A\left(g\right)+B\left(g\right)$, is first order reaction with a rate constant $k=4\times {10}^{-4}{s}^{-1}$ at $318$ $K$. If $AB$ has $26664.5$ $Pa$ pressure at the initial stage, what will be the partial pressure of $AB$ after half an hour?

Answer 1

Ans : The decomposition reaction is

$AB\left(g\right)\to A\left(g\right)+B\left(g\right)$

Given:

Rate can stant $k=4\times {10}^{-4}{s}^{-1}$

Temperature $T=318K$

Time $t=30\times 360sec=1800sec$

Initial partial pressure

$P_0=26664.5Pa$

To find :

Partial pressure

after half an have $P=?$

Relation between rate constant and partial pressure for

first order reac${}^n$

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

$k=\frac{2.303}{1800}[log26664.5-logP]$

$\frac{4\times {10}^{-4}\times 1800}{2.303}=log26664.5-logP$

$logP=4.42-0.3125=4.1075Pa$

$P={10}^{4.1075}=1.28\times {10}^{4}Pa$

$P=\frac{1.28\times {10}^4}{1.013\times {10}^5}=0.126atm$

$P=0.126atm$

The partial pressure after half an hour is $0.126atm$