Question

The reaction given below, is observed to be a first order with rate constant $7.5\times {10}^{-3}{s}^{-1}$. Calculate the time required for the total pressure in a system containing A at an initial pressure of $0.1atm$ to rise to $0.145atm$.

$2A\left(g\right)\to 4B\left(g\right)+C\left(g\right)$

• A. $0.8min$
• B. $0.4min$
• C. $1.4min$
• D. $2.5min$

Answer 1

The correct option is A $0.8min$

$2A\left(g\right)\to 4B\left(g\right)+C\left(g\right)$
$\text{At}t=0:P_{0}00$
$\text{At}t=t:P_0-2P^{\prime }4P^{\prime }{P}^{\prime }$

$P_{total}=P_0-{2P}^{{}^{\prime }}+4P^{{}^{\prime }}+P^{\prime }=P_0+3P^{{}^{\prime }}$
$P^{\prime }=\frac{1}{3}(P_{total}-P_0)$
Given:
$P_{total}=0.145atm$
$P_0=0.1atm$
$\therefore$
$P^{{}^{\prime }}=\frac{1}{3}(0.145-0.1)=0.015$

For first order kinetics,
$k=\frac{2.303}{t}\log \frac{P_0}{P_0-2P^{\prime }}$

$t=\frac{2.303}{7.5\times {10}^{-3}}\log \left(\frac{0.1}{0.07}\right)$

$t\approx 48s$
$t=0.8min$