Question

For a first order gas phase reaction :
$A_{_{\left(g\right)}}\to\quad2B_{_{\left(g\right)}}+C_{_{\left(g\right)}}$
$P_{0}$ be initial pressure of A and $P_{t}$ the total pressure at time ‘t’. Integrated rate equation is :

• A. $\frac{2.303}{t}log\left(\frac{P_{0}}{P_{0}-P_{t}}\right)$
• B. $\frac{2.303}{t}log\left(\frac{2P_{0}}{3P_{0}-P_{t}}\right)$
• C. $\frac{2.303}{t}log\left(\frac{P_{0}}{2P_{0}-P_{t}}\right)$
• D. $\frac{2.303}{t}log\left(\frac{2P_{0}}{2P_{0}-P_{t}}\right)$

Answer 1

Answer: (B)

$A_{_{\left(g\right)}}\to2B_{_{\left(g\right)}}+C_{_{\left(g\right)}}$
Initial : $\begin{matrix}P_{0}&0&0\\ P_{0}-P&2P&P\end{matrix}$
Total pressure at time $\left(t\right) =P_{0}-P+2P+P=P_{1}$
$\Rightarrow P_{t}=P_{0}+2P$
$P_{t}-P_{0}=2P \Rightarrow P=\frac{P_{t}-P_{0}}{2}$
$k=\frac{2.303}{t} log \left[\frac{P_{0}}{P_{0}-P}\right]=\frac{2.303}{t}log \left[\frac{P_{0}}{P_{0}-\left(\frac{P_{t-P_0}}{2}\right)}\right]$
$=\frac{2.303}{t} log \left[\frac{2P_{0}}{2P_{0}-P_{t}+P_{0}}\right]=\frac{2.303}{t} log \left[\frac{2P_{0}}{3P_{0}-P_{t}}\right]$