Question

For a first order gas phase reaction :
$A_{{}_{\left(g\right)}}\to 2B_{{}_{\left(g\right)}}+C_{{}_{\left(g\right)}}$
$P_0$ 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)

$A_{{}_{\left(g\right)}}\to 2B_{{}_{\left(g\right)}}+C_{{}_{\left(g\right)}}$
Initial : $\begin{array}{ccc}{P}_0& 0& 0\\ P_0-P& 2P& P\end{array}$
Total pressure at time $\left(t\right)=P_0-P+2P+P=P_1$
$\Rightarrow P_t=P_0+2P$
$P_t-P_0=2P\Rightarrow P=\frac{P_t-P_0}{2}$
$k=\frac{2.303}{t}log\left[\frac{P_0}{P_0-P}\right]=\frac{2.303}{t}log\left[\frac{P_0}{P_0-\left(\frac{P_{t-P_0}}{2}\right)}\right]$
$=\frac{2.303}{t}log\left[\frac{2P_0}{2P_0-P_t+P_0}\right]=\frac{2.303}{t}log\left[\frac{2P_0}{3P_0-P_t}\right]$