Question

Consider a first order gas phase decomposition reaction given below:
$A_{\left(g\right)}\to B_{\left(g\right)}+C_{\left(g\right)}$
The initial pressure of the system before decomposition of $A$ was $p_i$. After a lapse of time ${}^{\prime }{t}^{\prime }$ total pressure of the system increased by $x$ units and became ${}^{\prime }{p}_t{}^{\prime }$.
The rate constant $k$ for the reaction is given as:

• A. $k=\frac{2.303}{t}\log \frac{p_i}{p_i-x}$
• B. $k=\frac{2.303}{t}\log \frac{p_i}{2p_i-p_t}$
• C. $k=\frac{2.303}{t}\log \frac{p_i}{2p_i+p_t}$
• D. $k=\frac{2.303}{t}\log \frac{p_i}{2p_i+x}$

Answer 1

Answer: (B)

$k=\frac{2.303}{t}\log \frac{p_i}{2p_i-p_t}$

If $xatm$ be the decrease in pressure of $A$ at time $t$ and one mole each of $B$ and $C$ is being formed, the increase in pressure of $B$ and $C$ will also be $x$ atm each.

$\underset{\underset{Attimet(p_i-x)atm}{Att=0p_{i}atm}}{A_{\left(g\right)}}\to \underset{\underset{xatm}{0atm}}{B_{\left(g\right)}}+\underset{\underset{xatm}{0atm}}{C_{\left(g\right)}}$

where, $p_i$ is the initial pressure at time $t=0$

$p_t=(p_i-x)+x+x=p_i+x$

$x=(p_t-p_i)$

where, $p_A=p_i-x=p_i-(p_t-p_i)=2p_i-p_t$

$k=\left(\frac{2.303}{t}\right)\left(\log \frac{p_i}{p_A}\right)=\frac{2.303}{t}\log \frac{p_i}{(2p_i-p_t)}$.

Hence, the correct answer is option $\text{B}$.