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 lapse of time ‘t’ total pressure of the system increased by $x$ units and became $‘p_t’.$ 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. $k=\frac{2.303}{t}log$ $\frac{p_i}{p_i+x}$

Answer 1

Answer: (B)

If $x$ atm 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.
$\begin{array}{ccccc}{A}_{\left(g\right)}& \to & B_{\left(g\right)}& +& C_{\left(g\right)}\end{array}$
$\begin{array}{cccc}Att=0& p_{i}atm& 0atm& 0atm\\ Attimet& \left(p_i-x\right)atm& xatm& xatm\end{array}$
where, $p_i$ is the initial pressure at time $t=0$
$p_t$ $=\left(p_i-x\right)+x+x=p_i+x$
$x=\left(p_t-p_i\right)$
where, $p_A=p_i-x=p_i-\left(p_t-p_i\right)=2p_i-p_t$
$k=\left(\frac{2.303}{t}\right)$ $\left(log\frac{p_i}{p_A}\right)$ $=\frac{2.303}{t}log\frac{p_i}{\left(2p_i-p_t\right)}$