For the first order reaction $\mathrm{A}\left(\mathrm{g}\right)\to 2\mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{g}\right)$, the initial pressure is ${\mathrm{P}}_{\mathrm{A}}=90$ mm Hg, the pressure after 10 minutes is found to be 180 mm Hg. The rate constant of the reaction is

The reaction is                        $\begin{array}{ccccc}{\mathrm{A}}_{\left(\mathrm{g}\right)}& \to & 2{\mathrm{B}}_{\left(\mathrm{g}\right)}& +& \mathrm{C}\left(\mathrm{g}\right)\end{array}$

at t = 0 pressure                       $\begin{array}{ccccc}{\mathrm{P}}_{\mathrm{A}}& & 0& & 0\end{array}$

Pressure after 10 mints               $\begin{array}{ccccc}{\mathrm{P}}_{\mathrm{A}}-\mathrm{x}& & 2\mathrm{x}& & \mathrm{x}\end{array}$

Total pressure after 10 minutes, ${\mathrm{P}}_{\mathrm{T}}={\mathrm{P}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}+{\mathrm{P}}_{\mathrm{C}}$

 ${\mathrm{P}}_{\mathrm{T}}=({\mathrm{P}}_{\mathrm{A}}-\mathrm{x})+2\mathrm{x}+\mathrm{x}$

$\begin{array}{l}={\mathrm{P}}_{\mathrm{A}}+2\mathrm{x}=180\\ \because {\mathrm{P}}_{\mathrm{A}}=90\mathrm{mm}\end{array}$

$\begin{array}{l}90+2\mathrm{x}=180 \mathrm{or}, \mathrm{X}=45\mathrm{mm}\\ \therefore {\mathrm{P}}_{\mathrm{A}}-\mathrm{x}=90-45=45\mathrm{mm}\end{array}$

Since, $\mathrm{kt}=2.303\log \frac{{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}-\mathrm{x}}$

 $\mathrm{k}\times 10=2.303\log \frac{90}{90-45}$

$\mathrm{k}\times 10=0.693$

 $\mathrm{k}=\frac{0.693}{10}=0.0693 {\min }^{-1}$

    $=\frac{0.693}{10\times 60\text{ sec}}=1.155\times {10}^{-3}{\sec }^{-1}$