Chemistry - Chemical Kinetics Question with Solution | TestHub

Given, Initial pressure of $\text{A,}$  ${\text{P}}_0=90\text{ torr}$

Let's assume there is a decrease in pressure of $\text{A}$ by ${\text{P}}_0 \text{torr}$ after $\text{10}$ mins. Therefore, after $\text{10}$ mins the reaction mixture will look like,

$\mathrm{A}\left(\mathrm{g}\right)⟶\mathrm{ }2\mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\mathrm{ }\left(\mathrm{g}\right)$

$\left({\text{P}}_0-\text{ p}\right) 2\text{p} \text{p}$

${\text{P}}_{\text{t}}={\text{P}}_0+2\text{p}=180\text{ torr}$

$\text{p}=45\text{ torr}$

Thus, after $10$ mins, the pressure of $\text{A}$ will be, ${\text{P}}_{\text{A}}={\text{P}}_0-\text{p} = 90-45=45 \text{torr}$

$\text{k}=\frac{1}{\text{t}}\ln \frac{{\text{P}}_0}{{\text{P}}_{\text{A}}}$

$\text{k}=\frac{1}{600}\ln \frac{90}{45}$

$=1.15\times {10}^{-3}{\text{ s}}^{-1}$