For the reaction $\mathrm{A}(\mathrm{g}) ightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$
rate law is $-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]$
At the start pressure is $100 \mathrm{~mm}$ and after 10 min, pressure is $120 \mathrm{~mm}$, hence, rate constant $\left(\mathrm{min}^{-1}ight)$ is

Question

Aicharya · Accepted Answer

Reactants and products are in gaseous state thus,
$\mathrm{A}(\mathrm{g}) ightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$
at $\mathrm{t}=\begin{array}{llll}0 & \phi_{\mathrm{I}} & 0 & 0\end{array}$
at $\mathrm{t}=\mathrm{t} \quad\left(\mathrm{p}_{\mathrm{i}}-\mathrm{x}ight) \quad \mathrm{x} \quad \mathrm{x}$
total pressure at time
$\mathrm{t}=\mathrm{p}_{\mathrm{t}}=\mathrm{p}_{\mathrm{i}}-\mathrm{x}+\mathrm{x}+\mathrm{x}=\mathrm{p}_{\mathrm{i}}+\mathrm{x}$
$\therefore \mathrm{x}=\left(\mathrm{p}_{\mathrm{t}}-\mathrm{p}_{\mathrm{i}}ight)$
$(a-x)=p_{i}-\left(p_{t}-p_{i}ight)$
$=2 \mathrm{p}_{\mathrm{i}}-\mathrm{p}_{\mathrm{t}}$
Given $\mathrm{p}_{\mathrm{i}}=\mathrm{a}=100 \mathrm{~mm}$
$\mathrm{p}_{\mathrm{t}}=120 \mathrm{~mm}$
$\therefore(a-x)=2 p_{i}-p_{t}=80$
$\therefore k=\frac{2.303}{10} \log \frac{100}{80}$ *

For the reaction $\mathrm{A}(\mathrm{g}) ightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ rate law…

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