The decomposition of $\mathrm{O}_3(g)$ follows first order kinetics and is given by $\mathrm{O}_3(9)…
The decomposition of $\mathrm{O}_3(g)$ follows first order kinetics and is given by
$\mathrm{O}_3(9) \longrightarrow \mathrm{O}_2(9)+\mathrm{O}(9)$
The rate constant for this reaction is $1.0 \times 10^{-3} \mathrm{~s}^{-1}$. The initial pressure of $\mathrm{O}_3(g)$ is $100 \mathrm{~atm}$.
What will be the partial pressure (in atm) of $\mathrm{O}_3, \mathrm{O}_2, \mathrm{O}$ respectively after 38.38 minutes?
$95,5,5$
$10,90,0$
$10,90,90$
$10,0,90$
Solution
Decomposition of $\mathrm{O}_3$,
$\mathrm{O}_3(g) \longrightarrow \mathrm{O}_2(g)+\mathrm{O}(g)$
Initial pressure $100 \mathrm{~atm} \quad 0 \quad 0$
Partial pressure $100-x \quad x \quad x$ after $38.38 \mathrm{~m}$ in
$\begin{aligned}
p_i(\text { initial pressure }) & =100 \mathrm{~atm} \\
\left.p_f \text { (final pressure }\right) & \text { after } 38.38 \mathrm{~m} . \\
& =100-x
\end{aligned}$
From first order reaction,
Given, $\quad k=1 \times 10^{-3} \mathrm{~s}^{-1}, t=38.38 \mathrm{~m}$.
$\begin{aligned}
k & =\frac{2.303}{t} \log \frac{p_i}{p_f} \\
1 \times 10^{-3} \mathrm{~s}^{-1} & =\frac{2.303}{38.38 \times 60} \log \frac{100(\mathrm{~atm})}{100-x(\mathrm{~atm})} \\
\log \frac{100}{100-x} & =\frac{1 \times 10^{-3} \mathrm{~s}^{-1} \times 38.38 \times 60}{2.303}
\end{aligned}$
$x=90 \mathrm{~atm}$
Pressure of $\mathrm{O}_3$ after $38.38 \mathrm{~m}$.
$p \mathrm{o}_3=p_f=100-x=100-90 \mathrm{~atm}=10 \mathrm{~atm}$
Pressure of $\mathrm{O}_2(g)(x)=90 \mathrm{~atm}$
Pressure of $\mathrm{O}(g)(x)=90 \mathrm{~atm}$
Hence, option (3) is correct.