\(\begin{array}{llcc} & \mathrm{SO}_2 \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow & \mathrm{SO}_2(\mathrm{~g})+& \mathrm{Cl}_2(\mathrm{~g}) \\ \text { Initial pressure } & p_0 & 0 & 0 \\ \text { After some time } t & p_0-p & p & p\end{array}\)
Total pressure after some time \(t\),
\(p_T=\left(p_0-p\right)+p+p=p_0+p\)
Let the initial conc. of \(\mathrm{SO}_2 \mathrm{Cl}_2\) be a concentration of \(\mathrm{SO}_2 \mathrm{Cl}_2\) after some time, \(t=a-x\) for first order reaction,
\(\begin{aligned}
k=\frac{2 \cdot 303}{t} \log \frac{a}{a-x} ...(I)\\
a & \propto p_0 \\
(a-\mathrm{x}) & \propto\left(p_0-p\right) \\
& \propto p_0-\left(p_{\mathrm{T}}-p_0\right) \\
(a-x) & \propto 2 p_0-p_T
\end{aligned}\)
Substituting these values in \((i)\), we get
\(k=\frac{2.303}{t} \log \frac{p_0}{2 p_0-p_t}\)
when \(t=100 \mathrm{~s}\),
\(k=\frac{2.303}{100} \log \frac{0.5}{2 \times 0.5-0.6}=\frac{2.303}{100} \log (1.25)\)
\(=\frac{2 \cdot 303}{100}(0.0969)=2.2316 \times 10^{-3} \mathrm{~s}^{-1} \text {. }\)
When \(p_T=0.65 \mathrm{~atm}\), i.e., \(p_0+p=0.65 \mathrm{~atm}\) \(\therefore p=0.65-p_0=0.65-0.50=0 \cdot 15 \mathrm{~atm}\)
\(\therefore\) Pressure of \(\mathrm{SO}_2 \mathrm{Cl}_2\) at time
\(\begin{aligned}
t &=p_0-p=0.50-0.15=0.35 \mathrm{~atm} \\
\text { Rate } &=k\left[p_{\mathrm{SO}_2 \mathrm{Cl}_2}\right] \\
&=\left(2.2316 \times 10^{-3} \mathrm{~s}^{-1}\right)(0.35 \mathrm{~atm}) \\
&=7.8 \times 10^{-4} \mathrm{~atm} \mathrm{~s}^{-1} .
\end{aligned}\)