$\begin{array}{lcccc} \mathrm{SO}_2\mathrm{Cl}_{2(g)} & \rightarrow & \mathrm{SO}_{2(g)} & + & \mathrm{Cl}_{2(g)} \\ \text{Let initial pressure} & P_0 & & 0 & 0 \\ \text{Pressure at time } t & P_0 - p & & p & p \end{array}$ Let initial pressure $P_0 \propto R_0$ Pressure at time $t, P_t = P_0 - p + p + p = P_0 + p$$\therefore\;$ Pressure of reactant at time $t = P_0 - p = 2P_0 - P_t \propto R$Using formula, $k = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}$When $t = 100 \, \text{s},$$k = \frac{2.303}{100} \log \frac{0.5}{2 \times 0.5 - 0.6}$$= \frac{2.303}{100} \log(1.25)$$= \frac{2.303}{100}(0.0969)$$= 2.2316 \times 10^{-3} \, \text{s}^{-1}$When $P_t = 0.65$ atm, $\therefore$ Pressure of $\mathrm{SO}_2\mathrm{Cl}_2$ at time $t (p_{\mathrm{SO}_2\mathrm{Cl}_2}),$$R = 2P_0 - p_t$$= 2 \times 0.50 - 0.65$ atm $$\\$\;\;\;\;\;\;\;=0.35$\text{atm}\\text{Rate at that time}$=k \× p_{\ {SO}_{2} \ {Cl}_{2}}$\\$= (2.2316 \× 10^{-3}\ ) ×(0.35)$\\$=7.8 × 10^{-4} \ {atm} {s}^{-1}$