The following data were obtained during the first order thermal decomposition of $SO_{2}Cl_{2}$ at a constant volume.
$SO_{2}Cl_{2(g)} \to SO_{2(g)} + Cl_{2(g)}$

Experiment	Time $/$ $s$	Total pressure $/$ $atm$
$1$	$0$	$0.5$
$2$	$100$	$0.6$

Calculate the rate of the reaction when total pressure is $0.65 \ atm$.

(D) The thermal decomposition of $SO_{2}Cl_{2}$ at a constant volume is represented by the following equation:

Time Phase	Reaction: $SO_{2}Cl_{2(g)} \longrightarrow SO_{2(g)} + Cl_{2(g)}$
At $t=0$	$P_{0} \longrightarrow 0 + 0$
At $t=t$	$(P_{0} - p) \longrightarrow p + p$

After time $t$,the total pressure $P_{t} = (P_{0} - p) + p + p = P_{0} + p$.
Thus,$p = P_{t} - P_{0}$.
The pressure of $SO_{2}Cl_{2}$ at time $t$ is $P_{SO_{2}Cl_{2}} = P_{0} - p = P_{0} - (P_{t} - P_{0}) = 2P_{0} - P_{t}$.
For a first order reaction,$k = \frac{2.303}{t} \log \frac{P_{0}}{2P_{0} - P_{t}}$.
At $t = 100 \ s$ and $P_{t} = 0.6 \ atm$,$k = \frac{2.303}{100} \log \frac{0.5}{2(0.5) - 0.6} = \frac{2.303}{100} \log \frac{0.5}{0.4} = 0.02303 \times \log(1.25) \approx 2.231 \times 10^{-3} \ s^{-1}$.
When $P_{t} = 0.65 \ atm$,the pressure of $SO_{2}Cl_{2}$ is $P_{SO_{2}Cl_{2}} = 2P_{0} - P_{t} = 2(0.5) - 0.65 = 1.0 - 0.65 = 0.35 \ atm$.
The rate of reaction is $Rate = k \times P_{SO_{2}Cl_{2}} = (2.231 \times 10^{-3} \ s^{-1}) \times (0.35 \ atm) = 7.81 \times 10^{-4} \ atm \ s^{-1}$.