The following data were obtained during the first order thermal decomposition of $$S{O_2}C{l_2}$$  at a constant volume.
$$S{O_2}C{l_{2\left( g \right)}} \to S{O_{2\left( g \right)}} + C{l_{2\left( g \right)}}$$

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

What is the rate of reaction when total pressure is $$0.65\,atm?$$

Solution :
\[\underset{\begin{smallmatrix} \text{Initial pressure} \\ \\ \text{Pressure at time }t \end{smallmatrix}}{\mathop{{}}}\,\underset{\begin{smallmatrix} {{p}_{0}} \\ {{p}_{0}}-p \end{smallmatrix}}{\mathop{S{{O}_{2}}C{{l}_{2}}}}\,\to \underset{\begin{smallmatrix} 0 \\ \\ p \end{smallmatrix}}{\mathop{S{{O}_{2}}}}\,+\underset{\begin{smallmatrix} 0 \\ \\ p \end{smallmatrix}}{\mathop{C{{l}_{2}}}}\,\]
Let initial pressure, $${p_0} \propto {R_0}$$
Total Pressure at time $$t,{P_t} = {p_0} - p + p + p = {p_0} + p$$
Pressure of reactants at time $$t,{p_0} - p = 2{p_0} - {P_t} \propto R$$
$$\eqalign{ & k = \frac{{2.303}}{t}\log \frac{{{p_0}}}{{2{p_0} - {P_t}}} \cr & \,\,\,\, = \frac{{2.303}}{{100}}\log \frac{{0.5}}{{2 \times 0.5 - 0.6}} \cr & \,\,\,\, = \frac{{2.303}}{{100}}\log 1.25 \cr & \,\,\,\, = 2.2318 \times {10^{ - 3}}{s^{ - 1}} \cr} $$
$${\text{Pressure of}}\,\,S{O_2}C{l_2}\,\,{\text{at}}\,\,{\text{time}}$$       $$t\left( {{p_{S{O_2}C{l_2}}}} \right)$$
$$\eqalign{ & = 2{p_0} - {P_t} \cr & = 2 \times 0.50 - 0.65\,atm \cr & = 0.35\,atm \cr & {\text{Rate at that time}} \cr & = k \times {p_{S{O_2}C{l_2}}} \cr & = \left( {2.2318 \times {{10}^{ - 3}}} \right) \times \left( {0.35} \right) \cr & = 7.8 \times {10^{ - 4}}\,atm\,{s^{ - 1}} \cr} $$