Chemistry - Chemical Kinetics Question with Solution | TestHub

$\mathrm{S}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_{2\left(\mathrm{g}\right)}\to \mathrm{S}{\mathrm{O}}_{2\left(\mathrm{g}\right)}+\mathrm{C}{\mathrm{l}}_{2\left(\mathrm{g}\right)}$
$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{p}}_0}{2{\mathrm{p}}_0-{\mathrm{P}}_{\mathrm{t}}}$
$=\frac{2.303}{100}\log \frac{0.5}{2\times 0.5-0.6}=\frac{2.303}{100}\log 1.25$
$=2.2318\times 10^{-3}{\mathrm{s}}^{-1}$
Pressure of$\mathrm{S}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_2$at time t$\left({\mathrm{p}}_{\mathrm{S}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_2}\right)$
$=2{\mathrm{p}}_0-{\mathrm{P}}_{\mathrm{t}}=2\times 0.50-0.65$atm
$=0.35$atm
Rate of that time$=\mathrm{k}\times {\mathrm{p}}_{\mathrm{S}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_2}$
$=\left(2.2318\times 10^{-3}\right)\times \left(0.35\right)=7.8\times 10^{-4}$atm${\mathrm{s}}^{-1}$