Question

The decomposition of ${Cl}_2{O}_7$ at $400$ $K$ in the gas phase to ${Cl}_2$ and $O_2$ is a first order reaction.
$\left(i\right)$ After $55$ seconds at $400$ $K$, the pressure of ${Cl}_2{O}_7$ falls from $0.062$ to $0.044atm$. Calculate the rate constant.
$\left(ii\right)$ Calculate the pressure of ${Cl}_2{O}_7$ after $100$ seconds of decomposition at this temperature.

Answer 1

$\left(i\right)$ As pressure $\propto$ concentration,
$k=\frac{2.303}{t}{\log }_{10}\frac{P_i\left(initialpressure\right)}{P_i\left(pressureaftertimet\right)}$
$=\frac{2.303}{55}{\log }_{10}\frac{0.062}{0.044}=6.2\times {10}^{-3}{s}^{-1}$
$\left(ii\right)$ Again applying the first order kinetic equation,
$k=\frac{2.303}{t}{\log }_{10}\frac{P_i\left(initialpressure\right)}{P_i\left(pressureaftertimet\right)}$
$6.2\times {10}^{-3}=\frac{2.303}{100}{\log }_{10}\frac{0.062}{P_i}$
$\frac{6.2\times {10}^{-3}\times 100}{2.303}={\log }_{10}0.062-{\log }_{10}\left(P_t\right)$
$0.2692={\log }_{10}0.062-{\log }_{10}\left(P_t\right)$
${\log }_{10}\left(P_t\right)={\log }_{10}0.062-0.2692$
$=(\overline{2}.7924-0.2692)$
$P_t=0.033$ atmosphere
Pressure after $100$ sec $=0.033$ atm