Question

The reaction $S{O}_2{Cl}_2\to S{O}_2+{Cl}_2$, is a first order gas reaction with $k=2.2\times {10}^{-5}{s}^{-1}$ at $573$ $K$. What percent of $S{O}_2{Cl}_2$ is decomposed on heating at $573$ $K$ for $90$ $min$?

Answer 1

The reaction $S{O}_{2}C{l}_2\to S{O}_2+C{l}_2$, is a first order gas reaction with $k=2.2\times {10}^{-5}{s}^{-1}$ at 573 K.

Thus by using first order reaction ,
$k=\frac{2.303}{t}\log \frac{[A{]}_0}{\left[A\right]}$

$\log \frac{100}{\left[A\right]}=\frac{2.2\times {10}^{-5}\times 60\times 90}{2.303}$

$\log \frac{100}{\left[A\right]}=\frac{118.8\times {10}^{-3}}{2.303}$

$\log \frac{100}{\left[A\right]}=51.6\times {10}^{-3}=0.0516$

$\frac{100}{\left[A\right]}=1.12$

$\left[A\right]=89.3\%$