Question

Time required to decompose $S{O}_{2}C{l}_2$ to half of its initial amount is $60$ minutes. If the decomposition is a first order reaction,calculate the rate constant of the reaction.

Answer 1

Given, $t_{\frac{1}{2}}=60$ minutes
For a first order reaction , $t_{\frac{1}{2}}=\frac{0.693}{k}$
$\Rightarrow k=\frac{0.693}{t_{\frac{1}{2}}}$
Putting the value of $t_{\frac{1}{2}}$ , we get,
$k=\frac{0.693}{60}=1.155\times {10}^{-2}{\text{min}}^{-1}$

Final answer:

The rate constant of the reaction is
$1.155\times {10}^{-2}{\text{min}}^{-1}$.