Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, the time required is given by the integrated rate equation:

$t=\frac{2.303}{k}\log \left(\frac{a}{a-x}\right)$

Given $k=2.2\times 10^{-5}{\text{ s}}^{-1}$. For $15\%$ completion, the remaining concentration is $100-15=85\%$.

$t=\frac{2.303}{2.2\times 10^{-5}}\log \left(\frac{100}{85}\right)$

$t=\frac{2.303}{2.2\times 10^{-5}}(\log 20-\log 17)$

Substituting the values $\log 20\approx 1.3010$ and $\log 17\approx 1.2304$:

$t=\frac{2.303}{2.2\times 10^{-5}}(1.3010-1.2304)$

$t=\frac{2.303\times 0.0706}{2.2\times 10^{-5}}$

$t=\frac{0.16259}{2.2\times 10^{-5}}$

$t\approx 7391\text{ s}$

Answer: $7391$ S