Question

The half life of a first order reaction is $1.7$ hours. How long will it take for $20\%$ of the reactant to disappear?

Answer 1

$t_{1/2}=\frac{0.693}{k}k=\frac{0.693}{1.7}perhour=0.4076{hour}^{-1}Let{\left[A\right]}_{0}be100{\left[A\right]}_t=100-20=80k=\frac{2.303}{t}\log \frac{\left[A_o\right]}{\left[A_t\right]}{t}_{20\%}=\frac{2.303}{0.4076}\log \frac{100}{80}=\frac{2.303\times 0.0969}{0.4076}{t}_{20\%}=0.5476hours{t}_{20\%}=32.86minutes$