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Q. For certain first order reaction m, 75% of the reaction complete in 30 min. How much time it require to complete 99.9% of the reaction?

KCETKCET 2001Chemical Kinetics

Solution:

Its a 1st order reaction, So

$\frac{t_{x \%}}{t_{y \%}}=\frac{\log \frac{100}{100-x}}{\log \frac{100}{100-y}}$

Here $x \%=75 \%$ and $y \%=99.9 \%$

Substituting the values, We get

$\frac{30}{t_{y} \%}=\frac{\log \frac{100}{100-75}}{\log \frac{100}{100-99.9}}$

$t _{ y \%}=\frac{30}{0.2}=150 min$