Question

If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes)

(Take : log 2 = 0.30; log 2.5 = 0.40)

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Answer 1

For a first order reaction, $t=\frac{2.303}{k}lo{g}_{10}\frac{[A{]}_a}{[A{]}_t}=\frac{2.303}{k}lo{g}_{10}\frac{a}{a-x}=\frac{2.303}{k}lo{g}_{10}\frac{1}{1-f}$
Here f is the fraction of a reaction completed in time, t

$t_{75\%}=\frac{2.303}{k}lo{g}_{10}\frac{1}{1-0.75}=\frac{2.303}{k}lo{g}_{10}4t_{75\%}=\frac{2.303}{k}lo{g}_{10}{2}^{2}k=2\times \frac{2.303}{90}lo{g}_{10}2$
Also,
$t_{60\%}=\frac{2.303}{k}lo{g}_{10}\frac{1}{1-0.6}{t}_{60\%}=\frac{2.303}{k}lo{g}_{10}2.5t_{60\%}=\frac{2.303}{2\times \frac{2.303}{90}lo{g}_{10}2}lo{g}_{10}2.5$
$t_{60\%}=\frac{45\times 0.4}{0.3}t=60min$