If 75% of a first order reaction was completed in 90 minutes,

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 : $\log 2=0.30 ; \log 2.5=0.40$ )

Solution:

$\mathrm{t}_{0.75}=2 \times \frac{\ln 2}{\mathrm{k}}=90$

$\mathrm{k}=\frac{\ln 2}{45} \mathrm{~min}^{-1}$

$\mathrm{kt}=\ln \frac{1}{1-0.6}=\ln 2.5$

$\frac{\ln 2}{45} \times \mathrm{t}=\ln 2.5$

$\mathrm{t}=45 \times \frac{\log 2.5}{\log 2}=45 \times \frac{0.4}{0.3}=60 \mathrm{~min}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now