Search any question & find its solution
Question:
Answered & Verified by Expert
If $60 \%$, of a first order reaction was completed in 60 minutes, $50 \%$ of the same reaction would be completed in approximately:
$$
(\log 4=0.60, \log 5=0.69)
$$
Options:
$$
(\log 4=0.60, \log 5=0.69)
$$
Solution:
1699 Upvotes
Verified Answer
The correct answer is:
45 minutes
For a first order reaction
$k=\frac{2.303}{t} \log \frac{a}{a-x}$
When $t=60$ and $x=60 \%$
$\begin{aligned}
k & =\frac{2.303}{60} \log \frac{100}{100-60} \\
& =\frac{2.303}{60} \log \frac{100}{40} \\
& =0.0153
\end{aligned}$
Now,
$\begin{aligned}
t^{1 / 2} & =\frac{2.303}{0.0153} \log \frac{100}{100-50} \\
& =\frac{2.303}{0.0153} \times \log 2 \\
& =\frac{2.303}{0.0153} \times 0.3010 \\
& =45.31 \mathrm{~min}
\end{aligned}$
$k=\frac{2.303}{t} \log \frac{a}{a-x}$
When $t=60$ and $x=60 \%$
$\begin{aligned}
k & =\frac{2.303}{60} \log \frac{100}{100-60} \\
& =\frac{2.303}{60} \log \frac{100}{40} \\
& =0.0153
\end{aligned}$
Now,
$\begin{aligned}
t^{1 / 2} & =\frac{2.303}{0.0153} \log \frac{100}{100-50} \\
& =\frac{2.303}{0.0153} \times \log 2 \\
& =\frac{2.303}{0.0153} \times 0.3010 \\
& =45.31 \mathrm{~min}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.