KCET · Chemistry · Chemical Kinetics
A first order reaction is \(60 \%\) complete in 20 minutes. How long will the reaction take to be \(84 \%\) complete?
- A \(68 \mathrm{~min}\)
- B \(40 \mathrm{~min}\)
- C \(76 \mathrm{~min}\)
- D \(54 \mathrm{~min}\)
Answer & Solution
Correct Answer
(B) \(40 \mathrm{~min}\)
Step-by-step Solution
Detailed explanation
For a first order reaction,
\[
k=\frac{2.303}{t} \log \frac{a}{a-x}
\]
I Case
\(k=\frac{2.303}{20} \log \frac{100}{100-60}\)
\[
=\frac{2.303}{20} \log \frac{100}{40}
\]
\[
\begin{aligned}
&=\frac{2.303}{20}(\log 10-\log 4) \\
&=\frac{2.303}{20}(1-0.6020)=0.046
\end{aligned}
\]
II Case
\[
\begin{aligned}
k &=\frac{2.303}{t} \log \frac{100}{100-84} \\
t=& \frac{2.303}{0.046} \log \frac{100}{16} \\
&=\frac{2.303}{0.046}(\log 100-\log 16) \\
&=\frac{2.303}{0.046}(2-1.2041) \\
&=39.84 \mathrm{~min}=40 \mathrm{~min}
\end{aligned}
\]
\[
k=\frac{2.303}{t} \log \frac{a}{a-x}
\]
I Case
\(k=\frac{2.303}{20} \log \frac{100}{100-60}\)
\[
=\frac{2.303}{20} \log \frac{100}{40}
\]
\[
\begin{aligned}
&=\frac{2.303}{20}(\log 10-\log 4) \\
&=\frac{2.303}{20}(1-0.6020)=0.046
\end{aligned}
\]
II Case
\[
\begin{aligned}
k &=\frac{2.303}{t} \log \frac{100}{100-84} \\
t=& \frac{2.303}{0.046} \log \frac{100}{16} \\
&=\frac{2.303}{0.046}(\log 100-\log 16) \\
&=\frac{2.303}{0.046}(2-1.2041) \\
&=39.84 \mathrm{~min}=40 \mathrm{~min}
\end{aligned}
\]
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