KCET · Chemistry · Chemical Kinetics
If the rate constant for a first order reaction is \(k\), the time \((t)\) required for the completion of \(99 \%\) of the reaction is given by
- A \(t=\frac{4.606}{k}\)
- B \(t=\frac{2303}{k}\)
- C \(t=\frac{0.693}{k}\)
- D \(t=\frac{6.909}{k}\)
Answer & Solution
Correct Answer
(A) \(t=\frac{4.606}{k}\)
Step-by-step Solution
Detailed explanation
Given, for a first order reaction, let initial concentration \(=\left[R_{0}\right]\)
After \(99 \%\) completion, concentration
\(\begin{aligned}
{[R] } &=R_{0}-0.99 R_{0} \\
&=0.1 R_{0}
\end{aligned}\)
Using formula, \(t=\frac{2303}{k} \log \frac{\left[R_{0}\right]}{[R]}\)
\(=\frac{2303}{k} \log \frac{100}{1}=\frac{4.606}{k}\)
After \(99 \%\) completion, concentration
\(\begin{aligned}
{[R] } &=R_{0}-0.99 R_{0} \\
&=0.1 R_{0}
\end{aligned}\)
Using formula, \(t=\frac{2303}{k} \log \frac{\left[R_{0}\right]}{[R]}\)
\(=\frac{2303}{k} \log \frac{100}{1}=\frac{4.606}{k}\)
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