Question

$\text{Time required for completion of 99.9\% of a first order reaction is \_\_\_\_\_\_ times of half life } (t_{1/2}) \text{ of the reaction.}$

Answer 1

Correct Answer: 10

For a first-order reaction, the relationship between the concentration of reactant and time is given by the equation: $N=N_0e^{-kt}$, where \(N_0\) is the initial concentration, \(N\) is the concentration at time \(t\), and \(k\) is the rate constant. The half-life \(t_{1/2}\) for a first-order reaction is derived from: $N = \frac{N_0}{2}$ leading to \(\frac{N_0}{2}=N_0e^{-kt_{1/2}}\). Solving for \(t_{1/2}\), we get:

\(t_{1/2}=\frac{\ln(2)}{k}\) 

 

To find the time required for 99.9% completion, we use: $N=0.1\%$ of \(N_0\), thus: $N=0.001N_0$\(. Set \(0.001N_0 = N_0e^{-kt_{99.9\%}}\), which simplifies to:

\(-kt_{99.9\%}=\ln(0.001)\)

 

From \(\ln(0.001)=-6.90775\), we rearrange to find:

\(t_{99.9\%}=\frac{6.90775}{k}\)

 

Now, to express \(t_{99.9\%}\) in terms of \(t_{1/2}\), note: \(t_{99.9\%}=\frac{6.90775}{k}\) and \(t_{1/2}=\frac{0.693}{k}\). Divide to yield:

\(\frac{t_{99.9\%}}{t_{1/2}}=\frac{6.90775}{0.693}\approx 9.9658\)

 

A rounding gives \(t_{99.9\%} \approx 10 \times t_{1/2}\).

Answer 2

For a first-order reaction, the time required for a certain percentage of reaction completion can be calculated using the formula:

\(t = \frac{2.303}{k} \log \frac{[A]_0}{[A]},\)

where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.  

For 99.9% completion, \(\frac{[A]}{[A]_0} = 0.001\):  

\(t = \frac{2.303}{k} \log \frac{1}{0.001} = \frac{2.303}{k} \log(10^3) = \frac{2.303}{k} \times 3 = 10 \times t_{1/2}.\)

Thus, the time required for 99.9% completion is 10 times the half-life.
The Correct answer is: 10