Question

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=0.693/k
• B. t=6.909/k
• C. t=4.606/k
• D. t=2.303/k

Answer 1

The Correct Option is C

To find the time required for 99% completion of a first-order reaction, we start by understanding the key formula for first-order kinetics:

The integrated rate law for a first-order reaction is given by: 

\(\ln{\left(\frac{[A]_0}{[A]}\right)} = kt\)

where:

• \([A]_0\) is the initial concentration of the reactant.
• \([A]\) is the concentration of the reactant at time \(t\).
• \(k\) is the rate constant.
• \(t\) is the time.

For a reaction to reach 99% completion, 99% of the reactant has been consumed, leaving 1% remaining. Therefore, \(\frac{[A]}{[A]_0} = \frac{1}{100} = 0.01\).

Substitute these values into the integrated rate law:

\(\ln{\left(\frac{[A]_0}{0.01 \times [A]_0}\right)} = kt\)

This simplifies to:

\(\ln{(100)} = kt\)

We know that \(\ln{(100)} = 4.6052\) (approx.).

Thus:

\(kt = 4.606\)

Therefore, the time \(t\) required for 99% completion is:

\(t = \frac{4.606}{k}\)

Hence, the correct answer is

t=4.606/k

. This is consistent with the first-order kinetics formula where to reach a specific completion percentage, the time is calculated using a multiple of \(k\) that corresponds to the percentage conversion.