Question

For a first order reaction, the time required for completion of 90% reaction is ‘x’ times the half life of the reaction. The value of ‘x’ is (Given: In 10 = 2.303 and log 2 = 0.3010)

• A. 1.12
• B. 2.43
• C. 3.32
• D. 33.31

Answer 1

The Correct Option is C

To solve this problem, we need to understand the kinetics of a first-order reaction. The key aspect of a first-order reaction is that the time required to complete any specific fraction of the reaction solely depends on the rate constant and the specific fraction.

The formula for the half-life t_{1/2} of a first-order reaction is:

t_{1/2} = \frac{0.693}{k}

The formula for the time required to complete a certain percentage of the reaction, t\right), is given as:

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

For 90% completion of the reaction:

\frac{[A]}{[A]_0} = 0.1 \rightarrow \log \left(\frac{[A]_0}{[A]}\right) = \log 10 = 1

Hence, the time required for 90% completion t_{90\%} is:

t_{90\%} = \frac{2.303}{k} \cdot 1 = \frac{2.303}{k}

To find the ratio x of the time required for 90% completion to the half-life, we divide t_{90\%} by t_{1/2}:

x = \frac{t_{90\%}}{t_{1/2}} = \frac{\frac{2.303}{k}}{\frac{0.693}{k}} = \frac{2.303}{0.693}

Calculating the above expression:

x = \frac{2.303}{0.693} \approx 3.32

Thus, the time required for 90% reaction completion is approximately 3.32 times the half-life of the reaction. Hence, the correct answer is 3.32.