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. \( t = \frac{4.606}{k} \)
• B. \( t = \frac{2.303}{k} \)
• C. \( t = \frac{0.693}{k} \)
• D. \( t = \frac{6.909}{k} \)

Hint:

In first-order reactions, half-life remains constant and depends only on the rate constant.

Answer 1

The Correct Option is A

For a first-order reaction, the fraction remaining after time \( t \) is given by: \[ \ln \left( \frac{[A]_0}{[A]} \right) = k t \] For 99% completion (\( [A] = 0.01[A]_0 \)): \[ t = \frac{\ln(100)}{k} = \frac{4.606}{k} \] Thus, the correct formula for 99% completion in first-order kinetics is: \[ t = \frac{4.606}{k} \]