Question

For a first-order reaction, prove that the time taken in completion of 3/4 part of the reaction is two times that of its half-life period.

Hint:

For a first-order reaction, the fraction of reactant remaining can be analyzed using the integrated rate equation \( t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \).

Answer 1

Step 1: The integrated rate law for a first-order reaction is: \[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \] Step 2: For half-life, at \( t = t_{1/2} \), \( [A] = \frac{[A]_0}{2} \): \[ t_{1/2} = \frac{0.693}{k} \] Step 3: When 3/4 of the reaction is complete, \( [A] = \frac{[A]_0}{4} \): \[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0/4} \] \[ t = \frac{2.303}{k} \log 4 \] \[ t = \frac{2.303}{k} \times 0.602 \] \[ t = 2 t_{1/2} \] Thus, the time taken for 3/4 completion is twice the half-life.