1. Given data

   • Reaction is first order.
   • Half-life: $t_{1/2} = 30\,\text{min}$.
   • We need the time for 75% completion.

2. Interpret 75% completion If 75% of $A$ has reacted, then 25% remains. So, $[A]_t = 0.25[A]_0 = \frac{[A]_0}{4}$

3. First-order integrated rate law For a first-order reaction, $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$ Substituting $[A]_t = [A]_0/4$, $t = \frac{2.303}{k} \log 4$

4. Use half-life relation for first order reaction For a first-order reaction, $t_{1/2} = \frac{0.693}{k}$ Given $t_{1/2}=30$ min, $\frac{0.693}{k} = 30$

   Instead of finding $k$ separately, note that reducing from $[A]_0$ to $[A]_0/4$ means:

   • one half-life takes concentration to $[A]_0/2$
   • second half-life takes concentration to $[A]_0/4$

   Therefore, time required is simply 2 half-lives: $t = 2 \times 30 = 60\,\text{min}$

5. Verification using logarithms Since, $\log 4 = \log(2^2)=2\log 2 = 2(0.3010)=0.6020$ and $t = \frac{2.303}{k}(0.6020)$ while $30 = \frac{0.693}{k} = \frac{2.303\times 0.3010}{k}$ So, $t = 2\times 30 = 60\,\text{min}$

6. Final answer $\boxed{60}$

The derived answer matches the stored correct answer.