For a first-order reaction, the time $t$ required for a certain extent of reaction is given by: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$

For 60% completion in 60 minutes: $[A]_t = 100\% - 60\% = 40\%$ $60 = \frac{2.303}{k} \log \left(\frac{100}{40}\right) = \frac{2.303}{k} \log(2.5)$ We know $\log(2.5) = \log\left(\frac{10}{4}\right) = \log 10 - \log 4 = 1 - 0.60 = 0.40$ So, $60 = \frac{2.303}{k} \times 0.40$ — (Equation 1)

For 50% completion ($t_{50\%}$): $t_{50\%} = \frac{2.303}{k} \log \left(\frac{100}{50}\right) = \frac{2.303}{k} \log(2)$ We know $\log 4 = 2 \log 2 = 0.60 \implies \log 2 = 0.30$ So, $t_{50\%} = \frac{2.303}{k} \times 0.30$ — (Equation 2)

Dividing Equation 2 by Equation 1: $\frac{t_{50\%}}{60} = \frac{0.30}{0.40} = \frac{3}{4}$ $t_{50\%} = 60 \times \frac{3}{4} = 45 \text{ min}$