A first order reaction is 40% complete in 20 minutes. Calculate the time required for 90%…

Formula for first-order reaction: $k = \frac{2\cdot303}{t} \log\frac{[\text{A}]_0}{[\text{A}]}$ For $40\%$ completion in $t = 20\text{ minutes}$: $[\text{A}]_0 = 100, [\text{A}] = 100 - 40 = 60$ $k = \frac{2\cdot303}{20\text{ min}} \log\left(\frac{100}{60}\right) = \frac{2\cdot303}{20} \times 0\cdot2219 = 0\cdot02555\text{ min}^{-1}$ For $90\%$ completion: $[\text{A}] = 100 - 90 = 10$ $t_{90\%} = \frac{2\cdot303}{k} \log\left(\frac{100}{10}\right) = \frac{2\cdot303}{0\cdot02555\text{ min}^{-1}} \times 1 = 90\cdot15\text{ minutes}$