show that for first order reaction t87.5%=3t50%

Dear Student,

For a first order reaction, we know that : 

$$\begin{gathered} \ln \frac{\left[A\right]}{\left[A_o\right]} = -kt \\ \left(a\right)When 87.5\% of initial reac\tan t is converted, \\ remaining \left[A\right] = 1 - 0.875 = 0.125 \left[A_o\right] \\ Thus, \\ \ln \frac{0.125\left[A_o\right]}{\left[A_o\right]}=-kt \\ -kt = -2.079 \\ {t}_{87.5}=\frac{2.079}{k} --\overline{)1} \\ \left(b\right) When 50\% of initial reac\tan t is converted, \\ remaining \left[A\right] = 1 - 0.5 = 0.5\left[A_o\right] \\ Thus, \\ \ln \frac{0.5\left[A_o\right]}{\left[A_o\right]}=-kt \\ -kt = -0.693 \\ {t}_{50}=\frac{0.693}{k} \\ Multiplying this by 3, \\ we get 3t_{50} = \frac{0.693}{k} \times 3 = \frac{2.079}{k} --\overline{)2} \\ Thus, \overline{)1}=\overline{)2} \\ So t_{87.5} = 3t_{50} \end{gathered}$$

Hope it helps.

Regards