For a first order reaction at 27 c the ratio of time required for 75% completion to 25% completion of reaction is

Dear Student,

For a first order reaction, order = 1.

Now,

$$\begin{gathered} K = \frac{2.303}{t}\log \frac{A_0}{A_t}, where K = rate cons\tan t, t = time required \\ {A}_0= Initial concentration, A_t = Change in concentration \end{gathered}$$

Now,
Let t₁ be the time required for 75% completion and t₂ be the time required for 25% completion.
Let the initial concentration be 100.
Now,
For 75% completion,
$$\begin{gathered} A_0 = 100 , A_t= 100-75 = 25 \\ K = \frac{2.303}{t_1}\log \frac{100}{25} \\ {t}_1 = \frac{2.303}{K}\log 4 \end{gathered}$$
For 25% completion,
$$\begin{gathered} A_0= 100, A_t= 100-25=75 \\ {t}_2= \frac{2.303}{K}\log \frac{100}{75} \\ {t}_2= \frac{2.303}{K}\log \frac{4}{3} \end{gathered}$$
Dividing t₁ by t₂, we get:
$\frac{t_1}{t_2}= \frac{\log {\displaystyle 4}}{\log {\displaystyle \frac{4}{3}}}= \frac{0.602}{0.124}= 4.81$
Thus, the ratio of time required for 75% completion to 25% completion is 4.81.

Hope it helps.

Regards