**A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. Given: log 2= 0.3010 log 3= 0.4771 log 4=0.6021 (CBSE 2017)**

Solution:

From the first statement, the information we obtain is that the reaction is of the first order, time is given, and we should first find out rate constant (k).

Therefore,

k= 2.303/t . log (a/a-x) **(i)**

where a is the initial concentration and (a-x) is the remaining concentration of the starting material.

If 25% decomposed, that is ¼ of a then remaining (a-x) would be ¾ of a.

Using this information in equation **(i)**, we get,

k = 2.303/20 . log (a/¾ a) = 2.303/20. log(4/3)= 2.303/20 . (log 4 - log 3)

= 2.303/20 x (0.6021-0.4771)

**= (2.303 x 0.1250)/20 min ^{-1}**

Similarly, for 75% (¾ of a) of the starting material to decompose, remaining starting material will be ¼ of a.

Therefore, the time taken will be,

t= 2.303/k . log (a/a-x) = 2.303/k . log (a/¼ a) = 2.303/k . log(4)

= 2.303/k . 0.6021

Using the value of k in the equation, we get,

**t** = (2.303 x 0.6021 x 20)/2.303 x 0.1250 = (0.6021 x 20)/ 0.1250 = **96.34 min**

**Using log to solve the equation,**

log(t) = log (6021 x 10^{-4}) + log 20 -log (1250 x 10^{-4})

= log 6021 + log 10^{-4} + log 20 – log 1250 – log 10^{-4} = 3.7797 -4 + 1.3010 – 3.0969 +4

= 3.7797-3.0969+ 1.3010 = 1.9838

**t **= antilog (1.9838) = **96.34 min**

**Therefore, the time when 75% of the reaction will be completed will be 96.34 min. **