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
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.