Calculation for the time taken for a first order reaction to complete to 75% (using log table)- Chemical Kinetics

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,

= 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.  

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