If the 60% of first order reaction was completed in 60 minutes, 50 % of the same reaction would be completed in approximately (log 4 = 0.6, log 5 =0.69)
1- 45 mins
2- 60 mins
3-40 mins
4- 50 mins.
Answers
The answer is 45 minutes.
The rate of a general first-order Reaction is:
-kt = 2.303 log Ao/At
where k is the rate of the reaction
t is the time taken for the partial reaction
Ao is the initial concentration of reactant
Is the concentration of reactant left
Now, it's given that 60% of the reaction is completed in 60minutes.
i.e.
-k = (2.303/60mins) log (A/ A- 0.6)
= (2.303/ 60 mins) log (1/ 0.4)
=> k = (2.303/60mins) log 0.4
(as minus sign went on the other side and -log 1/x = log x)
=> k = 0.0153/ min
Now, we have to find the approx time when 50% of the same reaction will be complete i.e. t 1/2 (half time of the reaction.
We know that formula for the half life of any first order reaction is
t(1/2) = 0.693/ k
=> t (1/2) = 0.693/ (0.0153 per minute)
=> t (1/2) = 45.3 minutes
i.e ~45minutes
Hence, the answer is 45 minutes.