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Question: Time for half change for a first order reaction is \(25{\text{ min}}\). What will be the time requir...

Time for half change for a first order reaction is 25 min25{\text{ min}}. What will be the time required for 99%99\% completion of reaction.

Explanation

Solution

To solve this question we must know the equation for the rate constant of first order reaction. A reaction in which the rate of the reaction is directly proportional to the concentration of one of the reactant species is known as the first order reaction. First calculate the rate constant using the equation of half-life of first order reaction. Then calculate the rate constant at 99%99\% completion.

Complete solution:
We know that the expression for half life of a first order reaction is as follows:
t1/2=0.693k{t_{{\text{1/2}}}} = \dfrac{{0.693}}{k}
Where t1/2{t_{{\text{1/2}}}} is the half-life time for the reaction,
kk is the rate constant for the reaction.
We are given that the time for half change for a first order reaction is 25 min25{\text{ min}}. Thus,
25 min=0.693k25{\text{ min}} = \dfrac{{0.693}}{k}
k=0.69325 mink = \dfrac{{0.693}}{{25{\text{ min}}}}
k=0.02772 min1k = 0.02772{\text{ mi}}{{\text{n}}^{ - 1}}
Thus, the rate constant for the reaction is 0.02772 min10.02772{\text{ mi}}{{\text{n}}^{ - 1}}.
We know the equation for the rate constant of a first order reaction is,
k=2.303tlog[a]0[a]k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}
Where kk is the rate constant of a first order reaction,
tt is time,
[a]0{\left[ a \right]^0} is the initial concentration of the reactant,
[a]\left[ a \right] is the final concentration of the reactant.
Rearrange the equation for the time as follows:
t=2.303klog[a]0[a]t = \dfrac{{2.303}}{k}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}
The expression when reaction is 90%90\% complete is as follows:
Let the initial concentration of the reactant be 100. When the reaction is 99%99\% complete the final concentration will be 10099=1100 - 99 = 1. Thus,
t99%=2.3030.02772 min1log(100)(1){t_{99\% }} = \dfrac{{2.303}}{{0.02772{\text{ mi}}{{\text{n}}^{ - 1}}}}\log \dfrac{{\left( {100} \right)}}{{\left( 1 \right)}}
t99%=2.3030.02772 min1×2{t_{99\% }} = \dfrac{{2.303}}{{0.02772{\text{ mi}}{{\text{n}}^{ - 1}}}} \times 2
t99%=166.16 min{t_{99\% }} = 166.16{\text{ min}}

Thus, the time required for 99%99\% completion of reaction is 166.16 min166.16{\text{ min}}.

Note: The unit of rate constant for first order reaction is min1{\text{mi}}{{\text{n}}^{ - 1}}. The units do not contain concentration terms. Thus, we can say that the rate constant of a first order reaction is independent of the concentration of the reactant.