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Question: A second order reaction where \(a = b\) is \(20\% \) completed in \(500{\text{ seconds}}\). How long...

A second order reaction where a=ba = b is 20%20\% completed in 500 seconds500{\text{ seconds}}. How long will the reaction take to be 60%60\% complete?

Explanation

Solution

To solve this question we must know the equation for the rate constant of second order reaction. A reaction in which the rate of the reaction is directly proportional to the concentration of each of the two reactant species is known as the second order reaction. Using the equation deduce the expression for rate constant at 20%20\% completion and 60%60\% completion.

Complete answer:
We know the equation for the rate constant of a second order reaction is,
k=1txa(a−x)k = \dfrac{1}{t}\dfrac{x}{{a\left( {a - x} \right)}}
Where kk is the rate constant of a second order reaction,
tt is time,
aa is the initial concentration of the reactant,
xx is the final concentration of the reactant.
The expression when reaction is 20%20\% complete is as follows:
Let the initial concentration of the reactant be 100. When the reaction is 20%20\% complete the final concentration will be 20. Thus,
k=1500 seconds×20100(100−20)k = \dfrac{1}{{500{\text{ seconds}}}} \times \dfrac{{20}}{{100\left( {100 - 20} \right)}} …… (1)
k=5×10−6 M−2 s−1k = 5 \times {10^{ - 6}}{\text{ }}{{\text{M}}^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}}
Thus, the rate constant when the reaction is 20%20\% complete is 5×10−6 M−2 s−15 \times {10^{ - 6}}{\text{ }}{{\text{M}}^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}}.
The expression when reaction is 60%60\% complete is as follows:
Let the initial concentration of the reactant be 100. When the reaction is 60%60\% complete the final concentration will be 60. Thus,
k=1t×60100(100−60)k = \dfrac{1}{t} \times \dfrac{{60}}{{100\left( {100 - 60} \right)}} …… (2)
Rearrange the equation for time as follows:
t=1k×60100(100−60)t = \dfrac{1}{k} \times \dfrac{{60}}{{100\left( {100 - 60} \right)}}
Substitute 5×10−6 M−2 s−15 \times {10^{ - 6}}{\text{ }}{{\text{M}}^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}} for the rate constant. Thus,
t=15×10−6 M−2 s−1×60100(100−60)t = \dfrac{1}{{5 \times {{10}^{ - 6}}{\text{ }}{{\text{M}}^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}}}} \times \dfrac{{60}}{{100\left( {100 - 60} \right)}}
t=3000 secondst = 3000{\text{ seconds}}

Thus, the reaction will be 60%60\% complete in 3000 seconds3000{\text{ seconds}}.

Note: The unit of rate constant for second order reaction is M−2 s−1{{\text{M}}^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}}. The unit contains concentration terms. Thus, we can say that the rate constant of a second order reaction is dependent on the concentration of the reactant.