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Question: For a first order reaction \({{\text{t}}_{99\% }} = x \times {{\text{t}}_{90\% }}\). The value of ‘x...

For a first order reaction t99%=x×t90%{{\text{t}}_{99\% }} = x \times {{\text{t}}_{90\% }}. The value of ‘x’ will be-
A. 10
B. 6
C. 3
D. 2

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. Using the equation deduce the expression for rate constant at 99%99\% completion and 90%90\% completion.

Formula Used:
k=2.303tlog⁡[a]0[a]k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}

Complete step-by-step answer: 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 99%99\% 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 100−99=1100 - 99 = 1. Thus,
t99%=2.303klog⁡(100)(1){t_{99\% }} = \dfrac{{2.303}}{k}\log \dfrac{{\left( {100} \right)}}{{\left( 1 \right)}}
⇒t99%=2.303k×2\Rightarrow {t_{99\% }} = \dfrac{{2.303}}{k} \times 2 …… (1)
The expression when reaction is 90%90\% complete is as follows:
Let the initial concentration of the reactant be 100. When the reaction is 90%90\% complete the final concentration will be 100−90=10100 - 90 = 10. Thus,
t90%=2.303klog⁡(100)(10){t_{90\% }} = \dfrac{{2.303}}{k}\log \dfrac{{\left( {100} \right)}}{{\left( {10} \right)}}
⇒t90%=2.303k×1\Rightarrow {t_{90\% }} = \dfrac{{2.303}}{k} \times 1 …… (2)
Now, divide equation (1) by equation (2) as follows:
t99%t90%=21\dfrac{{{t_{99\% }}}}{{{t_{90\% }}}} = \dfrac{2}{1}
⇒t99%=2×t90%\Rightarrow {t_{99\% }} = 2 \times {t_{90\% }} …… (3)
We are given the equation,
t99%=x×t90%{{\text{t}}_{99\% }} = x \times {{\text{t}}_{90\% }} …… (4)
Comparing equation (3) and equation (4),
x=2x = 2
Thus, the value of ‘x’ is 2.

Thus, the correct option is (D) 2.

Note: The unit of rate constant for first order reaction is sec−1{\text{se}}{{\text{c}}^{ - 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.