Question

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

Answer 1

Hint: 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\% $ completion and $90\% $ completion.

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

Complete step-by-step answer:We know the equation for the rate constant of a first order reaction is,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}$
Where $k$ is the rate constant of a first order reaction,
$t$ is time,
${\left[ a \right]^0}$ is the initial concentration of the reactant,
$\left[ a \right]$ is the final concentration of the reactant.
Rearrange the equation for the time as follows:
\[t = \dfrac{{2.303}}{k}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}\]
The expression when reaction is $99\% $ complete is as follows:
Let the initial concentration of the reactant be 100. When the reaction is $99\% $ complete the final concentration will be $100 - 99 = 1$. Thus,
\[{t_{99\% }} = \dfrac{{2.303}}{k}\log \dfrac{{\left( {100} \right)}}{{\left( 1 \right)}}\]
\[\Rightarrow {t_{99\% }} = \dfrac{{2.303}}{k} \times 2\] …… (1)
The expression when reaction is $90\% $ complete is as follows:
Let the initial concentration of the reactant be 100. When the reaction is $90\% $ complete the final concentration will be $100 - 90 = 10$. Thus,
\[{t_{90\% }} = \dfrac{{2.303}}{k}\log \dfrac{{\left( {100} \right)}}{{\left( {10} \right)}}\]
\[\Rightarrow {t_{90\% }} = \dfrac{{2.303}}{k} \times 1\] …… (2)
Now, divide equation (1) by equation (2) as follows:
\[\dfrac{{{t_{99\% }}}}{{{t_{90\% }}}} = \dfrac{2}{1}\]
\[\Rightarrow {t_{99\% }} = 2 \times {t_{90\% }}\] …… (3)
We are given the equation,
${{\text{t}}_{99\% }} = x \times {{\text{t}}_{90\% }}$ …… (4)
Comparing equation (3) and equation (4),
$x = 2$
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 ${\text{se}}{{\text{c}}^{ - 1}}$. 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.