Question

Consider the following two competing first order reactions:
$P\xrightarrow{{{K}_{1}}}A+B$
$Q\xrightarrow{{{K}_{2}}}C+D$
If $50\% $ of the reaction of P was completed when $96\% $ of Q was completed, then the ratio $\left( {\frac{{{K_2}}}{{{K_1}}}} \right)$ will be:
A) $4.6$
B) $4.06$
C) $1.123$
D) $2.303$

Answer 1

Hint: The first order reactions are the reactions in which the rate constant of the reaction depends only on the concentration of the single reactant. The time for two reactions in which the reactants are consumed at different concentrations are equal, which means we can compare the time required for both the reactions to find the ratio of the rate constants.

Formula used: Half-life period for first order reaction ${t_{\frac{1}{2}}} = \frac{{0.693}}{K}$
Rate constant for first order reaction: $t = \frac{{2.303}}{K}\log \frac{{{I_0}}}{{{I_i}}}$

Complete Step by Step Solution:
Consider the original concentration of the reactant for both the reactions to be 100. As the reactions were first order reactions and the time for consumption of $50\% $ P is equal to the time consumed for consumption of $96\% $ Q.

And we know that, time for the consumption of half of the reactant can be calculated as follows:
${t_{\frac{1}{2}}} = \frac{{0.693}}{K}$

Also, for the first order reactions, the relationship between time and rate constant as follows:
$t = \frac{{2.303}}{K}\log \frac{{{I_0}}}{{{I_i}}}$
As per the conditions given in the question, we can say that ${t_{P\left( {50\% } \right)}} = {t_{Q(96\% )}}$

Substituting values as per equation (1) and (2):
$\frac{{0.693}}{{{K_1}}} = \frac{{2.303}}{{{K_2}}}\log \frac{{100}}{4}$
$ \Rightarrow \frac{{0.693}}{{{K_1}}} = \frac{{2.303}}{{{K_2}}}\log 25$
$ \Rightarrow \frac{{{K_2}}}{{{K_1}}} = \frac{{2.303 \times 2\log 5}}{{0.693}}$
$ \Rightarrow \frac{{{K_2}}}{{{K_1}}} = \frac{{3.219}}{{0.693}}$
$ \Rightarrow \frac{{{K_2}}}{{{K_1}}} = 4.64$

Thus, the ratio of rate constant for the second to the first reaction i.e., $\frac{{{K_2}}}{{{K_1}}}$ is $4.6$. So, the correct answer is option (A).

Note: It is important to note that while putting the values in the equation of rate constant, we have to substitute the value of the concentration remaining after completion of reaction, not the amount of the reactant consumed. Therefore, we have taken the concentration ${I_i}$ as 4 because 96% of the reactant was consumed in reaction 2.