Question

A molecule undergoes two independent first order reactions whose respective half lives are 12 min and 3 min. If both the reactions are occurring then the time taken for the 50% consumption of the reactant is _____ min. (Nearest integer)

Hint:

Always combine the reciprocal contributions of individual half-lives to find the effective rate.

Answer 1

Correct Answer: 2

To solve the problem of determining the time taken for 50% consumption of a reactant undergoing two independent first-order reactions, we first need to understand the concept of effective half-life in such a scenario. Given the half-lives of two reactions, \(t_{1/2,1} = 12\) min and \(t_{1/2,2} = 3\) min, we calculate the effective rate constant \(k_{\text{eff}}\) as follows:

For a first-order reaction, the rate constant \(k\) is given by \(k = \frac{\ln 2}{t_{1/2}}\). Thus, for the two reactions, the rate constants are:
\(k_1 = \frac{\ln 2}{12}\)
\(k_2 = \frac{\ln 2}{3}\)
The effective rate constant \(k_{\text{eff}} = k_1 + k_2\). Therefore:
\(k_{\text{eff}} = \frac{\ln 2}{12} + \frac{\ln 2}{3} = \ln 2 \left(\frac{1}{12} + \frac{1}{3}\right)\)
Calculating the combined fraction in the parentheses:
\(\frac{1}{12} + \frac{1}{3} = \frac{1}{12} + \frac{4}{12} = \frac{5}{12}\)
Thus, \(k_{\text{eff}} = \ln 2 \times \frac{5}{12}\).

The effective half-life \(t_{1/2,\text{eff}}\) is then given by:
\(t_{1/2,\text{eff}} = \frac{\ln 2}{k_{\text{eff}}} = \frac{\ln 2}{\ln 2 \times \frac{5}{12}} = \frac{12}{5} = 2.4\) min
The nearest integer to 2.4 min is 2 min, placing it within the expected range of 2 to 2. Thus, the time taken for 50% consumption of the reactant is 2 min.