A molecule undergoes two independent first-order reactions whose respective half-lives are 12 min and 3 min. If both rea

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A molecule undergoes two independent first-order reactions whose respective half-lives are 12 min and 3 min. If both reactions are occurring, the time taken for 50 consumption of the reactant is:

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Given: The molecule undergoes two independent first-order reactions with half-lives $12 \, \text{min}$ and $3 \, \text{min}$ .

Find: The time taken for $50\%$ consumption of the reactant when both reactions occur simultaneously.

For independent first-order reactions, the effective rate constant is

k_{\text{eff}} = k_1 + k_2

The effective half-life is therefore related to the individual half-lives by

\frac{1}{t_{\text{eff}}} = \frac{1}{t_1} + \frac{1}{t_2}

Substitute $t_1 = 12 \, \text{min}$ and $t_2 = 3 \, \text{min}$ :

\frac{1}{t_{\text{eff}}} = \frac{1}{12} + \frac{1}{3}

Using a common denominator,

\frac{1}{t_{\text{eff}}} = \frac{1}{12} + \frac{4}{12} = \frac{5}{12}

Hence,

t_{\text{eff}} = \frac{12}{5} = 2.4 \, \text{min}

Rounding to the nearest integer gives

t_{\text{eff}} = 2 \, \text{min}

Therefore, the time taken for $50\%$ consumption of the reactant is $2 \, \text{min}$ .

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