Relation Between Completion Time and Half Life for Reactions

To determine the relation between $t_{100\%}$ and $t_{1/2}$ for zero-order and first-order reactions, we analyze each kinetics model individually.

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1. Zero-Order Reaction

For a zero-order reaction, the integrated rate law is given by: $[A]_t = [A]_0 - kt$

where $[A]_0$ is the initial concentration of the reactant, $[A]_t$ is the concentration at time $t$, and $k$ is the rate constant.

• Half-life ($t_{1/2}$): The time required for $50\%$ of the reaction to complete occurs when $[A]_t = \frac{[A]_0}{2}$. $\frac{[A]_0}{2} = [A]_0 - k t_{1/2}$ $k t_{1/2} = \frac{[A]_0}{2} \implies t_{1/2} = \frac{[A]_0}{2k}$

• Completion time ($t_{100\%}$): The time required for $100\%$ completion of the reaction occurs when $[A]_t = 0$. $0 = [A]_0 - k t_{100\%}$ $t_{100\%} = \frac{[A]_0}{k}$

• Relationship: $t_{100\%} = 2 \left(\frac{[A]_0}{2k}\right) = 2t_{1/2}$

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2. First-Order Reaction

For a first-order reaction, the integrated rate law is given by: $[A]_t = [A]_0 e^{-kt} \quad \text{or} \quad t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_t}\right)$

• Half-life ($t_{1/2}$): The time required for $50\%$ completion occurs when $[A]_t = \frac{[A]_0}{2}$. $t_{1/2} = \frac{1}{k} \ln \left(\frac{[A]_0}{\frac{[A]_0}{2}}\right) = \frac{\ln 2}{k}$

• Completion time ($t_{100\%}$): The time required for $100\%$ completion occurs when $[A]_t \to 0$. $t_{100\%} = \lim_{[A]_t \to 0} \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_t}\right) = \infty$

A first-order reaction theoretically takes infinite time to reach $100\%$ completion, which is symbolically represented as: $t_{100\%} = (t_{1/2})^\infty$

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Conclusion

For a zero-order reaction: $t_{100\%} = 2t_{1/2}$ For a first-order reaction: $t_{100\%} = (t_{1/2})^\infty$

Thus, the correct option is B.