Given the rate law rate = 1301 min^{-1} [A] for a particular reaction, what is the half-life (t_{1/2}) for this reaction?

• A. t_{1/2} = rac{1}{1301ext{ min}^{-1} [A]_0}
• B. t_{1/2} = rac{ext{ln} 2}{1301ext{ min}^{-1}}
• C. t_{1/2} = 1301ext{ min}
• D. t_{1/2} = rac{0.693}{[A]_0}

Answer: (B)

Verified step by step guidance

1. Identify the order of the reaction from the given rate law. Since the rate law is rate = 1301 min^{-1} [A], it is a first-order reaction because the rate depends linearly on the concentration of A.
2. Recall the general formula for the half-life of a first-order reaction, which is independent of the initial concentration and given by: \(t_{1/2} = \frac{\ln 2}{k}\), where \(k\) is the rate constant.
3. From the rate law, recognize that the rate constant \(k\) is 1301 min^{-1}. This is the coefficient multiplying the concentration term in the rate law.
4. Substitute the value of \(k\) into the half-life formula: \(t_{1/2} = \frac{\ln 2}{1301 \text{ min}^{-1}}\).
5. Interpret the result: since the half-life depends only on \(k\) for a first-order reaction, the half-life is a constant value and does not depend on the initial concentration \([A]_0\).