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For a FIRST-ORDER reaction with rate constant $k$, the half-life is given by
A$t_{1/2} = \dfrac{0.693}{k}$
B$t_{1/2} = \dfrac{1}{k[A]_0}$
C$t_{1/2} = \dfrac{[A]_0}{2k}$
D$t_{1/2} = k \cdot \ln 2$
Answer & Solution
Correct answer: A. $t_{1/2} = \dfrac{0.693}{k}$
1. Integrated first-order rate law: $[A] = [A]_0\,e^{-kt}$ (NCERT §3.3.2).
2. At the half-life, $[A] = [A]_0/2$ by definition.
3. Substitute: $[A]_0/2 = [A]_0\,e^{-k t_{1/2}}$, so $1/2 = e^{-k t_{1/2}}$.
4. Take the natural log: $\ln(1/2) = -k\,t_{1/2}$, i.e. $t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}$.
5. KEY FEATURE: first-order half-life is INDEPENDENT of initial concentration — a defining characteristic of first-order decay (also seen in radioactive decay).
6. Option B is the second-order half-life. Option C is the zero-order half-life. Option D inverts the formula (multiplied instead of divided).
_Source: NCERT Class 12 Chemistry Part 1, Ch 3, §3.3.2 (Integrated Rate Equation — first order half-life), p. 13–14._
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