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For a first-order reaction, the half-life period $t_{1/2}$ is related to the rate constant $k$ as:
A$t_{1/2} = \dfrac{2.303}{k}$
B$t_{1/2} = \dfrac{0.693}{k}$
C$t_{1/2} = \dfrac{[A]_0}{2k}$
D$t_{1/2} = \dfrac{1}{k[A]_0}$
Answer & Solution
Correct answer: B. $t_{1/2} = \dfrac{0.693}{k}$
For a first-order reaction $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$. Setting $[A] = [A]_0/2$ at $t = t_{1/2}$ gives $k t_{1/2} = \ln 2 = 0.693$, hence $t_{1/2} = 0.693/k$. Option B is the zero-order expression; option C is the second-order one.
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