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In a first-order reaction $A \rightarrow B$, if $k$ is rate constant and initial concentration of the reactant A is 0.5 M, then the half-life is
A
$\dfrac{\log 2}{k}$
B
$\dfrac{\log 2}{k\sqrt{0.5}}$
C
$\dfrac{\ln 2}{k}$
D
$\dfrac{0.693}{0.5k}$
Detailed Solution
For a first order reaction: $k = \dfrac{2.303}{t}\log_{10}\dfrac{a}{a - x}$
When $t = t_{1/2}$, $x = \dfrac{a}{2}$
$k = \dfrac{2.303}{t_{1/2}}\log_{10}\dfrac{a}{a - a/2}$
$k = \dfrac{2.303}{t_{1/2}}\log_{10}2$
$t_{1/2} = \dfrac{2.303}{k}\log_{10}2 = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}$
The initial concentration cancels out, so the half-life of a first order reaction does not depend on the initial concentration (0.5 M here).
So the half-life is $\dfrac{\ln 2}{k}$.
When $t = t_{1/2}$, $x = \dfrac{a}{2}$
$k = \dfrac{2.303}{t_{1/2}}\log_{10}\dfrac{a}{a - a/2}$
$k = \dfrac{2.303}{t_{1/2}}\log_{10}2$
$t_{1/2} = \dfrac{2.303}{k}\log_{10}2 = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}$
The initial concentration cancels out, so the half-life of a first order reaction does not depend on the initial concentration (0.5 M here).
So the half-life is $\dfrac{\ln 2}{k}$.