The half-life of a first order reaction varie | vedclass

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(A) For a first order reaction,$t_{1/2} = \frac{0.693}{k}$.Taking natural logarithm on both sides:$\ln (t_{1/2}) = \ln(0.693) - \ln k$ $(i)$.From the

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$\ln (t_{1/2}) \propto \frac{1}{T}$

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(A) For a first order reaction,$t_{1/2} = \frac{0.693}{k}$.Taking natural logarithm on both sides:$\ln (t_{1/2}) = \ln(0.693) - \ln k$ $(i)$.From the Arrhenius equation,$k = A e^{-E_a/RT}$,so $\ln k = \ln A - \frac{E_a}{RT}$.Substituting $\ln k$ into equation $(i)$:$\ln (t_{1/2}) = \ln(0.693) - (\ln A - \frac{E_a}{RT})$$\ln (t_{1/2}) = \ln(\frac{0.693}{A}) + \frac{E_a}{RT}$Since $\ln(\frac{0.693}{A})$ and $\frac{E_a}{R}$ are constants,we have $\ln (t_{1/2}) \propto \frac{1}{T}$.

The half-life of a first order reaction varies with temperature according to

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