$\begin{array}{1 1}(a)\;ln \;t_{1/2}=ln\; A'+\large\frac{2E_a}{RT}\\(b)\;ln\;t_{1/2}=ln\;A'+\large\frac{E_a}{RT}\\(c)\;ln\;t_{1/2}=ln \;A'-\large\frac{E_a}{RT}\\(d)\;None\end{array}$

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Answer: $ln\;t_{1/2}=ln\;A'+\large\frac{E_a}{RT}$ where $A'=\large\frac{2^{n-1}-1}{(n-1)a_0^{n-1}A}$

From Arrhenius equation, $k=Ae^{-\large\frac{E_a}{RT}}$

$\Rightarrow ln \;k=ln A-\large\frac{E_a}{RT}$

Also for $n^{th}$ order reaction

$t_{1/2}=\large\frac{2^{n-1}-1}{k_n(n-1)a_0^{n-1}}$

Taking natural log and substituting for $ln\; k$, $ln\;t_{1/2}=ln\large\frac{2^{n-1}-1}{(n-1)a_0^{n-1}}$$-ln \;A+\large\frac{E_a}{RT}$

$\Rightarrow ln A'+\large\frac{E_a}{RT}$

$ln\;t_{1/2}=ln A'+\large\frac{E_a}{RT}$, where $A'=\large\frac{2^{n-1}-1}{(n-1)a_0^{n-1}A}$

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