$\begin{array}{1 1} 186 \\ 93 \\ 47.5 \\ 33 \end{array}$

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Answer: 186

Given the activation energy, two temperatures, we have to find the ratio of the rate constants.

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

$\Rightarrow \large\frac{k_2}{k_1} $$ = \large\frac{Ae^{\large\frac{-E_a}{RT_1}}}{Ae^{\large\frac{-E_a}{RT_2}}}$

Using the fact that $\large\frac{e^x}{e^y}$$ = e^{x-y}$ and taking the natural log on both sides and simplifying, we arrive at:

$ ln\large\frac{k_2}{k_1}$$ = \large\frac{E_a}{R}$$ (\large\frac{1}{T_1}$$ - \large\frac{1}{T_2}$$)$

Substituting the values (note: convert $^\circ$C to $^\circ$K) , we get:

$\Rightarrow ln\large\frac{k_2}{k_1}$$ = \large\frac{76700\; Jmol^{-1}}{8.314 JK.mol^{-1}}$$ (\large\frac{1}{273 ^\circ K}$$ - \large\frac{1}{323 ^\circ K}$$)$

$\Rightarrow ln\large\frac{k_2}{k_1}$$ = \large\frac{76700 \times 0.0005670}{8.314}$$=5.231$

$\Rightarrow \large\frac{k_2}{k_1}$$ = e^{5.231} = 186$

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