$\begin{array}{1 1} 55\;kJmol^{-1} \\ 83\;kJmol^{-1} \\ 72\;kJmol^{-1} \\ 68\;kJmol^{-1} \end{array}$

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Answer: $ 55\;kJmol^{-1}$

We have to find the activation barrier, i.e, the activation energy, give two temperatures $T_1$ and $T_2$.

Its also mentioned that the rate constant triples, so $k_2 = 3k_1$

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 for $k_2 = 3k_1$ and $T_1, T_2$ (in $^\circ$K), we get:

$\Rightarrow ln 3 = 1.0986 = \large\frac{E_a}{8.314\;J\;K.mol^{-1}}$$ (\large\frac{1}{293.15}$$ - \large\frac{1}{308.15}$$)$

Solving for $E_a$, we get $E_a = 55006\;Jmol^{-1} = 55\;kJmol^{-1}$

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