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The activation energy for a certain reaction is 76.7 kJ mol$^{-1}$. How many times faster will the reaction occur at 50$^\circ$C compared to 0$^\circ$C?

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

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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$
answered Jul 25, 2014 by balaji.thirumalai
edited Jul 25, 2014 by balaji.thirumalai
 

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