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The rate constant of a reaction at 32$^\circ$C is 0.055 s$^{-1}$. If the frequency or pre-exponential factor is 1.2 $\times$ 10$^{13}$ s$^{-1}$, what is the activation energy?

$\begin{array}{1 1}83.721 \;kJmol^{-1} \\ 41.86\;kJmol^{-1} \\20.926 \;kJmol^{-1} \\ 104.624 \;kJmol^{-1} \end{array}$

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Answer: $83.721 \;kJmol^{-1}$
Substituting in Arrhenius equation, $k=Ae^{\large\frac{-E_a}{RT}}$, we get:
$\Rightarrow 0.055s^{-1} = 1.2\times10^{13}s^{-1} \times e^{\large\frac{-E_a}{8.314\;JK.mol^{-1} \times 305^{\circ}K}}$
$\Rightarrow 4.58 \times 10^{-18} = e^{\large\frac{-E_a}{2535.7\;Jmol^{-1}}}$
Taking natural log and simplifying, $E_a = 83721\;Jmol^{-1} = 83.721 \;kJmol^{-1}$
answered Jul 25, 2014 by balaji.thirumalai

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