Question

The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:

Temperature(K)	k(s-1)
293	0.054
298	0.100

Short answer

The energy of activation for the reaction is 89.48 kJ/mol.

Step-by-step solution

Using Arrhenius equation 

From the Arrhenius Equation, the rate of reaction at two different temperatures is given as

$\mathbf{l}\mathbf{o}\mathbf{g}\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{2}\mathbf{.303}\mathbf{\ast }\mathbf{R}}\left(\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}\right)$

where ${\mathbf{k}}_{\mathbf{1}}$and ${\mathbf{k}}_{\mathbf{2}}$are the rate constants at ${\mathbf{T}}_{\mathbf{1}}$ and ${\mathbf{T}}_{\mathbf{2}}$ where ${\mathbf{T}}_{\mathbf{1}}\mathbf{<}{\mathbf{T}}_{\mathbf{2}}$. ${\mathbf{E}}_{\mathbf{a}}$is the activation energy (in J) and R is the gas constant.

Step 2: From the question

Given,

The ratio of $\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}\mathbf{=}\mathbf{1}\mathbf{.47}$

${\mathbf{T}}_{\mathbf{1}}=298K$

${\mathbf{T}}_{\mathbf{2}}=293K$

Calculating the Activation energy

Replacing the values in the Arrhenius equation,

$\begin{array}{rl}\log \frac{0.1}{0.054}& =\frac{E_a}{2.303\times 8.314}\left(\frac{1}{293}-\frac{1}{298}\right)\\ E_a& =0.267\times 2.303\times 8.314\times \left(\frac{293\times 298}{298-293}\right)\\ E_a& =89477.5J/mol\\ E_a& =89.48kJ/mol\end{array}$