We’re being asked to **determine the activation energy (E _{a})** of a reaction. We’re given the rate constants at two different temperatures.

This means we need to use the ** two-point form of the Arrhenius Equation**:

$\overline{){\mathbf{ln}}{\mathbf{}}\frac{{\mathbf{k}}_{\mathbf{2}}}{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}\frac{{\mathbf{E}}_{\mathbf{a}}}{\mathbf{R}}{\mathbf{[}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{2}}}{\mathbf{-}}\frac{\mathbf{1}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{]}}}$

where

**k _{1}** = rate constant at T

**k _{2}** = rate constant at T

**E _{a}** = activation energy (in J/mol)

**R** = gas constant (8.314 J/mol•K)

**T _{1} and T_{2}** = temperature (in K)

The rate constant of a reaction is 4.50×10^{−5} L/mol•s at 195°C and 3.20 x 10^{-3} L/mol•s at 258°C. What is the activation

energy of the reaction?

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Our tutors have indicated that to solve this problem you will need to apply the Arrhenius Equation concept. You can view video lessons to learn Arrhenius Equation. Or if you need more Arrhenius Equation practice, you can also practice Arrhenius Equation practice problems.

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Based on our data, we think this problem is relevant for Professor Bertolini's class at USC.

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.