# Problem: A reaction has a rate constant of 1.13×10−2 /s at 400 K and 0.694 /s at 450. KPart A Determine the activation barrier for the reaction.Part B What is the value of the rate constant at 425 K?

###### FREE Expert Solution

For the first part of the problem, we’re being asked to determine the activation energy (Ea) of the 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:

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

where:

k1 = rate constant at T

k2 = rate constant at T

Ea = activation energy (in J/mol)

R = gas constant (8.314 J/mol•K)

T1 and T2 = temperature (in K).

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###### Problem Details

A reaction has a rate constant of 1.13×10−2 /s at 400 K and 0.694 /s at 450. K

Part A Determine the activation barrier for the reaction.

Part B What is the value of the rate constant at 425 K?