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In going from room temperature (25 C) to 10 C above room temperature, the rate of...

In going from room temperature (25 C) to 10 C above room temperature, the rate of reaction doubles. Calculate the activation energy for the reaction .
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Answer #1
Use the logarithmic form of the Arrhenius equation, rearrange to the form of a straight line equation, and solve for the slope given 2 different points:

-Ea/R=ln(k2/k1)/(1/T2-1/T1)

k2/k1=2 since the rate doubles
Ea is activation energy
R = ideal gas constant, 8.3145 J/mol-K
T2=35C+273.15=308.15K
T1=25C+273.15=298.15K

Ea = 52949 J = 52.9 kJ
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