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Activation Energy and Arrhenius Equation

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Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 49 to 59°C?

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Solution Summary

The solution includes a calculation of activation energy using the Arrhenius equation, given the change in temperature of a solution.

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49 deg = 49 + 273 = 322 K
59 deg = 59 + 273 = 332 K

We have the Arrhenius equation: k = A exp[-Ea/RT]

Take log
ln k = ln A - Ea/(RT)

Let k1 and k2 ...

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