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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 Preview

    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 ...

    Solution Summary

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