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activation energy question?

A rule of thumb in chemical kinetics states that for many reactions, the rate of reaction approximately doubles for a temperature rise of 10°C. What must be the activation energy of a reaction if the rate is indeed found to double between 25°C and 35°C?

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  • Anonymous
    1 decade ago
    Favorite Answer

    The Arrhenius equation is

    k = A exp( -Ea / RT )

    so

    k1 = A exp( -Ea / R(298K) )

    k2 = A exp( -Ea / R(308K) )

    k2 / k1 = 2

    = exp( -Ea / R(308K) ) / A exp( -Ea / R(298K) )

    = exp( -Ea / R(308K) - -Ea / R(298K) )

    ln 2 = Ea/R ( 1/298K - 1/308K )

    or about 100 kJ/mol

    m

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