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  • Ex: The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be: (R = 8.314 JK–1mol–1 and log 2 = 0.301)
    A) 48.6 kJ mol–1
    B) 58.5 kJ mol–1
    C) 60.5 kJ mol–1
    D) 53.6 kJ mol–1
    Explanation: The question is based on the effect of temperature change on rate of reaction and involves the concept of activation energy. So, Arrhenius equation should be used for solving this problem.
    Using Arrhenius Equation
    ln(k/ k1) = (Ea / R) (1/T- 1/T2)
    We get,
    2.303 log2 = -( Ea/8.314)(1/300 -1/310)
    Ea = 5359.59 Jmol-1 = 53.6 kJ mol-1
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