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NEET-XII-Chemistry

04: Chemical Kinetics

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  • #8
    The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
    Ans : It is given that T1 = 298 K

    ∴T2 = (298 + 10) K

    = 308 K

    We also know that the rate of the reaction doubles when temperature is increased by 10°.

    Therefore, let us take the value of k1 = k and that of k2 = 2k

    Also, R = 8.314 J K-1 mol-1

    Now, substituting these values in the equation:



    We get:





    = 52897.78 J mol-1

    = 52.9 kJ mol-1

    Note: There is a slight variation in this answer and the one given in the NCERT textbook.