The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
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:
`log k_2/k_1 = E_a/(2.303R) [(T_2-T_1)/(T_1T_2)]`
`log "2k"/k = E_a/(2.303xx8.314)[10/(298xx308)]`
`=>log 2 = E_a/(2.303xx8.314)[10/(298xx308)]`
`=> (E_a = 2.303xx8.314xx298xx308xxlog2)/10`
= 52897.78 J mol−1
= 52.9 kJ mol−1
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