|Class 12 Chemistry Chemical Kinetics||Activation energy|
PROBLEM. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate ESolution.
SOLUTION. 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 k2/k1 = Ea/2.303R [T2-T1/T1T2]
Log 2k/k = Ea/2.303 X 8.314 [10/298 X 308]
Ea= 2.303 X 8.314 X 298 X 308 X log2/10
= 52897.78 J mol - 1
= 52.9 kJ mol - 1