Question

the rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298 K calculate Ea.​

Answer 1

Answer:

It is given that T 1= 298 K

∴T = (298 + 10) K

= 308 K

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

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/2.303*R(1/T1-1/T2)

Ea=52.9kJ mol

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