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Question

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

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Solution

It is given that T1 = 298 K

∴T2 = (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:

logk2k1=Ea2.303×R(1T11T2)

log21=Ea2.303×8.314(12981308)

Ea=52.9kJ mol1

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