Question

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

Answer 1

Answer:

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:

logk1k2=2.303×REa(T11−T21)

log12=2.303×8.314Ea(2981−3081)

Ea=52.9kJ mol−1