The rate of reaction is doubled when the temperature is changed from 298 k to 308 k. Calculate the energy of activation of the reaction.(in kj per mole)
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Answer:
It is given that T
1
= 298 K
∴T
2
= (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
k
1
k
2
=
2.303×R
E
a
(
T
1
1
−
T
2
1
)
log
1
2
=
2.303×8.314
E
a
(
298
1
−
308
1
)
E
a
=52.9kJ mol
−1
Explanation:
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