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प्रश्न
The rate constant of a reaction quadruples when the temperature changes from 700 K to 720 K. Calculate the activation energy for this reaction.
[log 2 = 0.30, log 4 = 0.60, 2.303 R = 19.15 JK−1 mol−1]
संख्यात्मक
उत्तर
Given: T1 = 700 K, T2 = 720 K, `k_1/k_2 = 4`
2.303 × R = 19.15 JK−1mol−1, log 4 = 0.60
`log k_2/k_1 = E_a/(2.303 R)[(T_2 - T_1)/(T_1T_2)]`
`log 4 = E_1/19.15[(720 - 700)/(720 xx 700)]`
`0.6 = E_a/19.15[20/504000]`
`E_a = (0.6 xx 19.15 xx 504000)/(20)`
= 289.548 J mol−1
= 289.548 K J mol−1
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