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How will you determine activation energy from rate constants at two different temperatures?

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Questions

How will you determine activation energy from rate constants at two different temperatures?

Derive an expression to determine activation energy for two different temperatures T1 and T2.

Derivation
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Solution

Arrhenius equation is k = `Ae^((-E_a)/(RT))`

From two different temperatures T1 and T2

`log_10 k_1 = log_10A - E_a/(2.303  RT_1)`  ....(1)

`log_10 k_2 = log_10A -E_a/(2.303  RT_2)`  ....(2)

Where k1 and k2 are the rate constants at temperatures T1 and T2 respectively. Subtracting equation (1) from equation (2),

`log_10 k_2 - log_10k_1 = -E_a/(2.303  R) 1/T_2 + E_a/(2.303  R) 1/T_1`

Hence, `log_10  k_2/k_1 = E_a/(2.303  R) (1/T_1 - 1/T_2)`

= `E_a/(2.303  R) ((T_2 - T_1)/(T_1T_2))`

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Chapter 6: Chemical Kinetics - Exercises [Page 137]

APPEARS IN

Balbharati Chemistry [English] Standard 12 Maharashtra State Board
Chapter 6 Chemical Kinetics
Exercises | Q 3. xiv. (b) | Page 137

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