Activation energy ( E_a ) and rate constants ( k_1 and k_2 ) of a chemical reaction at two different temperatures ( T_1 and T_2 ) are related by: - Chemistry MCQ Chemistry | ExamTest.live

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Activation energy (E_{a}) and rate constants (k_{1} and k_{2}) of a chemical reaction at two different temperatures (T_{1} and T_{2}) are related by:

A $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$
B $ln \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$
C $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} [\frac{1}{T _{2}} - \frac{1}{T _{1}}]$
D $ln \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} [\frac{1}{T _{2}} + \frac{1}{T _{1}}]$

Solution & Step-by-step Explanation

The integrated form of the Arrhenius equation for two temperatures is: ln k_{2} - ln k_{1} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} [\frac{T _{2} - T _{1}}{T _{1} T _{2}}] This matches option A.

Try it yourself before checking the explanation above.

Activation energy (E_{a}) and rate constants (k_{1} and k_{2}) of a chemical reaction at two different temperatures (T_{1} and T_{2}) are related by:

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]

ln \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} [\frac{1}{T _{2}} - \frac{1}{T _{1}}]

ln \frac{k _{2}}{k _{1}} = - \frac{E _{a}}{R} [\frac{1}{T _{2}} + \frac{1}{T _{1}}]

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