The reaction rate constant at –5. 0 ºc is k2 is 24.3 KJ.
Activation energy is the minimum amount of energy required to carry out a reaction.
The formula is [tex]log \dfrac{k_2}{k_1} = \dfrac{Ea}{2.3R} \;( \dfrac{1}{T_1}- \dfrac{1}{T_2} )[/tex]
Given, K1 is [tex]4. 1 \times 10^3 m^-^1 s^-^1[/tex]
Temperature is 32
k2 is to find?
r = 8.314
Energy is 63.9
[tex]log \dfrac{k_2}{4. 1 \times 10^3 } = \dfrac{63.9 }{2.3\times 8.31} ( \dfrac{1}{32.0}- \dfrac{1}{-5.0} ) = 24.3 kj[/tex]
Thus, the correct option is k2 is 24.3 kJ
Learn more about activation energy
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