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The rate constant 'k'. For a reaction va...

The rate constant 'k'. For a reaction varies with temperature 'T' according to the question. `log k = log A-(E_(a))/(2.303R)(1/T)`
Where `E_(a)` is the activation energy. When a graph is plotted for `log k vs 1//T`, a straight line with a slope of `-4250` K is obtained. Calcualte `E_(a)` for this reaction.

Text Solution

Verified by Experts

Slope =`E_a/(2.303R)`=-4250 K
So, `E_a=-2.303xxRxx"Slope"`
`=-2.303xx8.314 J K^(-1) "mol"^(-1) xx4250`
`=81375.3 "J mol"^(-1)`
`=81.375 "KJ mol"^(-1)`
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