Difficult
The rate constant of a reaction varies with temperature according to the equation: $\log K = \text{constant} - \frac{E_a}{2.303 RT}$. If a plot of $\log K$ versus $1/T$ yields a straight line with a slope of $-5632$,then the activation energy of the reaction is .......... $kJ \ mol^{-1}$.
- A$127.67$
- B$107.84$
- C$86$
- D$246.8$
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Medium
View SolutionFor a reaction,the value of rate constant at $300 \ K$ is $6.0 \times 10^5 \ s^{-1}$. The value of Arrhenius factor $A$ at infinitely high temperature is
MediumKCET 2023
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