Medium
It has been found that for a chemical reaction with a rise in temperature by $9 \, K$,the rate constant gets doubled. Assuming a reaction to be occurring at $300 \, K$,the value of activation energy is found to be $...... \, kJ \, mol^{-1}$. [nearest integer]
(Given $\ln 10 = 2.3, R = 8.3 \, J \, K^{-1} \, mol^{-1}, \log 2 = 0.30$ )
(Given $\ln 10 = 2.3, R = 8.3 \, J \, K^{-1} \, mol^{-1}, \log 2 = 0.30$ )
- A$66$
- B$12$
- C$59$
- D$78$
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Similar Questions
At $27^{\circ}C$ in the presence of a catalyst, the activation energy of a reaction is lowered by $10 \ kJ \ mol^{-1}$. The logarithm ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is .... (Consider that the frequency factor for both the reactions is the same)
DifficultJEE MAIN 2026
View SolutionWhich graph shows zero activation energy?
Medium
View SolutionThe first order rate constant for a certain reaction increases from $1.667 \times 10^{-6} \ s^{-1}$ at $727 \ ^oC$ to $1.667 \times 10^{-4} \ s^{-1}$ at $1571 \ ^oC$. The rate constant at $1150 \ ^oC$,assuming constancy of activation energy over the given temperature range is [Given : $\log \ 19.9 = 1.299$ ]
The activation energy is ..........
Easy
View SolutionCatalysts are used to increase the rate of a chemical reaction. Because it
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