Difficult
Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, $T (\text{K})$.
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k$ is $\text{s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$)
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k$ is $\text{s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$)
- A$12.42$
- B$18.63$
- C$14.34$
- D$24.84$
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Similar Questions
The activation energy for a simple chemical reaction $A \to B$ is ${E_a}$ in the forward direction. The activation energy for the reverse reaction:
Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}.$ If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300 \, K,$ then $\ln (k_2/k_1)$ is equal to :
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$
DifficultJEE MAIN 2017
View SolutionThe equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)
The activation energy of a certain reaction is $87 \ kJ \ mol^{-1}$. When the temperature is decreased from $37^{\circ} C$ to $15^{\circ} C$,what is the ratio of the rate constant at $37^{\circ} C$ to that at $15^{\circ} C$ for this reaction (in $/ 1$)?
Medium
View SolutionThe activation energy of a reaction can be determined from the slope of which of the following graphs?
MediumNEET 2015
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