Medium
The graph obtained between $\ln k$ ($k=$ Rate constant) on $y$-axis and $1/T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$
- A$166$
- B$332$
- C$765$
- D$382$
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Similar Questions
The temperature-dependent equation for the rate constant is written as:
Easy
View SolutionCalculate the activation energy of a reaction,whose rate constant doubles on raising the temperature from $300 \ K$ to $600 \ K$.
DifficultTS EAMCET 2021
View SolutionThe 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 Solution$A$ reaction takes place in three steps with individual rate constant and activation energy,
Step Rate constant and Activation energy $Step \ 1$ $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$ $Step \ 2$ $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$ $Step \ 3$ $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$
Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.
| Step | Rate constant and Activation energy |
| $Step \ 1$ | $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$ |
| $Step \ 2$ | $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$ |
| $Step \ 3$ | $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$ |
Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.
Difficult
View SolutionAt $298 \text{ K}$ temperature,the activation energy for the reaction $x_2 + y_2 \rightarrow 2xy + 20 \text{ kJ}$ is $15 \text{ kJ}$. What will be the activation energy for the reaction $2xy \rightarrow x_2 + y_2$?
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