According to the Arrhenius equation,which of the following statements are correct?
$(A)$ $A$ high activation energy usually implies a fast reaction.
$(B)$ Rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy.
$(C)$ Higher the magnitude of activation energy,stronger is the temperature dependence of the rate constant.
$(D)$ The pre-exponential factor is a measure of the rate at which collisions occur,irrespective of their energy.
$(A)$ $A$ high activation energy usually implies a fast reaction.
$(B)$ Rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy.
$(C)$ Higher the magnitude of activation energy,stronger is the temperature dependence of the rate constant.
$(D)$ The pre-exponential factor is a measure of the rate at which collisions occur,irrespective of their energy.
- A$B, C, A$
- B$B, C, D$
- C$B, A, D$
- D$B, C$
Explore More
Similar Questions
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 .......... $kJ \ mol^{-1}$.
Difficult
View SolutionReactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$
Easy
View SolutionConsider the following transformation involving first order elementary reactions in each step at constant temperature as shown below.
$A + B \underset{\text{Step } 3}{\overset{\text{Step } 1}{\rightleftharpoons}} C \xrightarrow{\text{Step } 2} P$
Some details of the above reaction are listed below.
Step Rate constant $(s^{-1})$ Activation energy $(kJ \ mol^{-1})$ $1$ $k_1$ $300$ $2$ $k_2$ $200$ $3$ $k_3$ $Ea_3$
If the overall rate constant of the above transformation $(k)$ is given as $k = \frac{k_1 k_2}{k_3}$ and the overall activation energy $(E_a)$ is $400 \ kJ \ mol^{-1}$,then the value of $Ea_3$ is $\qquad$ $kJ \ mol^{-1}$ (nearest integer).
$A + B \underset{\text{Step } 3}{\overset{\text{Step } 1}{\rightleftharpoons}} C \xrightarrow{\text{Step } 2} P$
Some details of the above reaction are listed below.
| Step | Rate constant $(s^{-1})$ | Activation energy $(kJ \ mol^{-1})$ |
|---|---|---|
| $1$ | $k_1$ | $300$ |
| $2$ | $k_2$ | $200$ |
| $3$ | $k_3$ | $Ea_3$ |
If the overall rate constant of the above transformation $(k)$ is given as $k = \frac{k_1 k_2}{k_3}$ and the overall activation energy $(E_a)$ is $400 \ kJ \ mol^{-1}$,then the value of $Ea_3$ is $\qquad$ $kJ \ mol^{-1}$ (nearest integer).
The reaction $X \to Y$ is an exothermic reaction. The activation energy of the forward reaction $X \to Y$ is $150\,kJ\,mol^{-1}$. The enthalpy of the reaction is $-135\,kJ\,mol^{-1}$. The activation energy for the reverse reaction,$Y \to X$,will be $.......\,kJ\,mol^{-1}$.
DifficultJEE MAIN 2013
View SolutionFor a first order reaction,a plot of $\ln k$ ($y$-axis) and $\frac{1}{T}$ ($x$-axis) gave a straight line with a slope equal to $-10^3 \ K$ and an intercept equal to $2.303$ (on $y$-axis). What is the activation energy ($E_a$ in $kJ \ mol^{-1}$) of the reaction? (Given $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo