Explain the rate of reaction with energy of activation and temperature with the help of the Arrhenius equation and state its importance.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The Arrhenius equation describes the relationship between the rate constant of a chemical reaction and temperature:
$k = A e^{-\frac{E_{a}}{RT}}$
Where:
$k = \text{Rate constant of reaction (proportional to rate)}$
$E_{a} = \text{Activation energy (J mol}^{-1})$
$T = \text{Absolute temperature (K)}$
$R = \text{Gas constant (8.314 J K}^{-1} \text{ mol}^{-1})$
$A = \text{Arrhenius frequency factor (or pre-exponential factor)}$
Taking the natural logarithm of both sides:
$\ln k = -\frac{E_{a}}{RT} + \ln A$
This equation follows the linear form $y = mx + c$,where $\ln k$ is $y$,$-\frac{E_{a}}{R}$ is the slope $m$,$\frac{1}{T}$ is $x$,and $\ln A$ is the intercept $c$.
$A$ plot of $\ln k$ versus $\frac{1}{T}$ yields a straight line with a negative slope equal to $-\frac{E_{a}}{R}$ and a y-intercept equal to $\ln A$.
Importance:
$1$. It allows for the calculation of activation energy $(E_{a})$ and the frequency factor $(A)$ experimentally.
$2$. It explains why reaction rates increase with temperature,as the fraction of molecules with energy greater than $E_{a}$ increases exponentially with $T$.

Explore More

Similar Questions

For the reaction $A_2 + B_2 \rightleftarrows 2AB$,the activation energy for the forward and backward reactions are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$,respectively. If a catalyst lowers the activation energy of both forward and backward reactions by $100 \ kJ \ mol^{-1}$,what will be the enthalpy change $(\Delta H)$ for the reaction $A_2 + B_2 \rightarrow 2AB$ in the presence of the catalyst?

The decomposition of $N_2O$ into $N_2$ and $O_2$ in the presence of argon follows first-order kinetics. If the rate constant is given by $K = 5.0 \times 10^{11} \, e^{-2000/T}$,what is the activation energy of the reaction?

The rate constant of a reaction is $2 \times 10^{-3} \ min^{-1}$ at $300 \ K$. By increasing the temperature by $10 \ K$,its value becomes double. Calculate the energy of activation $(E_a)$ and the rate constant at $320 \ K$.

The rate of the process doubles with every $10 \ K$ increase in temperature. When the temperature is increased from $303 \ K$ to $353 \ K$,how much will the rate of the process increase?

The Arrhenius plots of two reactions,$I$ and $II$ are shown graphically. The graph suggests that

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo