Arrhenius Equation

Arrhenius proposed an equation to calculate the activation energy ($E_a$) of a chemical equation having rate constant ($K$) and temperature ($T$) in 1889.

$$K = A e^{-E_a/RT} \quad \text{——————— eq-1}$$ Where—

$K =$ Rate Constant
$A =$ Frequency Factor
$e =$ Exponent $= 2.718$
$E_a =$ Activation Energy
$R =$ Gas Constant
$T =$ Absolute Temperature

Taking $\ln$ on both side in equation-1 we get—

$$\ln K = \ln A - \frac{E_a}{RT}$$ For two reactions— $$\ln K_1 = \ln A - \frac{E_a}{RT_1} \quad \text{——————— eq-2}$$ and $$\ln K_2 = \ln A - \frac{E_a}{RT_2} \quad \text{——————— eq-3}$$ $$\text{or, } \ln A = \ln K_2 + \frac{E_a}{RT_2} \quad \text{——————— eq-4}$$

Putting the value of $\ln A$ from equation-4 into equation-2 we get—

$$\ln K_1 = \ln K_2 + \frac{E_a}{RT_2} - \frac{E_a}{RT_1}$$ $$\text{or, } \ln K_1 - \ln K_2 = \frac{E_a}{R} \left[ \frac{1}{T_2} - \frac{1}{T_1} \right]$$ $$\text{or, } \ln K_2 - \ln K_1 = \frac{E_a}{R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$$ $$\text{or, } \ln \frac{K_2}{K_1} = \frac{E_a}{R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$$ $$\text{or, } \log \frac{K_2}{K_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right] \quad \text{——————— eq-5}$$

Now knowing the value of K₂, K₁, T₂ and T₁, Activation Energy (E_a) can easily be calculated.

Arrhenius Equation Calculator

k₁ (e.g., 1.234e-12): k₂ (e.g., 5.678e-10): Activation Energy (Ea) (J/mol): Gas constant (R) (J/mol*K): Temperature T₁ (K): Temperature T₂ (K):

Result:

NEET-UG Exam Date: 5.5.2024

The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation.
Given 𝑅 = 8.314 𝐽𝐾⁻¹ 𝑚𝑜𝑙⁻¹, 𝑙𝑜𝑔4 = 0.6021

1. 380.4 kJ/mol
2. 3.80 kJ/mol
3. 3804 kJ/mol
4. 38.04 kJ/mol

Solution:
Option 4 is correct answer.
Using the equation-5, you can solve easily and you will find the answer 38.04 kJ/mol.

NEET-UG Exam Date: 5.5.2024

Activation energy of any chemical reaction can be calculated if one knows the value of
1. Probability of collision
2. Orientation of reactant molecules during collision
3. Rate constant at two different temperatures
4. Rate constant at standard temperature

Solution:
Option 3 is correct answer.
According to equation-5, activation energy of any chemical reaction can be calculated only if rate constant at two different temperatures are known.

Applications of Arrhenius Equation

1. Arrhenius Equation can be used to find out the optimum temperature at which maximum product is obtained.
2. Arrhenius Equation also used to calculate the activation energy based on the rate of the reaction.

Graph of Arrhenius equation

Characteristics of Arrhenius Equation

The activation energy of a chemical reaction is 100kJ/mol and its Frequency factor is 10 M⁻¹s⁻¹. What would be the rate constant at 300 K temperature.

[Hints:
k = Ea = 100KJ/mol
A = 10 M/s
T = 300K
As we know that-
ln K = ln A − Ea/RT
So, putting the values in the above equation we get-
ln K = ln 10 − 100000/(8.314 x 300)
or, ln K = − 37.8
or, K = 3.834 x 10⁻¹⁷]

MCQs Based on Arrhenius Equation

1. Arrhenius equation shows the variation of --- with temperature?

a. Reaction rate
b. Rate constant
c. Energy of activation
d. Frequency factor

View Answer

Correct Answer is B. Rate constant

Arrhenius equation quantitatively shows how the rate constant (k) increases exponentially with absolute temperature (T).

2. Arrhenius equation is

a. k = Ae^Ea/RT
b. k = Ae^−Ea/RT
c. k = Ae^RT/Ea
d. k = Ae^−RT/Ea

View Answer

Correct Answer is B. k = Ae^−Ea/RT

The standard mathematical form includes a negative exponent, representing the fraction of molecules possessing energy equal to or greater than the activation energy ($E_a$).

3. The activation energy of a reaction can be determined from the slope of which of the following graph

a. ln K vs 1/T
b. ln K vs T
c. ln K vs ln T
d. all the above

View Answer

Correct Answer is A. ln K vs 1/T

Taking the natural log of the Arrhenius equation yields $\ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right)$. A plot of $\ln k$ versus $\frac{1}{T}$ gives a straight line with a slope related to $E_a$.

4. If we plot a graph between log K and 1/T by Arrhenius equation , the slope is

a. −Ea/R
b. +Ea/R
c. −Ea/2.303R
d. +Ea/2.303R

View Answer

Correct Answer is C. −Ea/2.303R

When converting from natural log ($\ln$) to base-10 log ($\log$), the equation becomes $\log k = \log A - \frac{E_a}{2.303RT}$. Thus, the slope of the line is $-\frac{E_a}{2.303R}$.

5. In the Arrhenius plot of ln k vs 1/T, a linear plot is obtained with a slope of -2 x 10⁴K. The energy of activation of the reaction (in kJ mole^-1) is (R value is 8.3 JK^-1 mol^-1)

a. 83 KJmol^-1
b. 166 KJmol^-1
c. 249 KJmol^-1
d. 332 KJmol^-1

View Answer

Correct Answer is B. 166 KJmol^-1

Since $\text{Slope} = -\frac{E_a}{R}$, we have $-2 \times 10^4 = -\frac{E_a}{8.3}$. Solving for $E_a$ gives $166,000 \text{ J/mol}$, which equals $166 \text{ kJ/mol}$.