Arrhenius Equation
Arrhenius equation
Arrhenius proposed an equation to calculate the activation energy(Ea) of a chemical equation having rate constant 'K' and temperature 'T' in 1889.Now knowing the value of K2, K1, T2 and T1, Activation Energy (Ea) can easily be calculated.
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
Arrhenius equation shows the variation of --- with temperature?
a. Reaction rate
b. Rate constant
c. Energy of activation
d. Frequency factor
Arrhenius equation is
a. k = AeEa/RT
b. k = Ae−Ea/RT
c. k = AeRT/Ea
d. k = Ae−RT/Ea
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
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
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
Answer: 1-b, 2-b, 3-a, 4-c, 5-b
