Arrhenius Equation & Activation Energy | Physical Chemistry PDF Download

Arrhenius Equation (Expression)

The following empirical relationship between temperature (T), rate constant (k) and activation energy (E_a) is known as the Arrhenius expression:

K = Ae^{-E_a / RT}

A is constant known as the frequency factor or Arrhenius pre-exponential factor.
A & E_a is temperature independent.
The unit of A is always equal to the unit of rate constant (k) k = Ae^{-E_a / RT} ..................(1)
taking natural log of this equation, we get ln

…(2)

or          …(3)
 

Problem. Prove that on increasing the activation energy, the rate constant will be decreasing and on increasing the temperature, the rate constant will be increasing.
 Sol:
k = Ae^{-E_a / RT}

when E_a increase then the value of ncreases and the value of ln k i.e. k will be decreases.
i.e.

When T increase then the value of decreases and the value of ln k i.e. k will be increase.
i.e.  

The graph of ln k vs  is given below 

y = –m. x + C

Problem.  Using the given equation find the value of A & E_a.

Sol. We know that 

i.e.           ln A = 2.3
&        A = e^2.3 = 9.974

E_a = 100 × R = 100 × 8.314

E_a = 831.4 J mol^–1
 

Problem.  When temperature is increased then t_1/2 of reaction will be 

(a) Remains constant               (b) Increased 

(c) Decreased                           (d) First increase and then decrease
 

Sol. We know that

&               k = Ae^{-E_a / RT}

i.e. on increasing T, eEa / RT decrease then we can say that on increasing temperature (T), the t _1/2 of the reaction will decrease. i.e.  

i.e. The correct answer is (C).

Variation of rate constant with temperature 

We know that k = Ae^{-E_a / RT} 

If k₁ and k₂ be the value of rate constant at temperature T₁ and T₂, we can derive

or 

Temperature Coefficient.

The ratio of rate constant of a reaction at two different temperature differing by 10 degree is know as temperature coefficient.
i.e. Temperature coefficient = 

Standard Temperature coefficient = 

                                                         = 2 to 3

Problem.  In the reaction mechanism

Find the overall rate constant (k_overall) and Activation energy E_a (overall).
 Sol. From the above reaction, the ate of format ion of product is 

                       ......(1)

 = k₁[X][Y] – k₂[Z] – k₃[Z]
= k₁[X][Y] – (k₂ + k₃)[Z]
0 = k₁[X][Y] – k₂[Z]                                                             [∵ k2 >> k3]

SSA on intermediate.
then

then we find,

 


i.e.                     …(2)


i.e. 
and 
i.e. 

E_overall = E₁ + E₃ – E₂

Problem.  What is the energy of activation of the reaction if it rate doubles when temperature is raised .290 to 300 K.
 Sol. We know that

E_a = 50145.617 J
E_a = 50.145 kJ

Problem. A plot of log k versus 1/T gave a straight line of which the slope was found to be –1.2 × 10⁴ K. What is the activation energy of the reaction.
 Sol.

y = m x + C

where             m = slope of line
then

E_a = –2.303 R (slope) = –2.303 × 8.314 × (–1.2 × 10⁴ K)
= 1.0 × 10⁵ J mol^–1

Fast Reaction.

Fast reactions are studies by following methods
(1)  Stopped-Flow technique: For reaction that occur on timescales as short at 1 ms (10^–3 s)
(2) Flash photolysis technique: Reaction that can be triggered by light are studied using flash photolysis.
(3)  Perturbation-relaxation methods: A chemical system initially at equilibrium is perturbed such that the system is not longer at equilibrium. By following the relaxation of the system back toward equilibrium, the rate constant for the reaction can be determined.
The temperature perturbation or T-jump are most important type of perturbation.

Problem. Using the T-jump method find out the relaxation time (ζ) of the following reaction,

Sol.  Let a be the total concentration of (A + B) and x the concentration of B at any instant. Then rate 

If x_e is the equilibrium concentration, then
Δx = x - x_e or x = Δx + x_e
Since

 
at equilibrium, = 0 and x = x_e. Hence

k₁(a – x_e) = k_–1 x_e 

  then             
                          = -k_rΔx
where              k_r = k₁ + k_-1
                    = relaxat ion rate constant
then

Then reciprocal of kr i.e. k_r^-1 is called relaxat ion time.
It is represented by ζ

Problem.  Find the relaxation time for the following reaction.

Sol. Let a be the total concentration and x the concentration of B which is equal to the concentration of C. Then, the rate law is given by

 Now             Δx = x – x_e

x_e = equilibrium concentration of x

at equilibrium,

 = 0, hence

we get  

Δx is very small than (Δx)² is neglected,

where               kr = k₁ + 2k_–1 x_e
is the relaxation rate constant.
and 

The relaxat ion time ζ in this case is

Problem.  The relaxation time for the fast reaction       is 10 µs and the equilibrium constant is 1.0 × 10^–3. Calculate the rate constant for the forward and the reverse reactions.
 Sol.

 
K = equilibrium constant = 1.0 × 10^–3 =

∴ k₁ = 1.0 × 10^–3 k_–1
∵