Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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Arrhenius Equation (Expression)

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

K = Ae-Ea / RT

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

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev…(2)

or       Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev   …(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-Ea / RT

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

when Ea increase then the value of Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRevncreases and the value of ln k i.e. k will be decreases.
i.e. Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

When T increase then the value of Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRevdecreases and the value of ln k i.e. k will be increase.
i.e.  Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

The graph of ln k vs Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev is given below 

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

y = –m. x + C

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

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Sol. We know that 

  Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

i.e.           ln A = 2.3
&        A = e2.3 = 9.974

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

E= 100 × R = 100 × 8.314

Ea = 831.4 J mol–1
 

Problem.  When temperature is increased then t1/2 of reaction will be 

(a) Remains constant               (b) Increased 

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

Sol. We know that

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

&               k = Ae-Ea / RT

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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.  Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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

Variation of rate constant with temperature 

We know that k = Ae-Ea / RT 

  Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

or Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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 = Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Standard Temperature coefficient = Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

                                                         = 2 to 3

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Problem.  In the reaction mechanism

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev                      ......(1)

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev = 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

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

then we find,

  Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
i.e. Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev                    …(2)
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
i.e. Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
and Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
i.e. Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Eoverall = 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

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Ea = 50145.617 J
Ea = 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.

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

y = m x + C

where             m = slope of line
then

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

E= –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,

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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

  Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev 

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

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
at equilibrium, Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev= 0 and x = xe. Hence

k1(a – xe) = k–1 xe 

  then              Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
                          = -krΔx
where              k= k1 + k-1
                    = relaxat ion rate constant
then

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Problem.  Find the relaxation time for the following reaction.

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

 Now             Δx = x – xe

xe = equilibrium concentration of x

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

at equilibrium,

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev= 0, hence

we get  Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

where               kr = k1 + 2k–1 xe
is the relaxation rate constant.
and 

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

The relaxat ion time ζ in this case is

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

Problem.  The relaxation time for the fast reaction   Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev    is 10 µs and the equilibrium constant is 1.0 × 10–3. Calculate the rate constant for the forward and the reverse reactions.
 Sol.

 Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
K = equilibrium constant = 1.0 × 10–3 =Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

∴ k1 = 1.0 × 10–3 k–1

Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev
Arrhenius Equation - Chemical Kinetics Chemistry Notes | EduRev

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