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^{Ea / RT}
A is constant known as the frequency factor or Arrhenius preexponential factor.
A & E_{a} 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
…(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^{Ea / 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^{Ea / 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^{Ea / RT }
If k_{1} and k_{2} be the value of rate constant at temperature T_{1 }and T_{2}, 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)
= k_{1}[X][Y] – k_{2}[Z] – k_{3}[Z]
= k_{1}[X][Y] – (k_{2} + k_{3})[Z]
0 = k_{1}[X][Y] – k_{2}[Z] [∵ k2 >> k3]
SSA on intermediate.
then
then we find,
i.e. …(2)
i.e.
and
i.e.
E_{overall }= E_{1} + E_{3} – E_{2}
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 × 10^{4} 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 × 10^{4} K)
= 1.0 × 10^{5} J mol^{–1}
Fast Reaction.
Fast reactions are studies by following methods
(1) StoppedFlow 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) Perturbationrelaxation 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 Tjump are most important type of perturbation.
Problem. Using the Tjump 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
k_{1}(a – x_{e}) = k_{–1 }x_{e}
then
= k_{r}Δx
where k_{r }= k_{1} + 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 = k_{1} + 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} =
∴ k_{1} = 1.0 × 10^{–3} k_{–1}
∵
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