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**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 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^{-Ea / RT} ..................(1)

taking natural log of this equation, we get ln

â€¦(2)

or â€¦(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

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 = â€“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 Ã— R = 100 Ã— 8.314

E_{a} = 831.4 J mol^{â€“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).** From the above reaction, the ate of format ion of product is

Sol.

......(1)

= k_{1}[X][Y] â€“ k_{2}[Z] â€“ k_{3}[Z]

= k_{1}[X][Y] â€“ (k_{2} + k_{3})[Z]

0 = k_{1}[X][Y] â€“ k_{2}[Z] [âˆµ k2 >> k3]

SSA on intermediate.

then

then we find,

i.e. â€¦(2)

i.e.

and

i.e.

E_{overall }= E_{1} + E_{3} â€“ 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 â€“1.2 Ã— 10 ^{4} K. What is the activation energy of the reaction. Sol.**

y = m x + C

where m = slope of line

then

E_{a }= â€“2.303 R (slope) = â€“2.303 Ã— 8.314 Ã— (â€“1.2 Ã— 10^{4} K)

= 1.0 Ã— 10^{5} J mol^{â€“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^{â€“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 (Î¶) 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

Î”x = x - x_{e} or x = Î”x + x_{e}

Since

at equilibrium, = 0 and x = x_{e}. Hence

k_{1}(a â€“ x_{e}) = k_{â€“1 }x_{e}

then

= -k_{r}Î”x

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 Î¶

**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 Î”x = x â€“ x_{e}

x_{e} = equilibrium concentration of x

at equilibrium,

= 0, hence

we get

Î”x is very small than (Î”x)^{2} is neglected,

where kr = k_{1} + 2k_{â€“1} x_{e}

is the relaxation rate constant.

and

The relaxat ion time Î¶ in this case is

**Problem. The relaxation time for the fast reaction **** is 10 Âµs and the equilibrium constant is 1.0 Ã— 10 ^{â€“3}. Calculate the rate constant for the forward and the reverse reactions. Sol.**

K = equilibrium constant = 1.0 Ã— 10^{â€“3} =

âˆ´ k_{1} = 1.0 Ã— 10^{â€“3} k_{â€“1}

âˆµ

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