- Explain activation energy.
- Describe how energy vary during a chemical reaction.
- Plot chemical potential energy of the system as a function of the reaction coordinate.
- Explain the model
*k*= A exp (*E*_{a}/*RT*) for chemical kinetics. - Plot
*k*versus 1/*T*when*k*is measured at several*T*. - Calculate
*E*_{a}when a series of*k*and*T*are given. - Calculate
*k*at certain*T*.

Chemical potential energies of reactants and products. The difference is the enthalpy of reaction. |

This potential energy difference is the driving force for a chemical reaction to take place. After the reaction energy is released. The products are at a more stable energy level than the level the reactants are.

Activation Energy and Enthalpy of Chemical Reaction |

The rate constant

**
Example 1
**

(a) Calculate

(b) What is the rate constant at 308 K?

*Solution*

Let *k*_{1} and *k*_{2} be the rate constants determined at
*T*_{1} and *T*_{2}, respectively.
Then you have two equations:

lnFrom these, you should be able to derive the following relationships,k_{1}= lnA -E_{a}/(RT_{1}) lnk_{2}= lnA -E_{a}/(RT_{2})

(a) Further, you should givek_{2}E_{a}/ 1 1 \ ln ---- = - --- | --- - --- |k_{1}R \T_{2}T_{1}/

It is a good idea to manipulate the formula with symbols until you have obtained the desirable form before you substitute numerical values into it. The necessary units are included here to show you the derivation of units forT_{1}T_{2}Rk_{2}E_{a}= -------- ln --T_{2}-T_{1}k_{1}(350 K)(298 K) 8.314 J/(K mol) 0.0394 = ------------------------------- ln ------- (350 - 298)k0.00123 = 57811 J/mol = 57.8 kJ/mol

(b) To calculate *k* at 308 K,

57811 lnk= ln (0.00123) - ----- (1/308 - 1/298) 8.314 = -6.70 + 0.758 = -5.94k= 0.00263

*Discussion*

An increase of 10 *k* doubles the rate constant in this case.

If *E*_{a} is positive, increasing temperature always leads to an
increase in the rate constant.

**
Example 2
**

*Solution*

The statement of the problem is equivalent to the condition given:

*k*_{310} = 2 k_{300}

or *k*_{1} = 2 *k*_{0}, then using the same equation as you have used in the previous
example, you have

T_{1}T_{0}Rk_{0}E_{a}= -------- ln --T_{0}-T_{1}k_{1}310 300 Rk_{0}E_{a}= ------------ ln ---- 300 - 310 2k_{0}93000k8.314 (J / (mol K) = --------------------------- ln (0.5) - 10 = 53594 J / mol = 53.6 kJ / mol

*Discussion*

This question is intended to show the general magnitude of *E*_{a}
for the rule of thumb: For every 10 K increase in temperature, the reaction
rate doubles.

First order gas phase reactions:

N

OH + H

**The effect of temperature on chemical reaction is a manifestation of**

(a) the kinetics energies of molecules

(b) the activation energy of the reaction

(c) heat content of reactants and products

(d) the size of molecules.

**Discussion -**

Only a fraction of molecules have sufficient energy to react at a specified temperature. The distribution of kinetic energy amongst molecules and the required energy for molecules to react give rise to the activation energy. The activation energy quantitatively describes the effect of temperature on reaction rates.**If the activation energy is positive, an increase in temperature will always lead to**

(a) an increase in*reaction rate*

(b) a decrease in*reaction rate*

(c) no change in*reaction rate*

**Discussion -**

A positive*E*_{a}means increase rate at higher temperature.**For a spontaneous reaction the activation energy for the reverse reaction***E*_{ar}is,

(a) greater than

(b) the same as

(c) smaller than,

the activation energy*E*_{a}of the forward reaction.

**Discussion -**

Usually,*E*_{ar}>*E*_{a}.**For a reaction,***k*_{1}= 0.0503 /(M s) at 289 K and*k*_{2}= 6.71 /(M s) at 333 K. Calculate the activation energy of the reaction?

(R = 8.314 J/(mol K), give your answer in kJ/mol)

**Discussion -**

R

The data were given for the reaction*T*_{1}*T*_{2}*E*_{a}= ------- ln(*k*_{2}/*k*_{1})*T*_{2}-*T*_{1}

C_{2}H_{5}I + OH^{-}--> C_{2}H_{5}OH + I^{-}**For a reaction,***k*_{1}= 0.0503 /(M s) at 289 K and*k*_{2}= 6.71 /(M s) at 333 K.

What is the rate constant at 305 K?

(Use your results from the previous problem)**Discussion -**

*E*_{a}(*T*_{1}-*T*_{2}) ln*k*_{2}= ln*k*_{1}- ------------ R*T*_{1}*T*_{2}**If the rate for a particular reaction at 373 K is four (4) times faster than it was at 323 K, what is the activation energy? (R = 8.314 J/(mol K))**

(a) 27.8 kJ/mol

(b) -27.8 kJ/mol

(c) 27.8 J/mol

**Discussion -**

R

*T*_{1}*T*_{2}*E*_{a}= ------- ln(*k*_{2}/*k*_{1})*T*_{2}-*T*_{1}*E*_{a}must be positive if the rate increases at higher temperature.*E*_{a}= (ln(4)*8.314*373*323)/(50)