Skills to develop
- Explain activation energy.
- Describe how energy vary during a chemical reaction.
- Plot chemical potential energy of the system as a function of the
- Explain the model k = A exp (Ea/RT) for chemical kinetics.
- Plot k versus 1/T when k is measured at several T.
- Calculate Ea when a series of k and T are given.
- Calculate k at certain T.
Energy of a Chemical System as the Reaction Proceeds
A spontaneous reaction usually releases energy. The mixture of
reactants has more energy than that of the products. This energy is
referred to as the chemical potential energy. The difference
between the potential energies is called the enthalpy of reaction.
A simple diagram illustrating this relationship is given below.
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.
The Activation Energy, Ea
We all know that a mixture of H2 and O2 will not react until its
temperature has reached the ignition point, despite the large amount of
energy released in the oxidation reaction. This phenomenon is best described
by the requirement of an
activation energy, Ea.
The relation between Ea and chemical potential energy in a reaction is given
Activation Energy and Enthalpy of Chemical Reaction
The rate constant k is affected by the temperature and this dependence may
be represented by the Arrhenius equation:
k = A e
where the pre-exponential factor A is assumed to be independent of
temperature, R is the gas constant, and T the temperature in K.
Taking the natural logarithm of this equation gives:
ln k = ln A - Ea/(RT)
ln k = -Ea/(RT) + constant
ln k = -(Ea/R)(1/T) + constant
These equations indicate that the plot of ln k vs. 1/T is a
straight line, with a slope of -Ea/R. These
equations provide the basis for the experimental determination of
The reaction constants k determined at 298 K and 350 K are
0.00123 /(M s) and 0.0394 /(M s) respectively.
(a) Calculate Ea.
(b) What is the rate constant at 308 K?
Let k1 and k2 be the rate constants determined at
T1 and T2, respectively.
Then you have two equations:
ln k1 = lnA - Ea/(R T1)
ln k2 = lnA - Ea/(R T2)
From these, you should be able to derive the following relationships,
k2 Ea / 1 1 \
ln ---- = - --- | --- - --- |
k1 R \ T2 T1 /
(a) Further, you should give
T1 T2 R k2
Ea = -------- ln --
T2 - T1 k1
(350 K)(298 K) 8.314 J/(K mol) 0.0394
= ------------------------------- ln -------
(350 - 298) k 0.00123
= 57811 J/mol
= 57.8 kJ/mol
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 for Ea.
(b) To calculate k at 308 K,
ln k = ln (0.00123) - ----- (1/308 - 1/298)
= -6.70 + 0.758
k = 0.00263
An increase of 10 k doubles the rate constant in this case.
If Ea is positive, increasing temperature always leads to an
increase in the rate constant.
For a particular reaction the rate constant doubles when the temperature
is raised by 10 K from 300 K. Calculate the activation energy.
The statement of the problem is equivalent to the condition given:
k310 = 2 k300
or k1 = 2 k0, then using the same equation as you have used in the previous
example, you have
T1 T0 R k0
Ea = -------- ln --
T0 - T1 k1
310 300 R k0
Ea = ------------ ln ----
300 - 310 2 k0
93000 k 8.314 (J / (mol K)
= --------------------------- ln (0.5)
= 53594 J / mol
= 53.6 kJ / mol
This question is intended to show the general magnitude of Ea
for the rule of thumb: For every 10 K increase in temperature, the reaction
Professor Frank L. Lambert of Occidental College has given an interesting
view of the chemical ideas
related to activation energy in his website about second law of
Arhenius Parameters for Some Reactions:
Not all chemical reactions agrees with the Arhenius model in terms of
temperature dependence, but the following reactions are known to agree
well with the Arhenius equation. Parameters in the Arhenius equation for
these reactions have been determined. (Note that 3e11 means 3x1011)
First order gas phase reactions:
N2O -> N2 + O; A = 8e11, Ea = 251 kJ/mol
N2O5 -> 2 NO + O2; A = 6e14, Ea = 88 kJ/mol
Second order gas phase reactions:
N2 + O -> N + NO; A = 1e11, Ea = 315 kJ/mol
OH + H2 -> 2 H2O + H; A = 1e11, Ea = 42 kJ/mol
Second order reactions in aqueous solution:
CO2 + OH- -> HCO3; A = 1e11, Ea = 315 kJ/mol
Confidence Building Questions
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.
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
A positive Ea means increase rate at higher temperature.
For a spontaneous reaction the activation energy for the reverse reaction
(a) greater than
(b) the same as
(c) smaller than,
the activation energy Ea of the forward reaction.
Usually, Ear > Ea.
For a reaction, k1 = 0.0503 /(M s) at 289 K and
k2 = 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)
R T1 T2
Ea = ------- ln(k2/k1)
T2 - T1
The data were given for the reaction
C2H5I + OH- --> C2H5OH + I-
For a reaction, k1 = 0.0503 /(M s) at 289 K
and k2 = 6.71 /(M s) at 333 K.
What is the rate constant at 305 K?
(Use your results from the previous problem)
Ea (T1 - T2)
ln k2 = ln k1 - ------------
R T1 T2
The experimental value is 0.37 /(M s)
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
R T1 T2
Ea = ------- ln(k2/k1)
T2 - T1
Ea must be positive if the rate increases at higher temperature.
Ea = (ln(4)*8.314*373*323)/(50)