16.3 Activation Energy
By the 1800's chemists had enough experimental evidence to show
that only a small fraction of the actual collisions that took place
produced a reaction. The Swedish chemist Svante Arrhenius
proposed an explanation called activation energy, Ea the minimum
amount of energy required by the reactants before collision will
cause the products to form.
The activation energy is determined by measuring the effect of temperature on the rate of the
reaction. At a higher temperature there is a greater proportion of reactants with the required
activation energy (E ≥ Ea), increasing the rate of the reaction. Changing the temperature does not
however change the activation energy. It only changes the frequency of collisions and the
proportion of reactants with the kinetic energy, E that is greater than or equal to the activation
energy, Ea (E ≥ Ea). Arrhenius proposed an equation to represent the proportion of molecules with E
≥ Ea.
Activation energy = Proportion of molecules = e ( -Ea/RT )
Total area under curve
This mathematical model developed into now what is called the Arrhenius equation.
k = A e ( -Ea/RT )
this equation is in the data booklet
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Where:
A is the Arrhenius constant or frequency factor. It is related to the
collision frequency and the probability that the molecules have the
correct orientation/geometry at the point of collision
Ea is the activation energy in Jmol-1
R is the gas constant (8.314 JK-1mol-1)
T is the absolute temperature (K)
k is the rate constant (s-1)
As a general rule each 10°C change in temperature doubles the rate of the reaction and the rate
constant, k. For example if the activation energy for a reaction is 50 000 Jmol-1 the rate constant, k
will double if the temperature increases by 10°C from 15°C to 25°C.
To prove this, first the temperature needs to be converted to Kelvin. The Ea needs to be in Jmol-1 by
R the gas constant is measured in JK-1mol-1.
15°C = 15 + 273
= 288 K
25°C = 25 + 273
= 298 K
at 10°C
at 20°C
k
k
=
A e ( -Ea/RT )
=
A e (-50000 Jmol-1 ÷ (8.314 JK-1mol-1 x 288K)
=
A e ( -Ea/RT )
=
A e (-50000 Jmol-1 ÷ (8.314 JK-1mol-1 x 293K)
Ratio k(20°C) / k(10°C)
=
A e (-50000 Jmol-1 ÷ (8.314 JK-1mol-1 x 298K)
A e (-50000 Jmol-1 ÷ (8.314 JK-1mol-1 x 288K)
=
e (-20.18)
e (-20.88)
=
2.00
(A cancels because it is constant)
When the temperature doubles the rate constant k doubles. This can be explained using
deduction:
Premise 1: When the temperature increases, RT increases
Premise 2: As RT increases, Ea/RT decreases,
Premise 3: As Ea/RT decreases, –Ea/RT becomes less negative
Premise 4: As –Ea/RT becomes less negative, e ( -Ea/RT ) increases
Therefore since k ∝ e ( -Ea/RT ), as e ( -Ea/RT ) increases so does k proportionally
Alternatively when the temperature increases more of the reactants have E ≥ Ea therefore the rate
increases. Since rate = k [ ] and the concentration remains constant, k must increase to increase
the rate of the reaction.
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The calculus form of the equation is often used. If you take the natural log of the Arrhenius
equation:
ln k = - Ea + ln A
RT
The equation is in the data booklet
And then plot ln k versus 1/T (called an Arrhenuis plot) a straight line is obtained.
Slope of line = ∆ ln k (K)
∆ 1/T
The activation energy can be determined from the slope/gradient of the line using:
Slope of line = - Ea
R
The intercept of the y axis (ln k) when the Arrhenius plot is extrapolated back to the y axis gives ln
A. The arrhenius constant, A can the be found from the inverse of ln A or in other words A = eintercept
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Exercises
1.
Which one of the following is not true about the activation energy of a reaction?
A
It can never be negative
B
It is reduced by the addition of a catalyst
C
It is the minimum amount of energy that the reactants must have in order to form
the products
D
The greater the activation energy the greater the rate of the reaction
2.
The activation energy of a chemical reaction can be determined by measuring the effect on
reaction rate of varying the:
A
temperature
B
concentration of the reactants
C
concentration of the catalyst
D
surface area in contact
3.
(N98) Dinitrogen oxide is moderately stable compound used as an anesthetic. However, at
high temperature it decomposes according to the following equation.
N2O (g) N2 (g) + ½ O2 (g)
The rate expression for the reaction is
∆H = - 76 kJmol-1
R = k [N2O(g)]2
Explain qualitatively why the value of the rate constant, k increases with increasing
temperature.
[2]
4.
N99/420/(H)2M
(a)
The rate constant, k, for any reaction is related to the activation energy, Ea, by the
Arrhenius equation.
i)
Explain the meaning of the terms: rate constant and activation energy. [2]
ii)
A particular reaction is found to be second order overall. Specify the units for
both the rate of the reaction and the rate constant. [2]
(b)
The rate constant, k was determined for the decomposition of hydrogen iodide at
various temperatures. The results given in lnk for the range of temperatures are
given below.
Temperature T / K
550
600
650
700
750
i)
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lnk
- 15.6
- 12.2
- 9.4
- 7.0
- 4.9
Temperature-1 (1/T) / K-1
1.82 x 10-3
1.67 x 10-3
1.54 x 10-3
1.43 x 10-3
1.33 x 10-3
Plot a graph of lnk against T-1 (1/T). (Take the lnk axis from -20 to 0 and the T1
axis from 1.2 x 10-3 to 1.9 x 10-3 K-1)
[4]
5.
ii)
Calculate the gradient (slope) of the graph and its units and use it to
determine a value for the activation energy, Ea, stating its units. [5]
iii)
Without obtaining the actual value, state two different ways in which the
value of A in the Arrhenius equation could be determined.
[2]
Describe how you could determine the activation energy for the decomposition of
hydrogen peroxide given experimental data on how the rate constant changes with
increasing temperature. State two different ways in which the value of A in the
Arrhenius equation could be determined.
Bibliography
Clark, Jim. Chem Guide. 2008. <http://www.chemguide.co.uk/>.
Clugston, Michael and Rosalind Flemming. Advanced Chemistry. Oxford: Oxford University Press, 2000.
Derry, Lanna, et al. Chemistry for use with the IB Diploma Programme Higher Level. Melbourne: Pearson
Heinemann, 2009.
Derry, Lanna, Maria Connor and Carol Jordan. Chemistry for use for the IB Diploma Standard level.
Melbourne: Pearson Education, 2008.
Green, John and Sadru Damji. Chemistry for use with the International Baccalaureate Programme.
Melbourne: IBID Press, 2007.
Neuss, Geoffrey. IB Diploma Programme Chemistry Course Companion. Oxford: Oxford University Press,
2007.
—. IB Study Guides, Chemistry for the IB Diploma. Oxford: Oxford University Press, 2007.
Organisation, International Baccalaureate. Online Curriculum Centre.
<http://occ.ibo.org/ibis/occ/guest/home.cfm>.
—. "Chemistry Data Booklet." International Baccalaureate Organisation, March 2007.
—. "Chemistry Guide." International Baccalaureate Organisation, March 2007.
—. "IB Chemistry Examination Papers ." Cardiff: International Baccalaureate Organisation, 1999-2008.
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16.3
1.
Activation Energy
D
ANSWERS
2.
A
3.
When the temperature increases e-Ea/RT in the equation k = A e ( -Ea/RT ) increases;
When the temperature increases by 10°C e-Ea/RT in the equation
k = A e ( -Ea/RT ) increases causing k to double.
causing k to increase;
or
Rate = k [N2O (g) ]2
For the same concentration , as T increases, the rate increases;
Therefore k must have increased;
4. a)
(i)
rate constant, k - It relates the rate of the reaction to the concentration of the reactants in a
rate expression. The rate constant of a reaction increases with increasing temperature.
Rate of reaction, R = k [ A]m x [ B]n
Rate expression / rate law
/ rate equation
Activation energy - the minimum amount of energy required by the reactants before
collision will cause the products to form.
(ii)
For a second order reaction the rate expression could be:
R = k [ A]2 or
Units for rate of reaction – moldm-3s-1
Units for k – mol-1dm3s-1
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R = k [ A] x [ B]
4 b)
(i)
(ii)
Slope = ∆ ln k =
∆ 1/T
- (15.0 – 5.3)
=
(1.79x10-3 – 1.35x10-3)
2.20 x 104 K
unit of the slope
= ∆ ln k
∆ 1/T
= K
=
1
K-1
Slope of line = - Ea
R
- Ea
=
slope x R
- Ea
=
- 2.20 x 104 K x 8.314 JK-1mol-1
- Ea
=
- 182980 Jmol-1
Ea
=
182980 Jmol-1
Ea
=
183 kJmol-1 (range of answers: Ea = 182, 180, 181, 179, 178)
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5.
A number of separate experiments with different initial concentrations of each reactant and
the initial rate of each reaction determined
k is found from the rate expression: R = k [ A]m x [ B]n
The order of the reaction is determined from the experimental data. Rate =
[concentration of the reactant]order
Method 1:
Once k has been determined
Plot ln k versus 1/T.
Extrapolate the line back to the y axis. ln k = ln taking the inverse of ln A will give you A.
Slope of line = ∆ ln k
∆ 1/T
The activation energy can be determined using:
Slope of line = - Ea
R
method 2
If the temperature and k are known
Plot ln k versus 1/T.
Extrapolate the line back to the y axis. ln k = ln taking the inverse of ln A will give you A.
Rearrange
ln k = - Ea + ln A to find Ea
RT
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