The Arrhenius equation relates the rate constant, k, to the activation
energy of a reaction, Ea. It affords a means of determining activation
energies by experimentally finding the rate constant at different temperatures.
Syllabus reference R2.2.12
Reactivity 2.2.12 - The Arrhenius equation uses the temperature dependence of the rate constant to determine the activation energy. (HL)
- Describe the qualitative relationship between temperature and the rate constant.
- Analyse graphical representations of the Arrhenius equation, including its linear form.
Guidance
- The Arrhenius equation and its linear form are given in the data booklet.
The Arrhenius equation
The actual dependence of the rate constant on temperature is given by the Arrhenius
equation.
Where:
- k is the rate constant
- A is the Arrhenius factor (different for every reaction)
- e is the natural log base
- Ea is the minimum energy required for a reaction to take place (known as
the activation energy)
- R is the universal gas constant (8.31 J kg-1 ºC-1)
- T is the absolute temperature in Kelvin
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Arrhenius' constant, A - the pre-exponential factor
It is possible for colliding particles to possess enough energy for reaction,
but still not have a successful collision (one that results in reaction).
This is accounted for by the Arrhenius constant 'A' , also called the pre-exponential
or frequency factor.
Imagine a collision between two cars; clearly more damage is going to be
caused by a head on collision than a glancing scrape.
The Arrhenius constant (pre-exponential or frequency factor) is a number
between 0 and 1, that reflects the proportion of successful collisions amongst
those particles with enough energy for reaction.
For example, when A is very small, only a small proportion of collisions
lead to reaction, regardless of the energy. When A = 1, all collisions with
sufficient energy cause reaction.
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Using the Arrhenius equation
In reality, the basic form of the Arrhenius equation is not very convenient
for graphing or analysing date. To analyse experiments at different temperatures
we usually use the natural log form of the equation:
k = Ae-Ea/RT
taking natural logs throughout this gives:
lnk = lnA - Ea/RT
Thus a plot of lnk against 1/RT, 1/T or any variation, will allow us to find
the activation energy of a specific reaction as a function of the gradient,
and the Arrhenius constant as a function of the intercept to the y axis.
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A typical plot used to calculate the activation energy from the Arrhenius
equation.
In this graph the gradient of the line is equal to -Ea/R
Extrapolation of the line to the y axis gives an intercept value of lnA
When the temperature is increased the term Ea/RT gets smaller. This means
in turn, that the term e-Ea/RT gets bigger.
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Simultaneous equations
Alternatively two results may be analysed simultaneously to obtain values for
Ea, the activation energy and the Orientation factor, A. This is not particularly
reliable as only two values for the rate constant are used at two different
temperatures. This can introduce large errors because of too little data.
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Example: Calculate
the rate constant when T = 300K (A = 0.3, Ea = 50kJ mol-1)
k = Ae-Ea/RT
Ea/RT = 50000/(8.314 x 300) = 20.05
e-Ea/RT = 1.97 x 10-9
k = Ae-Ea/RT
k = 5.90 x 10-10
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Example: The rate of a reaction A(g) + B(g)
→ C(g) + D(g) has
been studied as a function of temperature between 5000 and 18000 K. The
following data were obtained for the rate constant:
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T (K)
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k(mol dm-3 /sec)
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5000
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5.49 x 106
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10000
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9.86 x 108
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15000
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5.57 x 109
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18000
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9.92 x 109
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Calculate the activation energy for the reaction.
k = Ae-Ea/RT and lnk = lnA - Ea/RT
from expt 1: →
ln(5.49 x 106) = lnA - Ea/ (8.314x5000)
from expt 2: →
ln(9.86 x 108) = lnA - Ea/ (8.314x10000)
Subtract one from the other (to remove lnA)
ln(9.86 x 108) - ln(5.49 x 106) = -Ea/ (8.314x10000) + Ea/ (8.314x5000)
10000[ln(9.86 x 108) - ln(5.49 x 106)] = -Ea/8.314 + 2Ea/8.314
8.314 x 10000[ln(9.86 x 108) - ln(5.49 x 106)] = Ea
8.314 x 10000 [20.709 - 15.518] = Ea
8.314 x 10000 x 5.191 = Ea
431579.74 = Ea
Ea = 432 kJ
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Note: Use of the simultaneous equation method is no longer required for first examinations 2025
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Worked examples
Q641-01 The first-order reaction:
2N
2O(g)
→ 2N
2(g)
+ O
2(g), Has a rate constant of 1.3 x 10
-11 s
-1 at
270°C, and 4.5 x 10
-10 s
-1 at 350°C. What is the
activation energy for this reaction?
Answer
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k = Ae-Ea/RT and lnk = lnA - Ea/RT
ln(1.3 x 10-11) = lnA - Ea/(8.314 x 543)
and ln(4.5 x 10-10) = lnA - Ea/(8.314 x 623)
Subtract one from the other (to remove the terms lnA)
ln(4.5 x 10-10) - ln(1.3 x 10-11) = - Ea/(8.314 x 623) +
Ea/(8.314 x 543)
8.314[ln(4.5 x 10-10) - ln(1.3 x 10-11)] = Ea/543 - Ea/623
8.314[ln(4.5 x 10-10) - ln(1.3 x 10-11)] = (623Ea - 543Ea)/
(623 x 543)
8.314 x 623 x 543 [ln(4.5 x 10-10) - ln(1.3 x 10-11)]
= 80Ea
2812534.7[-21.52 + 25.07] = 80Ea
(2812534.7 x 3.5449)/80 = Ea
Ea = 124806
Ea = 124.8kJ mol-1
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Q641-02 What is the activation
energy for a reaction, if its rate doubles when the temperature is raised from
20°C to 35°C?
- 342 kJ mol-1
- 269 kJ mol-1
- 34.7 kJ mol-1
- 15.1 kJ mol-1
Answer
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If the rate doubles the rate constant doubles
when T = 293K the value for k is half that for T = 308K
thus: Ae-Ea/308R = 2 x Ae-Ea/293R
lnA - Ea/308R = ln2 + lnA -Ea/293R
rearrange
ln2 = Ea/293R - Ea/308R
ln2 x 8.314 = Ea/293 - Ea/308
ln2 x 8.314 x 308 x 293 = 15Ea
519950 = 15 Ea
Ea = 34663J
Ea = 34.7kJ
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Q641-03 Two reactions with different
activation energies have the same rate at room temperature. Which statement correctly
describes the rates of these two reactions at the same higher temperature?
- The reaction with the greater activation energy will be faster.
- The reaction with the smaller activation energy will be faster.
- The two reactions will have the same rates.
- A prediction cannot be made without further information.
Answer
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The question is asking you to consider the effect of changing T on
the rate constant in the Arrhenius equation.
By considering the Maxwell - Boltzmann distribution:
Increasing the temperature has a greater effect on reactions with lower
activation energy
Shown mathematically:
k1 = A1e-Ea1/RT
k2 = A2e-Ea2/RT
The term e-Ea/RT has a smaller value when Ea/RT is larger
For different values of Ea, the smaller Ea will give the smaller term
Ea/RT for any given temperature and hence a larger value for the term
e-Ea/RT. This in turn provides a higher value for the rate
constant.
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Q641-04 The rate of an elementary
reaction: A(g) + B(g)
→ C(g)
+ D(g) has been studied as a function of temperature between 5000 and 18000 K.
The following data were obtained for the rate constant:
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T (K)
|
k (mol dm-3/sec)
|
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5000
|
5.49 x 106
|
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10000
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9.86 x 108
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15000
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5.57 x 109
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18000
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9.92 x 109
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Calculate the activation energy for the reaction.
- 25.9 kJ/mol
- 52.0 kJ/mol
- 359 kJ/mol
- 432 kJ/mol
Answer
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Using Arrhenius k = Ae-Ea/RT
For reaction 1: 5.49 x 106 = Ae-Ea/5000R, therefore
A = 5.49 x 106 /e-Ea/5000R
For reaction 2: 9.86 x 108 = Ae-Ea/10000R, therefore
A = 9.86 x 108 / e-Ea/10000R
As A is a constant, 5.49 x 106 /e-Ea/5000R =
9.86 x 108 / e-Ea/10000R
Rearrange: (5.49 x 106)/(9.86 x 108) = (e-Ea/5000R)/(e-Ea/10000R)
Take natural logs: -5.19072 = (-Ea/5000R) - (-Ea/10000R)
Rearrange: -5.19072 = -Ea/10000R
Rearrange: Ea = 431557 = 432 kJ
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Q641-05 What happens to the
rate constant (k) and the activation energy (Ea) of a reaction when the temperature
is increased?
- k increases and Ea is unaffected
- k decreases and Ea is unaffected
- Ea increases and k is unaffected
- Ea decreases and k is unaffected
Answer
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Increasing temperature has no effect on the activation energy, but
it does increase the rate constant. Therefore k
increases and Ea is unaffected
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Q641-06 For a given reaction
in which the activation energies of the forward and reverse reactions are equal
then:
- the equilibrium constant must equal one.
- the rate law can be determined from the stoichiometric equation.
- the overall order must be zero.
- ΔH must equal zero.
Answer
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If the forward and back activation energies are equal and the value
is equal to the difference between the energy level of the reactants
(or products) and the highest energy state of the reaction pathway,
then the enthalpy of reactants must equal the enthalpy of products.
This is probably easier to see on a reaction profile graph.
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The only way for Eaf to equal Eab in the graph is
for the enthalpy values of reactants to be the same as the products. |
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Q641-07 To what does 'A' in
the Arrhenius equation k = Ae
-Ea/RT refer?
- Activation energy
- Rate constant
- Gas constant
- Collision geometry
Answer
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The constant 'A' in the Arrhenius equation is the orientation factor,
or collision geometry, a number from
0 to 1 representing the probablility of a collision being successful
(assuming sufficient energy).
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Q641-08 Values of a rate constant,
k, and absolute temperature, T, can be used to determine the activation energy
of a reaction by a graphical method. Which graph produces a straight line?
- k versus T
- k versus 1/T
- ln k versus T
- ln k versus 1/T
Answer
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Using the log form of the Arrhenius equation ln k = ln A -Ea/RT, and
plotting ln k against 1/T gives a
graph of the form y = mx + c. This is a straight line graph of gradient
= -Ea/R and y intercept = ln A.
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Q641-09 A student carried out
a series of experiments to find the rate constant, k, for a reaction at 300K,
400K, 500K and 600K. He plotted the following graph as data processing. Using
the graph, which of the following is the most probable activation energy for this
reaction?
- 5.15 kJ mol-1
- 1.51 kJ mol-1
- 11.5 kJ mol-1
- 55.1 kJ mol-1
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From the graph of ln k against 1/T gives a gradient = -Ea/R.
The gradient can be estimated to equal approximately 1/0.0007 = -1250
Therefore Ea = 1250 x 8.314 = approximately 10 kJ
Therefore the most likely answer is 11.5
kJ mol-1
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Q641-10 What is the activation
energy (in kJ) of a reaction whose rate constant increases by a factor of 100
upon increasing the temperature from 300 K to 360 K?
- 27
- 35
- 42
- 53
- 69
Answer
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Using the Arrhenius equation, k = Ae-Ea/RT
If k(at 360K) = 100 x k(at 300K), then Ae-Ea/RT(at 360K)
= 100 x Ae-Ea/RT(at 300K)
Take log base e: ln A - Ea/360R = ln 100 + ln A - Ea/300R
Ea/300R - Ea/360R = ln100 = 4.605
Multiply through by 300R: Ea - 5Ea/6 = 4.605 x 300 x 8.314
Multiply through by 6: 6Ea - 5Ea = 68915 J
Therefore Ea = 68.9 kJ
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