1. Chemical Kinetics Temperature Dependence of Reaction Rates

2. This is the behaviour of k with temperature for many reactions.
Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent.
“k versus T”
This is the behaviour of k with temperature for many reactions.
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3. The curve can be fitted to a
Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent.
The curve can be fitted to a
y=aex
relationship.
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4. “k versus Temperature”
In 1887, Svante Arrhenius suggested that rate constants vary exponentially with the reciprocal of the absolute temperature.
“k versus Temperature”
“ln k versus 1/T”
k = b eaT
It turns out all reactions have rate constants with this equation, but with different constants a and b for each reaction.
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5. In 1887, Svante Arrhenius suggested that rate constants vary exponentially with the reciprocal of the absolute temperature.
Arrhenius’ Equation
(The above equation is purely empirical.)
Note:
R is the gas constant (8.314 J mole-1 K-1, 1.98 cal mole-1 K-1)
T is the temperature in Kelvin
Ea is the activation energy
2. RT is in units of energy per mole, thus,
Ea is also in units of energy per mole.
A has the same units as the rate constant, k.
A is a number that represents the likelihood that collisions would occur in a reaction.
A is the frequency factor
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6. Other forms of Arrhenius’ Equation
1. Take the natural logarithm of both sides of the equation
Linear form
y = m x b
When k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot of “ln k vs. 1/T”
Slope = -Ea / R
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7. Other forms of Arrhenius’ Equation
Example
Determine the activation energy for the
decomposition of N2O5 from the following
temperature dependence rate constant data.
k (s1)
Temperature (°C)
4.8x104
45.0
8.8x104
50.0
1.6x103
55.0
2.8x103
60.0
N2O5 (g)  2 NO2 (g) + ½ O2 (g)
Recall: Rate = k [N2O5]
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8. Other forms of Arrhenius’ Equation
2. For two temperatures,
Subtract the two equations
Two k and temperatures form
Note: A drops out of the equation when it is in this form.
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9. Other forms of Arrhenius’ Equation
Example
Determine the rate constant at 35°C for the hydrolysis of sucrose, given
that at 37°C it is 0.91 mL mole-1 sec-1. The activation energy of this reaction
is 108 kJ/mol.
Calculation shows that:
k2/k1 = 1.31 An increase of 2oC increases the reaction rate by 31%.
k1, the rate constant at 35oC is L mole-1 sec-1.
Rate constant increases when T2>T1
k2 = 0.91 mL mole-1 sec-1
= 9.1 x 10-4 L mole-1 sec-1
T2 = 37oC = K
T1 = 35oC = K
Ea = 108 kJ mole-1
Calculate k1
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10. Arrhenius’ Equation
What else can we get from the slope of Arrhenius’ plot?
High activation energy corresponds to a reaction rate that is very sensitive to temperature (slope of Arrhenius graph is steep)
Slope = -Ea / R
Small activation energy indicates a rate that varies only slightly with temperature.
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11. “k versus Temperature”
Why does the temperature dependence obey the relationship shown below?
“k versus Temperature”
“ln k versus 1/T”
k = b eaT
Is there any physical insights into the order of the reaction or the meaning of the constants (Ea/R) and A in this dependency?
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12. Second Order Reactions
Magnitudes of A and Ea
First Order Reactions A
(s-1) Ea
kJ/mole)
cyclopropene  propane
1.58 x 1015
272
2 N2O5  4 NO2 + O2
4.94 x 1013
103
N2O  N2 + O
7.94 x 1011
250
Second Order Reactions A
(L mole-1s-1) Ea
kJ/mole)
Sucrose in acidic water
1.50 x 1015
107.9
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13. As Ea increases, k decreases.
Summary
Arrhenius Equation
In natural logarithmic form,
y = m x b
Determine Ea from the slope.
Determine A from the intercept.
As Ea increases, k decreases.
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