16.3
- Activation energy
16.3.1: Describe qualitatively the relationship between
the rate constant (k) and temperature (T).
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Effect of temperature on the reaction rate
Direct observations make it clear that increasing the temperature increases
the rate of a chemical reaction. In approximate terms, most reactions
double in rate for a ten degree increase in temperature.
This effect was first quantified by Arrhenius who produced the equation:
| rate constant k = Ae-Ea/RT |
| quantity |
description |
meaning |
| A - |
the Arrhenius constant |
This is a measure of the proportion of molecules that
collide with enough energy to react AND which have the correct orientation
for successful collision. |
| e - |
the natural number on which the natural logarithms is
based |
2.303 |
| Ea - |
the activation energy |
This is the minimum energy that a molecular collision
must have before it can be successful and lead to reaction - units
kJ mol-1 |
| R - |
the universal gas constant |
8.314 in SI units |
| T - |
the Absolute temperature in Kelvin (K) |
Equal to the temperature in Celsius + 273 |
16.3.2 Determine activation energy values from the
Arrhenius equation by a graphical method.
The Arrhenius equation: and its logarithmic form are provided in the chemistry
data booklet . Use of simultaneous equations will not be assessed.
Using the Arrhenius equation
Although it is not easy to see the relationship between the rate constant
and the absolute temperature from the equation, if we break it down into
steps perhaps it will help.
- The temperature appears in the term Ea/RT
- If T increases then the term Ea/RT gets smaller
- However in the Arrhenius equation Ea/RT has a negative value, therefore
as T increases Ea/RT gets LESS negative.
- So as -Ea/RT is the power to which 'e' is raised then the term e-Ea/RT
gets larger (as the power gets less negative) as the temperature increases.
- The rate constant is directly proportional to the term e-Ea/RT
and so the rate constant gets larger as T gets larger.
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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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