R2.2.12 – Activation
Energy
Essential idea: The activation energy of a
reaction can be determined from the effect of
temperature on reaction rate.
Activation Energy
• The rate constant k is temperature-dependent
• The relationship between temperature and the rate
constant is described by the Arrhenius equation.
• This equation can be found in section 1 of the IB data booklet:
•
•
•
•
•
k=Ae−Ea/RT
A is the Arrhenius constant, also known as the frequency factor.
Ea is the activation energy (J mol−1).
T is the absolute temperature in kelvin (K).
R is the universal gas constant (8.31 J K−1 mol−1).
e is a mathematical constant, also known as Euler’s number.
Activation Energy
•
−Ea/RT
k=Ae
The Arrhenius constant, A, takes into account the
frequency of collisions and the probability that they
have the correct orientation (or geometry).
• The expression (e−Ea /RT) (the exponential factor) is the
fraction of molecules that have sufficient kinetic
energy to react at a certain temperature.
• Using this expression, we can see that an increase in
temperature will increase the value of the rate
constant k and therefore increase the rate of reaction.
Graphical determination of Ea
• We can rewrite the Arrhenius equation by taking the natural logarithm
(ln) of both sides:
𝑬𝒂
𝐥𝐧 𝒌 = 𝐥𝐧 𝑨 −
𝑹𝑻
• This equation can be rewritten to show even more clearly the
relationship to the equation for a straight line (y = mx + b):
𝒍𝒏 𝒌 =
−𝑬𝒂
𝑹
𝟏
𝑻
𝒚=
−𝑬𝒂
𝑹
𝒙 + 𝒍𝒏 𝑨
+ 𝒍𝒏 𝑨
Graphical determination of Ea
• A graph of ln k against
𝒚=
1/T gives a straight line.
• The slope of the line is
equal to –Ea/R.
• The y-intercept is equal
to ln A.
−𝑬𝒂
𝑹
𝒙 + 𝒍𝒏 𝑨
Graphical determination of Ea
• The activation energy can be calculated as follows:
slope = −Ea/R
Ea = - slope × R
• Note that if the units of temperature are kelvin (K) and the
units of the gas constant are J K−1 mol−1, the Arrhenius plot
gives the activation energy in J mol−1.
• The slope of the line has the units of kelvin, K.
• Once we have determined Ea for the reaction, we can
calculate the Arrhenius constant, A, by substituting Ea into
the Arrhenius equation.
Solving Simultaneous Equations
• The activation energy for a reaction can also be
determined from the value of the rate constant, k1
and k2, at two different temperatures T1 and T2.
• The equation in this format can also be found in
section 1 of the IB chemistry data booklet.
Solving Simultaneous Equations
• The activation energy for a reaction can also be
determined from the value of the rate constant, k1
and k2, at two different temperatures T1 and T2.
• The equation in this format can also be found in
section 1 of the IB chemistry data booklet.
Solving Simultaneous Equations
327 °C: k = 2.75 × 10−8 mol−1 dm3 s−1
527 °C: k = 1.95 × 10−7 mol−1 dm3 s−1
2.75𝑥10 − 8
𝐸𝑎 1
1
ln
=
(
−
)
1.95𝑥10 − 7 8.31 800 600
𝑬𝒂 = 𝟑𝟗, 𝟎𝟓𝟒 𝑱 = 𝟑𝟗. 𝟏 𝒌𝑱