Chapter 6 – Principles of Reactivity: Energy
and Chemical Reactions
All atoms and molecules have energy:
• Atoms are constantly in motion, vibrating and rotating.
• Within each atom, electrons are constantly in motion.
This means that, in order to compare the energies of two
systems, both systems must contain exactly the same number
and types of atoms. (The exception to this rule is in nuclear
chemistry where we can compare systems with different types of
atoms as long as they have the same total mass and charge.)
Recall that when filling orbitals, we always put the electrons in
the lowest energy orbitals first. Atoms and molecules will
always try to attain the lowest energy possible:
LOWER ENERGY
= MORE STABLE ATOM OR MOLECULE
Often, we will be working with negative numbers for energies.
Recognize that –100 kJ is a lower energy than -10 kJ!
Energy of Reactions
It is possible to perform a reaction in which the products are less
stable than the reactants, but this will require input of energy
from an external source. This type of reaction is called
__________________.
A reaction in which the products are more stable than the
reactants will release energy to its surroundings. This type of
reaction is called _________________.
Note that, in both pictures, there is an ‘in the middle’ state that is
less stable (higher in energy) than either reactants or products.
This is the transition state.
The energy input required to get from reactants to the transition
state is called the _______________________.
The First Law of Thermodynamics states that the total energy
of the universe is constant:
Σ Ereactants
+
∆E
=
Σ Eproducts
where ∆E is energy added to the system from the surroundings.
In an endothermic reaction:
In an exothermic reaction:
Enthalpy
Enthalpy (H) is a major component of the energy of a substance.
It can be defined as the heat content of a substance at constant
temperature and pressure.
Σ Hreactants
+
In an endothermic reaction:
In an exothermic reaction:
∆H
=
Σ Hproducts
As with energy, we use enthalpy to compare systems containing
the same number and type of atoms. e.g. reactants vs. products
In order for us to be able to do this, we need a starting point. By
definition, the enthalpy of any element in its most stable form at
25 ˚C and 1 bar is 0 kJ. This is the element’s standard state.
The enthalpy of any other substance is defined by its standard
molar enthalpy of formation (∆Hf˚). This is the enthalpy
change for the reaction in which one mole of the substance is
made from elements in their standard state.
e.g. O2(g) is an element in its standard state.
Therefore, ∆Hf˚(O2(g)) = 0 kJ/mol
Na(g) is an element, but not in its standard state.
Na(s) → Na(g)
∆H˚ = 107.3 kJ/mol
Therefore, ∆Hf˚(Na(g)) = 107.3 kJ/mol
FeCl2(s) can be made from Fe(s) and Cl2(g).
Fe(s) + Cl2(g) → FeCl2(s)
∆H˚ = -341.79 kJ/mol
Therefore ∆Hf˚(FeCl2(s)) = -341.79 kJ/mol
HF(g) can be made from H2(g) and F2(g).
½ H2(g) + ½ F2(g) → HF(g)
∆H˚ = -273.3 kJ/mol
Therefore ∆Hf˚(HF(g)) = -273.3 kJ/mol
Which of the reactions above are endothermic?
Which of the reactions above are exothermic?
Compare the two reactions below:
A ½ H2(g) + ½ F2(g) → HF(g)
∆Hf˚(HF) = -273.3 kJ/mol
B
H2(g) + F2(g) → 2 HF(g)
∆Hf˚(HF) = -273.3 kJ/mol
Note that, because the standard enthalpy of formation is reported
in kJ per mole, both enthalpy values appear the same.
If we look at the total enthalpy of reaction, this is no longer the
case. In reaction A, 1 mole of product is made. In reaction B, 2
moles of product are made. It is logical that twice as much heat
will be released if twice as much product is made.
A ∆H˚rxn = -273.3 kJ/mol × 1 mol = -273.3 kJ
B ∆H˚rxn = -273.3 kJ/mol × 2 mol = -546.6 kJ
Note that the number of moles is an exact number with an
infinite number of significant figures.
To calculate the enthalpy of a reaction, subtract the ∆Hf˚ for all
reactants from the ∆Hf˚ for all products, remembering to factor
in the number of moles of each:
∆H˚rxn =
Σ n∆Hf˚(products)
–
Σ n∆Hf˚(reactants)
Since ∆Hf˚ for H2(g) and F2(g) are both 0 kJ/mol, this is what we
just did in the example above.
e.g. Propane burns according the reaction equation below.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
(a) Predict whether the enthalpy of reaction will be positive or
negative given that propane is used as a fuel.
(b) Calculate the
burned.
∆Hf˚(C3H8(g))
∆Hf˚(CO2(g))
∆Hf˚(H2O(g))
∆Hf˚(H2O(l))
enthalpy of reaction if 1 mole of propane is
=
=
=
=
-104.7 kJ/mol
-393.509 kJ/mol
-241.83 kJ/mol
-285.83 kJ/mol
(c) Calculate the heat released by burning 10 kg of propane.
Hess’s Law
If an overall reaction can be written as the sum of two or more
reactions, the enthalpy for the overall reaction is equal to the
sum of the enthalpies for each component reaction.
∆H˚(overall) =
Σ ∆H˚(steps)
e.g. Hydrazine (N2H4(l)) has been used as a rocket fuel because
it reacts very exothermically with oxygen:
N2H4(l) + O2(g) → N2(g) + 2 H2O(g)
∆H˚rxn = -524 kJ
Reaction of the nitrogen with more oxygen gives N2O5(s):
2 N2(g) + 5 O2(g) → 2 N2O5(s)
∆H˚rxn = -86 kJ
Calculate the enthalpy for the reaction between hydrazine
and oxygen to give nitrogen pentoxide and water.
Important Concepts from Chapter 6
• relationship between energy and stability
• energy diagrams (reactants, products, transition states)
• exothermic vs. endothermic
• first law of thermodynamics
• enthalpy calculations
o elements in standard state
o standard molar enthalpy of formation
o Hess’s law