Chapter 6 – Principles of Reactivity: Energy
and Chemical Reactions
Chemical Thermodynamics
Thermodynamics is defined as the study of energy. It reaches
across the different physical sciences, and you will no doubt
encounter thermodynamics in physics as well. We will be
focusing on thermodynamics from the chemist’s perspective.
Chemical thermodynamics is the study of how far reactions will
progress. Equilibrium is a key aspect of thermodynamics – the
“measuring how far” part. In Chapters 6 and 19, we will be
looking at the “explaining why the equilibrium lies where it
does” part of thermodynamics. We do this by comparing the
energy of the reactants with the energy of the products (“lower
is better”). In high school, you focused your study of energy on
enthalpy. In Chemistry 2000, we will also factor in entropy,
allowing us to calculate and compare free energy at different
temperatures.
The term spontaneous is used to indicate the direction in which
a reaction is favoured. A product-favoured reaction is called
“spontaneous” while a reactant-favoured reaction is called
“nonspontaneous”. ***This term gives no information about a
reaction’s speed/kinetics; it just means the reaction can occur
without a net input of outside energy.***
e.g. Formation of green copper salts on brass statues/roofs is a
spontaneous process by the chemical definition – but it can
take months! Reaction between NaOH and HCl in a
titration is also spontaneous – but much faster.
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!
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
+
∆H
=
Σ Hproducts
In an endothermic reaction:
In an exothermic reaction:
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.
Heat Flow and Calorimetry
Calorimetry is the science of heat measurement. There are two
different types of calorimetry:
• constant-pressure calorimetry (in “coffee cup calorimeter”)
• constant-volume calorimetry (in “bomb calorimeter”)
In both types, the heat produced/consumed by a reaction is
determined by measuring temperature change. The temperature
change can be related to heat produced/consumed using the
specific heat capacity of the heat-absorbing medium:
q = m C ∆T
where q is heat (in kJ), m is mass of absorbing medium (in kg),
C is specific heat capacity (in kJ·kg-1·K-1) and ∆T is temperature
change (in K).
Specific heat capacity is a measure of how easily a compound
changes temperature. A compound with a high heat capacity
must absorb a lot of energy to increase its temperature by 1 K
(and therefore lose a lot of energy to decrease by 1 K). A
compound with a low heat capacity need only absorb a small
amount of energy to increase its temperature by 1 K (and
therefore lose a small amount of energy to decrease by 1 K).
Heat capacities are known for many common substances:
e.g. copper 0.385 kJ·kg-1·K-1
glass
0.8 kJ·kg-1·K-1
air
~1.0 kJ·kg-1·K-1
water
4.184 kJ·kg-1·K-1
Alternately, it is sometimes easier to calibrate a calorimeter to
find its heat capacity (and use that value for m·C in calculations).
By convention, we view energy changes from the perspective of
the system.
• If a system gains energy from its surroundings, ∆E > 0
• If a system loses energy to its surroundings, ∆E < 0
In calorimetry, the system is self-contained, so there is no
transfer of energy to the surroundings. As such:
q absorbed by calorimeter + q released by reaction = 0
q absorbed by calorimeter = - q released by reaction
• If the calorimeter absorbs heat (q absorbed > 0), the reaction
inside was exothermic (q released < 0).
• If the calorimeter releases heat (q absorbed < 0), the reaction
inside was endothermic (q released > 0).
If there are multiple substances absorbing heat, the heat
absorbed by each must be calculated. Adding them all together
gives q absorbed.
e.g. Sulfur (S8(s), 2.56 g) is burned in a bomb calorimeter with
excess oxygen (O2(g)). The temperature increases from
21.25 ˚C to 26.72 ˚C. The bomb has a heat capacity of 923
J/K, and it contains 815 g water. Calculate the heat
evolved per mole of SO2(g) formed (“heat of formation”!).
In constant volume calorimetry, all of the energy produced by
a reaction is transferred as heat:
∆E = qv
In constant pressure calorimetry, some of the energy
produced by a reaction is used to do work; the rest is transferred
as heat:
∆E = qp + w
Work is the application of a force over a distance. Increasing
the volume of a coffee-cup calorimeter qualifies as work
because a force moves the lid. The greater the external pressure,
the greater the force that must be applied to do this work.
w = - P∆V
Therefore:
∆E = qp - P∆V
Again, we view work from the perspective of the system’s total
energy.
• If a system does work on its surroundings, its energy
decreases: ∆E < 0 therefore w < 0
• If a system has work done on it by its surroundings, its
energy increases: ∆E > 0 therefore w > 0
By definition, heat change at a constant pressure is enthalpy:
qp = ∆H
Therefore:
∆E = ∆H – P∆V
or
∆H = ∆E + P∆V
• At the introductory level, we generally consider that ∆H ≈
∆E and don’t correct for pressure-volume work.
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 dinitrogen pentoxide and water.
e.g. The reactions below are two of the key reactions in the
processing of uranium for use as fuel in nuclear power
plants:
UO2(s) + 4 HF(g) → UF4(s) + 2 H2O(g)
UF4(s) + F2(g) → UF6(g)
.
(a) Calculate the enthalpy change for each reaction.
(b) Calculate the enthalpy change for the overall process:
UO2(s) + 4 HF(g) + F2(g) → UF6(g)
+ 2 H2O(g)
compound
UO2(s)
UF4(s)
UF6(g)
HF
H2O(g)
∆Hf˚
(kJ/mol)
-1085
-1914
-2147
-271.1
-241.818
Important Concepts from Chapter 6
• relationship between energy and stability
• energy diagrams (reactants, products, transition states)
• exothermic vs. endothermic
• system vs. surroundings (including sign conventions for
energy and work)
• calorimetry and specific heat capacity
• constant-volume vs. constant-pressure calorimetry
• enthalpy calculations
o elements in standard state
o standard molar enthalpy of formation
o Hess’s law