Spontaneity & Entropy
Unit 3 - Thermodynamics
Spontaneity
• In chemical terms, a spontaneous reaction is a
reaction that occurs on its own.
• Speed is not an issue (that is it can happen
fast or slowly).
Spontaneity
• Here are some examples of
spontaneous chemical
reactions:
Na(s) + ½Cl2(g) → NaCl(s) + 411.2 kJ
H2(g) + ½O2(g) → H2O(g) + 242 kJ
• Note that these are all
exothermic. Energy, like
boulders, tend to run
downhill.
• The lower energy state of the
products is more stable,
which is why exothermic
reactions tend to be
spontaneous.
Spontaneity
• Some endothermic reactions are also
spontaneous:
Br2(l) + Cl2(g) + 29.3 kJ → 2 BrCl(g)
H2(s) + ½O2 + 6.01 kJ → H2O(l)
• However, the energy needed for these two
reactions is very little.
• If a reaction is highly endothermic (a lot of
energy is required) the reaction will not be
spontaneous.
Stability of Compounds
• The stability of a compound refers to how
readily it breaks down.
• Therefore, the reverse of a heat of formation
reaction will tell us how readily a compound
breaks down.
• That is, we can tell how stable a compound is
by looking at the ΔH°f value.
Stability of Compounds –
Situation 1
• If ΔHf is large and negative
• These compounds give off a lot of energy when they
form.
• For example:
• Mg(s) + ½O2 → MgO(s) + 601.7 kJ
• The decomposition of MgO(s) requires a lot of heat
to break down into its component elements.
• MgO(s) + 601.7 kJ → Mg(s) + ½O2
• Because of this, we say that MgO(s) is stable.
Stability of Compounds –
Situation 2
• If ΔHf is small and negative
• These compounds give off little energy when
formed. Hence, when they decompose they require
little energy to do so.
• For example, the formation of HBr(s) is:
• ½ H2(g) + ½ Br2(g) → HBr(g) + 36.4 kJ
• Therefore, the decomposition of HBr(s) is:
• HBr(g) + 36.4 kJ → ½ H2(g) + ½ Br2(g)
• Since little energy is required for HBr to break down
spontaneously, we call it unstable.
Stability of Compounds –
Situation 3
• If ΔHf is positive
• If a compound requires heat to form, then its
decomposition will give heat off.
• For example:
• C(s) + 2 S(s) + 117.4 kJ → CS2(g)
• Reversing this to see the decomposition reaction:
• CS2(g) → C(s) + 2 S(s) + 117.4 kJ
• Since exothermic reactions tend to be spontaneous,
we say these types of compounds are also unstable.
Entropy
• Entropy (S) is the measure of randomness in an object. The
units for entropy are J/K•mol.
• All substances, whether it be an atom or a compound, contains
a certain degree of disorder due to the constant motion of
particles.
• Thus, entropy is always a positive value.
Entropy
• The second law of thermodynamics states
that spontaneous systems always proceed in
the direction of more entropy. This is also
known as the law of disorder.
• That is, over time, systems tend to become
more random instead of more ordered.
• An analogy for entropy:
• Your room at home tends to become more messy
over time as compared to more clean.
Entropy Example
• Entropy can be very hard to explain and often harder to
understand. Here are a couple of examples that hopefully will
help clarify the topic of entropy.
• Imagine you are throwing bricks into a pile
• When you throw the bricks, they will
likely land into a random pile
• This is a HIGHLY DISORDERED
• This is likely what will happen because
it is HIGH IN ENTROPY
• Imagine you start through bricks into a pile
again
• When you throw these bricks, they land
it a perfectly stacked pile.
• This is HIGHLY ORDERED
• This likely WILL NEVER HAPPEN because
it is LOW IN ENTROPY
Entropy Example
• Place a gas the container flask. That flask is
connected to an empty flask.
• If you open the gap between the flasks, the gas will
spread to fill both containers
• This is MORE DISORDERED so it is HIGHER IN
ENTROPY
• If you open the gap, the gas stays only on one side.
• This is MORE ORDERED so it is LOW IN ENTROPY
Can entropy (S) ever be equal to
0?
• The third law of thermodynamics says that as
the temperature approaches absolute zero all
processes cease and entropy approaches a
minimum value (0 J/mol*K).
Can entropy (S) be a negative
number?
• Concerning chemical reactions, we will focus
on the change in entropy (ΔS):
• As a system goes from a state of high disorder (a
gaseous state for example) to a state of less
disorder (a liquid state for example) ΔS will be a
negative value.
• That is, a negative number indicates more
order in the system.
• Likewise, a positive number indicates an
increase in randomness.
Predicting Entropy Changes
• The following suggest an increase in entropy:
• i. Changes in state:
•
•
•
•
a. solid  liquid
b. liquid  gas
c. solid  gas
d. solid or liquid  aqueous state (dissolving)
• ii. An increase in the number of moles in the
products compared to the reactants. That is, more
particles moving about imply more disorder.
• iii. Increase in temperature
Calculating Entropy Changes
• To calculate the change in entropy we will use the same formula
as Hess's law. However, we will be using entropy values instead
of enthalpy values.
ΔS = ΣSproducts – ΣSreactants
• ΔS values for compounds can be found on thermochemical data
sheet.
• ΔS values that are positive indicate less order (spontaneous).
• ΔS values that are negative indicate more order (not
spontaneous).
• Note ΔS values must be multiplied by balancing coefficients.
Example:
• Predict whether the following chemical
reaction will become more or less random.
• Then, using ΔS values, calculate the change in
entropy.
2 NO(g) + O2(g)  N2O4(g)