# Standard Absolute Entropy Change Calculations (ΔS°) Chemistry Tutorial

## Key Concepts

• Entropy is given the symbol S
• The third law of thermodynamics states that at absolute zero (0 K)(1) the entropy of a pure, perfect crystalline solid (S0) is zero (0)(2):

S0 = 0

• The absolute entropy(3) of a substance, ST, is the increase in entropy when a substance is heated from 0 K to a temperature of T K.
• The standard absolute entropy of a substance, S°, is the absolute entropy of a substance in its standard state (298.15 K, 100 kPa).
• Standard absolute entropy values are given in units of joules per kelvin per mole; J K-1 mol-1
• Values of standard absolute entropy (S°) for many substances have been tabulated.
• The change in standard absolute entropy for a chemical reaction (ΔS°) can be calculated using these tabulated values:

ΔS°reaction = ΣS°products - ΣS°reactants

• For the general reaction in which reactants A and B react to produce products C and D:

aA + bB → cC + dD

ΔS°reaction = ΣS°products - ΣS°reactants

ΔS°reaction = [c(C) + d(D)] - [a(A) + b(B)]

• If ΔS°reaction is positive (ΔS°reaction > 0), the entropy of the system increased.

If ΔS°reaction is negative (ΔS°reaction < 0), the entropy of the system decreased.

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## Defining Standard Absolute Entropy (S°)

Recall that entropy (S) can be thought of as the:

• degree of randomness in the system
• relative distribution of energy within the system(4)

A perfect crystal at absolute zero (0 K) is in a state of minimum entropy because:

• each atom in the crystal is locked into a fixed position in the crystal lattice (minimum randomness)
• each atom has the same minimum energy (minimum distribution of energy)

The third law of thermodynamics states that the entropy (S) of a pure, perfect crystalline solid is zero at absolute zero:

S = 0 (perfect crystal at 0 K)

If we heat this perfect crystal to a temperature above 0 K the atoms gain energy such as kinetic energy so the atoms making up the crystal no longer have minimum energy, and, because the atoms are moving there is an increase in the "randomness" of the system.
The entropy (S) of this heated crystal has increased, it is no longer zero:

for a crystal at T > 0 K then S > 0

If we let:

S0 = entropy of crystal at 0 K

ST = entropy of crystal at T K

Then, the increase in the entropy of the crystal when heated from 0 K to a higher temperature of T K is:

S(0 → T) = ST - S0

Since the entropy of a perfect crystal at 0 K is zero:

S0 = entropy of crystal at 0 K = 0

We can substitute 0 for S0 in the equation to get:

S(0 → T) = ST - 0 = ST

ST is then referred to as the absolute entropy of this crystal at temperature T K.

Standard absolute entropy refers to the absolute entropy of a substance in its standard state (that is, its state at 298.15 K and atmospheric pressure).
Standard absolute entropy is given the symbol S°
The entropy of a substance reflects the energy distribution (joules, J) at a specific temperature (kelvin, K) for a specific amount of substance (moles, mol), so the units of standard absolute entropy are J K-1 mol-1.
The values of standard absolute entropy (S°) have been tabulated for many substances. Some examples are given in the table below:

Standard Absolute Entropies at 298.15 K
Solid Substance
J K-1 mol-1
Liquid Substance
J K-1 mol-1
Gaseous Substance
J K-1 mol-1
C (diamond) 2.38   H2O(l) 69.9   H2(g) 130.6
C (graphite) 5.74   Hg(l) 77.4   HCl(g) 186.8
MgO(s) 26.8   H2O2(l) 92.0   CH4(g) 187.9
Fe(s) 27.2   CH3OH(l) 126.3   H2O(g) 188.7
Al(s) 28.3   H2SO4(l) 156.9   N2(g) 191.5
S8(s) (rhombic) 31.9   C2H5OH(l) 160.7   NH3(g) 192.3
Mg(s) 32.5   CCl4(l) 214.4   CO(g) 197.6
Cu(s) 33.3         C2H2(g) 200.8
SiO2(s) (quartz) 41.8         O2(g) 205.1
P4(s) (white) 44.0         H2S(g) 205.7
Na(s) 51.0         CO2(g) 213.6
NaCl(s) 72.4         C2H4(g) 219.5
Na2O(s) 72.8         Cl2(g) 223.0
KCl(s) 82.7         C2H6(g) 229.5
MgCl2(s) 89.5         SO2(g) 248.1
NH4Cl(s) 94.6         SO3(g) 256.6

Note the (generally) large positive values of S° for gaseous substances in which the molecules are chaotically and randomly distributed:

(H2O(g)) = 188.7 J K-1 mol-1
(H2S(g)) = 205.7 J K-1 mol-1
(CO2(g)) = 213.6 J K-1 mol-1

and the (generally) smaller positive values of S° for solid substances in which intermolecular forces act to keep in the particles in a more structured and ordered array:

(NaCl(g)) = 72.4 J K-1 mol-1 (3-dimensional ionic lattice)
(graphite) = 5.74 J K-1 mol-1 (2-dimensional covalent lattice)
(diamond) = 2.38 J K-1 mol-1 (3-dimensional covalent lattice)

Note that water exists in two different states at 298.15 K and atmospheric pressure, as a liquid (H2O(l)) and as a gas (H2O(g)).
Note that the value for the standard absolute entropy for gaseous water (188.7 J K-1 mol-1) is greater than for liquid water (69.9 J K-1 mol-1).

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## Calculating the Change in Standard Absolute Entropy for Chemical Reactions (ΔS°)

If you leave a small amount of water on a watch glass in the lab overnight it will probably have evaporated by the time you get back to it in the morning. Similarly if you have poured an aqueous suspension of a precipitate through a filter paper, you have probably left the wet filter paper on a watch glass on the lab bench, or even "hanging up", overnight to dry.
This is because water molecules can escape from the liquid phase and enter the gaseous phase. When they do this, the entropy of the water molecule system increases, that is, the entropy of the system in which water molecules are present in the gaseous phase is greater than the entropy of the system in which water molecules are present in the liquid phase:

 word equation: balanced chemical equation: relative entropy: liquid water → gaseous water H2O(l) → H2O(g) lower entropy → higher entropy

Because water can exist in both states, gas and liquid, under standard conditions (298.15 K, 100 kPa), the values for the standard absolute entropy for each state are available:

S°(H2O(l)) = 69.9 J K-1 mol-1

S°(H2O(g)) = 188.7 J K-1 mol-1

This allows us to calculate the change in the standard absolute entropy for this physical change:

 word equation: balanced chemical equation relative entropy liquid water → gaseous water H2O(l) → H2O(g) lower entropy → higher entropy 69.9 → 188.7

The change in standard absolute entropy (ΔS°) for this reaction is:

ΔS°(reaction) = S°(H2O(g)) - S°(H2O(g))

ΔS°(reaction) = 188.7 - 69.9 = 118.8 J K-1 mol-1

When water molecules escape from the liquid phase and enter the gas phase the entropy of this system increases by 118.8 J K-1 mol-1.

Similarly, we might watch droplets of water condense on a glass, in this case water molecules are leaving the gas phase and entering the liquid phase so we expect the entropy of this system to decrease:

 word equation: balanced chemical equation relative entropy gaseous water → liquid water H2O(g) → H2O(l) higher entropy → lower entropy 188.7 → 69.9

The change in standard absolute entropy (ΔS°) for this reaction is:

ΔS°(reaction) = S°(H2O(l)) - S°(H2O(g))

ΔS°(reaction) = 69.9 - 188.7 = −118.8 J K-1 mol-1

Note the negative sign in front of 118.8, this means the entropy of this system has decreased by 118.8 J K-1 mol-1.

Let's consider 2 moles of gaseous water condensing to form liquid water at 298.15 K

 balanced chemical equation: for 2 moles H2O(g): H2O(g) → H2O(l) ΔS°(reaction) = −118.8 J K-1 mol-1 2H2O(g) → 2H2O(l) 2 × ΔS°(reaction)     = 2 mol × −118.8 J K-1 mol-1     = −237.6 J K-1

Because there are more water molecules in the system, the entropy of the system has decreased even more when these water molecules condense to form the liquid!

Another way to look at this is to consider the stoichiometric coefficient of each species in the equation (mole ratio) and determine the total entropy of each reactant and each product:

 balanced chemical equation: S° (J K-1 mol-1) 2H2O(g) → 2H2O(l) (2 × 188.7) → (2 × 69.9) 377.4 → 139.8

We then use these values to determine the value of the change in standard absolute entropy for the reaction (ΔS°):

ΔS°(reaction) = S°(all products) - S°(all reactants) = 139.8 - 377.4 = -237.6 J K-1

For the general reaction in which reactants A and B react to produce products C and D:

aA + bB → cC + dD

ΔS°reaction = ΣS°products - ΣS°reactants

ΔS°reaction = [c(C) + d(D)] - [a(A) + b(B)]

If ΔS°(reaction) is positive (ΔS°(reaction) > 0), the entropy of the system increased.

If ΔS°(reaction) is negative (ΔS°(reaction) < 0), the entropy of the system decreased.

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## Worked Example of Standard Absolute Entropy Change Calculations for Chemical Reactions (ΔS°)

Question: When liquid methanol (CH3OH(l)) combusts it reacts with oxygen gas (O2(g)) from the air to form carbon dioxide gas (CO2(g)) and water vapor (H2O(g)) according to the balanced chemical equation given below:

2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)

Given the values of the standard absolute entropy of each species below, calculate the change in the standard absolute entropy (ΔS°) for the reaction:

Substance S° (J K-1 mol-1)
CH3OH(l) 126.3
O2(g) 205.1
CO2(g) 213.6
H2O(g) 188.7

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate ΔS°(reaction)
ΔS°(reaction) = ? J K-1 mol-1

2. What data (information) have you been given in the question?

Extract the data from the question:

(i) Balanced chemical equation: 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)

(ii) Values of S° for each species in the equation:

Substance S° (J K-1 mol-1)
CH3OH(l) 126.3
O2(g) 205.1
CO2(g) 213.6
H2O(g) 188.7

3. What is the relationship between what you know and what you need to find out?
ΔS°(reaction) = ΣS°(products) - ΣS°(reactants)

ΔS°(reaction) = [2S°(CO2(g)) + 4S°(H2O(g))] - [2S°(CH3OH(l)) + 3S°(O2(g))]

4. Substitute in the values for S°(products) and S°(reactants) and solve for ΔS°(reaction):
ΔS°(reaction) = [2S°(CO2(g)) + 4S°(H2O(g))] - [2S°(CH3OH(l)) + 3S°(O2(g))]

ΔS°(reaction) = [(2 × 213.6) + (4 × 188.7)] - [(2 × 126.3) + (3 × 205.1)]

ΔS°(reaction) = [427.2 + 754.8] - [252.6 + 615.3]

ΔS°(reaction) = [1182] - [867.9]

ΔS°(reaction) = 314.1 J K-1 mol-1

The balanced chemical equation has 3 moles of gas on the left hand side and 2 + 4 = 6 moles of gas on the right hand side, that is, we expect the entropy of the system to increase (ΔS°(reaction) will be positive). Our value for ΔS° is positive.
Next we can perform a "rough" calculation to make sure our value is in the "right ball park":
ΔS°(reaction) ≈ [(2 × 200) + (4 × 200)] - [(2 × 100) + (3 × 200)] = [400 + 800] - [200 + 600] = 1200 - 800 = 400 J K-1 mol-1
Since 314.1 J K-1 mol-1 is of the same order of magnitude and sign as 400 J K-1 mol-1, we are reasonably confident our answer is plausible.
6. State your solution to the problem "calculate ΔS°(reaction)":

ΔS°(reaction) = 314.1 J K-1 mol-1

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Footnotes

(1) Absolute zero is a temperature of 0 K or -273.15°C
You can convert temperature from Kelvin (K) to degrees Celsius (°C) and vice versa as shown below:
T (K) = T (°C) + 273.15
T (°C) = T (K) - 273.15
The SI unit for temperature is kelvin (K).

(2) At abdsolute zero all quenchable energy has been quenched. This means that for a perfect crystal at 0 K all the atoms are arranged in a uniform, regular lattice. The absence of disorder and thermal chaos suggests that this perfect crystal has zero entropy.
Another generalised statement of the third law of thermodynamics is that it is impossible to reach absolute zero in a finite number of steps.

(3) Absolute entropies are also referred to as third law entropies.

(4) The ways that energy can be distributed and stored in a chemical system is referred to as a microstate.