CHM 101

(INTRODUCTORY CHEMISTRY I)

THERMOCHEMISTRY (2)
ORDER – DISORDER PHENOMENON
 ENTROPY
Entropy: A thermodynamic function that measures the degree of randomness or
disorder in a system. It is denoted as S. Its unit is J.K−1.mol−1
• Entropy is a state function and an extensive property.
(State function: property whose value doesn’t depend on the path taken to reach
that specific value is known to as state functions or point functions- they depend on
the state of the substance like temperature, pressure or the amount or type of the
substance. An extensive property is a property that depends on the amount of
matter in a sample.)
• During change of state, systems tend to a state of greatest disorder (high entropy)
– Second Law of Thermodynamics

↑entropy = ↑disorder
The greater the randomness, the higher the entropy. As the solid changes through
the liquid to the gaseous state. i.e., gaseous systems show much greater disorder
than liquid systems, which in turn show greater disorder than solid systems.
 To determine the direction of change in
entropy……
➢Second law of thermodynamics states that all closed system tend to
maximize entropy (Reversing the ever increasing tendency requires the
input of energy)
e.g. Solid → Liquid → Gas.
Lump of ice → liquid
: (heat is absorbed from the surroundings,
disorderliness increases, higher energy)
∆S = Sproduct – Sreactant
➢When Sproduct < Sreactant , ∆S = -ve
➢When Sproduct > Sreactant , ∆S = +ve

OR ΔS =S2-S1 where 1 and 2 refer to initial and final states

1. Change of State :
When a substance changes to a state of higher energy, the entropy
increases (∆S = +ve). But if the change is to a state of lower energy,
entropy decreases (∆S = -ve).
e.g. H2O(l) → H2O(g) ∆S = +ve (increase in disorder)
H2O(l) → H2O(s) ∆S = -ve (decrease in disorder)

…..Positive entropy therefore indicates spontaneity of the process

(following this law)
 Entropy Changes associated with changes of State
• At normal melting point or boiling point of a substance, the two states of
matter present at that temp and 1 atm are in equilibrium (isolated system).
Since change of state from solid to liquid is a reversible process, hence,

𝑞𝑟𝑒𝑣 ΔHfusion
ΔS = =
𝑇 𝑇
where qrev = ΔHfusion = energy required to melt 1 mol of solid at the
melting point
T = melting point in K
ΔHvap
Also for change from liquid to gas at the boiling point , ΔS =
𝑇𝑏𝑝
(in this case, qrev = ΔHvaporization and T= Boiling point)
 2. Decomposition or Composition:
When a substance decomposes into 2 or more substances, entropy
increases i.e. ∆S = +ve. But if two or more substances compose into a
single substance, entropy decreases i.e. ∆S = -ve

e.g. CaCO3(s) → CaO(s) +CO2(g) ∆S = +ve (disordered)

N2(g) + 3H2(g) → 2NH3(g) ∆S = -ve(ordered)
2HCl(s) → H2(g) + Cl2(g) ∆S = +ve (disordered)
3. Change in Volume Occupied:
when a gaseous substance decomposes into another gas of smaller volume,
entropy decreases; but entropy increases if the product is of higher volume.
• Example, N O 2 NO
2 4 2

Here, volume changes from 1 vol. in N2O4 to 2 vols. in NO2, therefore,

entropy increases. i.e. ∆S = +ve.
Note in summary:
➢The entropy change involved when a system changes from an
orderly state to a disorderly state is always positive (∆S = +ve).
➢From a disorderly state to an orderly state, entropy is always
negative (∆S = -ve).
 Entropy Changes
• First law of thermodynamics: The energy of the universe is constant
• Second law of thermodynamics: The entropy of the universe is constantly
increasing

• Change in entropy of the universe:

ΔSuniv = ΔSsys + ΔSsurr

If ΔSuniv = +ve, entropy of the universe increases, the process is spontaneous

If ΔSuniv = -ve, the process nonspontaneous (but spontaneous in the opposite direction)
If ΔSuniv = 0, the process is spontaneous has no tendency to occur, system is at equilibrium
 Entropy changes in the surrounding
• For entropy changes in the surrounding, consider:
1. The sign of ΔSsurr depends on the direction of heat flow
2. The magnitude of ΔSsurr depends on temperature
Hence,
𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑡(𝐽)
Exothermic process: ΔSsurr = +
𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑡(𝐽)
Endothermic process: ΔSsurr = −
𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝐾)
ΔSsurr for a reaction taking place under constant temp and pressure,
Δ𝐻
ΔSsurr = − Spontaneity often depends on T
𝑇
Determination of ΔSuniv form ΔSsys and ΔSsurr
Signs of Entropy Change
ΔSsys ΔSsurr ΔSuniv Process spontaneous?
+ + + Yes
- - - No (process will occur
in opposite direction)
+ - ? Yes, if ΔSsys has a larger
magnitude than ΔSsurr
- + ? Yes, if ΔSsurr has a
larger value than ΔSsys
 Change in Entropy of a Reaction
• Change in entropy of a reaction • Example: Calculate ∆So for the reduction of aluminum oxide
by hydrogen gas
can be determined using the
𝐴𝑙2𝑂3 𝑠 + 3𝐻2(𝑔) → 2𝐴𝑙 𝑠 + 3𝐻2𝑂(𝑔)
expression:
given that So (J K-1 mol-1) for 𝐴𝑙2𝑂3 𝑠 , 𝐻2(𝑔), 𝐴𝑙 𝑠 and
∆S = Σn∆S (products) – Σm∆S(reactants) 𝐻2𝑂(𝑔) are 51, 131, 28 and 189 respectively

• Solution:
At standard conditions, ∆So = Σn∆So (products) – Σm∆So(reactants)
= 2𝑆𝑜𝐴𝑙 𝑠 + 3𝑆𝑜𝐻2𝑂 𝑔 − {𝑆𝑜𝐴𝑙2𝑂3 𝑠 + 3𝑆𝑜𝐻2 𝑔 }
∆So = Σn∆So (products) – Σm∆So(reactants) = 2 𝑚𝑜𝑙 28 𝐽 𝐾 − 1 𝑚𝑜𝑙 − 1 + 3 𝑚𝑜𝑙(189 𝐽 𝐾 −
−
1 𝑚𝑜𝑙 1) − − 1 𝑚𝑜𝑙 1)
− 1 𝑚𝑜𝑙(51
−1
𝐽−𝐾 −
3 𝑚𝑜𝑙 131 𝐽 𝐾 𝑚𝑜𝑙 1
where n and m are coefficients of = (56 + 567 − 393 − 51) 𝐽/ 𝐾
the substances in the product(s) and = 𝟏𝟕𝟗 𝑱/𝑲
reactant(s) respectively.
FREE ENERGY
 FREE ENERGY CHANGES
• Free Energy (G): Thermodynamic function also related to spontaneity
𝐺 = 𝐻 − 𝑇𝑆
H = Enthalpy, T is Temperature (K), S = Entropy
• Free energy is called Gibbs free energy (G) after Josiah Willard Gibbs, the
scientist who developed the measurement.
• Every chemical reaction involves a change in free energy, called delta G (∆G).
• The change in free energy can be calculated for any system that undergoes a
change, such as a chemical reaction.
To calculate ∆G, subtract the amount of energy lost to entropy (denoted as ∆S)
from the total energy change of the system (∆H).
For a process occurring at constant temperature, Δ𝐺 = Δ𝐻 − 𝑇Δ𝑆
At standard conditions of reaction, Δ𝐺𝑜 = Δ𝐻𝑜 − 𝑇Δ𝑆𝑜
 Free Energy and Spontaneity:
• Gibb’s free energy (G) depends on the initial and final state of the system undergoing change (i.e.
Gibb’s free energy is also a state function).
Δ𝐺 = 𝐺𝑝𝑟𝑜𝑑𝑢𝑐𝑡 − 𝐺𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

• For any chemical reaction, Δ𝐺 = Σ𝑛𝐺𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − Σ𝑚𝐺𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

when,
∆G = -ve Reaction is spontaneous (the reaction will occur as written)
∆G = 0 Reaction is at equilibrium (i.e. two phases are in equilibrium, no useful work)
∆G = +ve Reaction is not spontaneous (reaction cannot do work)

e.g. ∆G = -ve for T > 273.16K i.e. ice changes to water ∆G= +ve for T < 273.16K i.e. reverse reaction of
freezing taken place.
• Under standard conditions (25 oC and 1 atmosphere pressure),
∆G⁰ = ∆H⁰ – T∆S⁰
• Free energy of formation of an element in its standard state is zero.
 Estimation of ∆G
• Using the equation ∆G = ∆H - T∆S
(a) At low temperatures: T∆S becomes small and ∆H controls the sign of ∆G.
Exothermic reactions are thus feasible at low temperature.
(b) (b) At high temperatures: T∆S becomes large and more important.
i. For Endothermic reactions, ∆S must large enough for ∆G to be –ve
ii. Endothermic reactions become explosive and favourable at high temperature

• Reactions that have a negative ∆G (∆G <0) are called exergonic reactions.
Exergonic means energy is exiting the system
• If a chemical reaction requires an input of energy rather than releasing energy,
then the ∆G for that reaction will be a positive value. These chemical reactions
are called endergonic reactions; they are non-spontaneous.
Spontaneity and Nonspontaneity (∆G values)
Examples:
Example 2
Example 1. Determine the standard free
energy change for the following reaction at
25 oC.
N2 + 3H2 → 2NH3
Given ΔH and ∆S are -81.5 kJ and -189.0 J/K
• Solution:
We have an equation, ΔG = ΔH – TΔS
Substitute the above values in this equation
ΔG = -81.5 kJ – (298 K) (-0.1890 kJ/K)
ΔG = -24.7 kJ
ENTHALPY
 ENTHALPY
Enthalpy (H) is expressed as
H = U + PV
U = internal energy
P = Presssure
V = Volume
• When a process occurs at constant
pressure, the heat flow (q or Q) (either
released or absorbed) for the process is
equal to the change in enthalpy.
i.e ΔH = q
• Enthalpy is a state function which
depends entirely on the state functions
Source: https://www.slideshare.net/mrtangextrahelp/tang- T , P and U .
01b-enthalpy-entropy-and-gibbs-free-energy • The unit of molar enthalpy is J.mol-1.
ENTHALPY CHANGES: Exothermic and endothermic reactions
• Enthalpy is usually expressed as the change in
enthalpy (ΔH ) for a process between initial
and final states

• When a system (reaction) loses heat to the

surrounding, the reaction is exothermic. i.e.
∆H is –ve (∆Hproduct < ∆Hreactant )

• When a system (reaction) absorbs heat from https://cdn1.byjus.com/wp-content/uploads/2016/06/Energy-

the surrounding, the reaction in endothermic
i.e. ∆H = +ve (∆Hproduct > ∆Hreactant ) Level-Diagram-of-an-Endothermic-Reaction-700x410.png
ENTHALPY OF REACTION
• When energy needs to be added • Enthalpy of reaction is the enthalpy
to a material to change charge accompanying a reaction
its phase from a liquid to a gas, represented by a balanced chemical
that amount of energy is called equation.
the enthalpy (or latent heat) of
vaporization
• Other phase transitions have • For any physical or chemical process
similar associated enthalpy where there is an enthalpy change,
changes, such as the enthalpy (or the reverse process will be
latent heat) of fusion for changes accompanied by the reverse enthalpy
from a solid to a liquid change (i.e. change in sign)
e.g.
C(s) + O2(g) →CO2(g) ∆H = -394 kJ mol-1
CO2(g) →C(s) + O2(g) ∆H = +394 kJ mol-1
 ∆H can be related to the number of moles of each
reactant…..
e.g. The phase (or state) of each substance (gas,
liquid or solid) taking part in the reaction must
be specified.
4Cr(s) + 3O2(g) →2Cr2O3(s) ∆H = -2256 kJ mol-1 (i)
C(graphite) + O2(g) → CO2(g) ∆H = -394 kJ mol-1
This should be written as:
3
2Cr(s) + O2(g) → Cr2O3(s) ∆H = -1128 kJ mol -1 (ii)
2 ∆H depends on the state of substances.

e.g.
Note: The stoichiometry coefficient of the
1
reaction in (ii) is a fraction of reaction (i). H2(g) + O2(g) →H2O(l) ∆H = -286 kJ mol-1___(i)
2
Hence, ∆H value changes 1
H2(g) + 2O2(g) →H2O(g) ∆H = -242 kJ mol-1___(ii)
Example 1:
How much heat is evolved when mass given = 500kg = 500 x 103g
500 kg of ammonia is produced 500×103 𝑔
჻ 𝑛𝑁𝐻3 = = 2.94 × 104 mol
according to the following equation? 17𝑔 𝑚𝑜𝑙 −1

N2(g) + 3H2(g) →2NH3(g) From the balanced chemical

equation,
∆H= -91.8 kJ mol-1
2 moles of NH3 gave -91.8 kJ mol-1.
[N = 14.0, H = 1.00g mol-1].
჻ 2.94 × 104 mol of NH3 will give:
Solution: −91.8 kJ mol−1
𝑚𝑎𝑠𝑠 𝑔𝑖𝑣𝑒𝑛 × 2.94 × 104 𝑚𝑜𝑙
No of moles of NH3 =
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑁𝐻3
2𝑚𝑜𝑙

= −1.35 × 106 𝑘𝐽 𝑚𝑜𝑙 −1

molar mass of NH3 = 14 + 3(1.00) = 17 g mol-1
ΔH, o
ΔH reaction and the Standard state
Δ = represents the change in the enthalpy; The Standard State:
O= signifies that the reaction is a standard • The standard state of a substance is the pure
substance in a specified state(solid or liquid or gas)at
enthalpy change, 1 atm pressure.
reaction = denotes that this change is the enthalpy
• Standard conditions are:
of reaction - gases at a pressure of 1 atm ( 105 Pa)
∆Hreaction = ∑n∆Hp – ∑m∆Hr - solutions in unit concentration
- substances in their standard state
where:
• The ΔHoreaction is the standard heat of reaction or
∑ represents summation standard enthalpy of a reaction, and like ΔH also
measures the enthalpy of a reaction. However,
n and m are the coefficients of the substances in ΔHoreaction takes place under "standard" conditions,
meaning that the reaction takes place at 25o C and 1
the product(s) and reactant(s) respectively atm.
Hp =enthalpies of the product(s) • The benefit of a measuring ΔH under standard
conditions lies in the ability to relate one value of
Hr = enthalpies of the reactant(s) ΔHo to another, since they occur under the same
conditions.
Enthalpy Change Accompanying a Change in State
A change in state involves a change in enthalpy
3. Enthalpy of vaporization (ΔHvap )
1. Enthalpy of fusion or enthalpy of melting -heat required to vaporize one mole of a
(ΔH𝑓𝑢𝑠𝑖𝑜𝑛 ) substance.
- the enthalpy change when one mole of a
substance melts. ΔHvap = ΔHvapor - ΔHliquid
e.g For example, for one mole of ice the enthalpy is given as: e.g for one mole of water, ΔHvap = 44.0 kJ at 298 K
ΔHfusion = 6.01 kJ at 273.15 K 4. Enthalpy of condensation (ΔHcondensation )
ΔHcondensation = -ΔHvap

2. Enthalpy of freezing (ΔH𝑓𝑟𝑒𝑒𝑧𝑖𝑛𝑔 ) 5. Enthalpy of sublimation

- Enthalpy change when one mole of a substance ΔHsublimation= ΔHfussion+ ΔHvap
freezes.
ΔHfreezing = - ΔHfusion Specific temperatures the change takes place are
stated in all enthalpies of phase changes
Relationship between Temperature and Heat
• When the temperature Specific heat capacity ( c ) is the heat required
increases, the amount of to raise the temperature of a unit mass of
molecular interactions also substance by 1o C (unit is J K-1 g-1 or J oC-1g-1)
increases.
ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦
• As the number of Specific heat capacity, c =
𝑚𝑎𝑠𝑠 (𝑚)
(1)
interactions increase, the
internal energy of the system 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑡 (𝑄)
rises. 𝐻𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = (2)
𝑅𝑖𝑠𝑒 𝑖𝑛 𝑡𝑒𝑚𝑝 (∆𝑇)
• if the internal energy (U)
increases then, the ΔH Combining equations (1) and (2) we obtain,
increases as temperature
rises. (since H = U +PV) 𝑄 = 𝑚𝑐∆𝑇 (3)
Examples
(1) Water has specific heat 4.18 J K-1g-1. 100 g of water rise in temperature by
2.0 K. Calculate the heat quantity absorbed by water.
Solution: q = mc∆T
= 100g x 4.18 J K-1g-1 x 2.0 K
= 9.36 J
(2) 293.7 J of heat is removed from 5g of Aluminum causing the temperature
to drop from 85⁰C to 19⁰C. What is the specific heat capacity of Aluminum?
Solution: q = mc∆T
-293.7 J = (5g) (c)(292-358 K)
-293.7 J = -330 c (g. K)
c = 0.89 J K-1 g-1
3. A 50 g of an unknown material at 200 ⁰C was added to 100 g of water at 25 ⁰C. The final
temp of the mixture was 41.8 ⁰C. What is the specific heat capacity of the unknown material?
(Water has the sp. heat capacity of 4.18 J oC-1g-1)
𝑚𝑒𝑡𝑎𝑙
41.8⁰C (q = -ve)
200⁰𝐶

𝐻2 𝑂
41.8⁰C (q = +ve).
25⁰𝐶

But, Qm = QH20
჻ - Qm = QH20
- mc∆T = - mc∆T
-mc (Tf – Ti) = -mc (Tf – Ti).
-50(c) (25 – 200 oC) = 100 (4.184) (41.8 – 25 ⁰C)
7910(c) = 7029.12
c = 0.8895 J.
specific heat capacity of the material is 0.8895 J oC-1g-1
Practice Questions
• For practice questions, go to
https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_
1202/Unit_7:_Principles_of_Thermodynamics/7.E:_Exercises_on_Ent
ropy_and_Gibbs_Energy

• https://www.chemguide.co.uk/physical/entropymenu.html