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PRINCIPLES OF CHEMISTRY II
CHEM 1212
CHAPTER 17
DR. AUGUSTINE OFORI AGYEMAN
Assistant professor of chemistry
Department of natural sciences
Clayton state university
CHAPTER 17
CHEMICAL THERMODYNAMICS
CHEMICAL THERMODYNAMICS
Used to
- Describe heat transfer in chemical systems
- Evaluate the heat evolved or absorbed by a reaction
- Predict the maximum energy that can be produced by a reaction
- Predict whether a proposed reaction is feasible
- Predict whether a process is spontaneous or not
CHEMICAL THERMODYNAMICS
Spontaneous Process
- Takes place with no apparent cause
Nonspontaneous Process
- Requires something to be applied in order for it to occur
(usually in the form of energy)
ENERGY
- The ability to do work or to transfer heat
- Energy is necessary for life: humans, plants, animals, cars
- Forms of energy are interconvertible
- Units: kg∙m2/s2 or Joules (J)
SYSTEM AND SURROUNDINGS
System
- The limited and well-defined portion of the
universe under study
Surroundings
- Everything else in the universe
Studying energy changes in a chemical reaction
- The reactants and products make up the system
- The reaction container and everything else make up
the surroundings
SYSTEM AND SURROUNDINGS
Open System
- Matter and energy can be exchanged with the surroundings
(water boiling on a stove without a lid)
Closed System
- Energy but not matter can be exchanged with the surroundings
(two reactants in a closed cylinder reacting to produce energy)
Isolated System
- Neither matter nor energy can be exchanged with the surroundings
(insulated flask containing hot tea)
WORK (w)
- The energy transferred when a force moves an object
- The product of force (F) and distance (d) through which
the object moves
w=Fxd
- Units: kg∙m2/s2 or Joules (J)
HEAT (q)
- The energy transferred between a system and its surroundings
due to difference in temperature
- A form of energy necessary to raise the temperature
of a substance
- Units: kg∙m2/s2 or Joules (J)
INTERNAL ENERGY (E)
- Sum of all potential and kinetic energies of all components
- Change in internal energy = final energy minus initial energy
E = Efinal - Einitial
- Energy can neither be created nor destroyed
- Energy is conserved
INTERNAL ENERGY
E = Efinal - Einitial
If Efinal > Einitial
E is positive and system has gained energy from its surroundings
If Efinal < Einitial
E is negative and system has lost energy to its surroundings
INTERNAL ENERGY
E = q + w
q = heat added to or liberated from a system
w = work done on or by a system
Internal energy of a system increases when
- Heat is added to the system from surroundings (positive q)
- Work is done on the system by surroundings (positive w)
+w
+q
system
INTERNAL ENERGY
E = q + w
q = heat added to or liberated from a system
w = work done on or by a system
Internal energy of a system decreases when
- Heat is lost by the system to the surroundings (negative q)
- Work is done by the system on the surroundings (negative w)
-w
-q
system
INTERNAL ENERGY
Endothermic Process
- Process in which system absorbs heat (endo- means ‘into’)
- Heat flows into system from its surroundings
(melting of ice - the reason why it feels cold)
Exothermic Process
- Process in which system loses heat
- Heat flows out of the system (exo- means ‘out of’)
(combustion of gasoline)
INTERNAL ENERGY
State Function
- Property that depends on initial and final states of the system
- Does not depend on path or how a change occurs
- Internal Energy depends on initial and final states
- Internal Energy is a state function
- q and w, on the other hand, are not state functions
INTERNAL ENERGY
Internal energy is influenced by
- Temperature
- Pressure
- Total quantity of matter
- Internal Energy is an extensive property
INTERNAL ENERGY
Calculate E for a system absorbing 22 kJ of heat from its
surroundings while doing 11 kJ of work on the surroundings.
State whether it is an endothermic or an exothermic process
q = +22 kJ (heat is added to the system from surroundings)
w = -11 kJ (work is done by the system on the surroundings)
E = q + w
E = (+ 22 – 11) kJ = +11 kJ
Endothermic
PRESSURE-VOLUME WORK
A gas against constant pressure
w = - PV
w = work and P = pressure
V = change in volume = Vfinal – Vinitial
w is positive when the gas contracts (negative V)
w is negative when the gas expands (positive V)
Units: L·atm (1 L·atm = 101.3 J)
PRESSURE-VOLUME WORK
Calculate the work associated with the expansion of a gas from
32 L to 58 L at a constant pressure of 12 atm
w = - PV
w = - (12 atm)(58 L - 32 L) = - 310 L·atm
Gas expands hence work is done by system on surroundings
PRESSURE-VOLUME WORK
Calculate the work associated with the compression of a gas from
58 L to 32 L at a constant pressure of 12 atm
w = - PV
w = - (12 atm)(32 L - 58 L) = + 310 L·atm
Gas contracts hence work is done on system by surroundings
PRESSURE-VOLUME (P-V) WORK
Expansion of Volume
- V is a positive quantity and w is a negative quantity
- Energy leaves the system as work
- Work is done by the system on the surroundings
Compression of Volume
- V is a negative quantity and w is a positive quantity
- Energy enters the system as work
- Work is done on the system by the surroundings
THE FIRST LAW OF THERMODYNAMICS
- The law of conservation of energy
- Energy can be neither created nor destroyed
- Concerned with change in energy
In an isolated system
- Neither matter nor energy can enter or leave
E = 0
In a closed system
- Energy can enter or leave in the form of heat and work
E = q + w
ENTHALPY (H)
- Heat flow in processes occurring at constant pressure
- Only P-V work are performed
H = E + PV
H, E, P, and V are all state functions
Change in Enthalpy
H = (E + PV)
ENTHALPY (H)
Change in Enthalpy at Constant Pressure
H = E + PV
E = q + w
PV = - w
Implies
H = (qp + w) - w = qp
qp = heat at constant pressure
ENTHALPY (H)
H = qp
Change in enthalpy = heat gained or lost at constant pressure
Positive H
- System gains heat from the surroundings
- Endothermic process
Negative H
- System releases heat to the surroundings
- Exothermic process
ENTHALPY (H)
H = E + PV
For reactions involving solids and liquids
V ≈ 0
H ≈ E
For gases
n = nfinal – ninitial
n = total gas moles of products – total gas moles of reactants
Implies
PV = (n)RT
H = E + (n)RT (R = 8.314 J/mol·K)
ENTHALPY (H)
E = q + w
At constant volume
V = 0
w = - PV = 0
and
E = qv
qv = heat gained or lost at constant volume
ENTROPY (S)
- The amount of disorder in a process
- Is a measure of randomness
- Many spontaneous reactions are accompanied by release of
energy (exothermic processes)
- Some endothermic processes, however, are spontaneous
(dissolution of some salts such as barium hydroxide)
- Disorder plays an important role in predicting the
spontaneity of a reaction
ENTROPY (S)
- Entropy is a state function
Change in entropy
ΔS = Sfinal – Sinitial
ENTROPY (S)
Some General Concepts
- The entropy of a substance increases as the substance
changes from solid to liquid to gas
- Due to increase in randomness of the molecules
Ssolid < Sliquid < Sgas
Generally
ΔS from liquid to gas > ΔS from solid to liquid
ENTROPY (S)
Some General Concepts
- The entropy of a substance increases when a molecular
solid or liquid dissolves in a solvent
Solid → Aqueous
- Increase in disorder when a solute dissolves in a solvent
- Increase in disorder of the solute is greater than the
increase in disorder of the solvent
- Net decrease in entropy occurs with some ionic solids
(solids with highly charged ions due to hydration)
ENTROPY (S)
Some General Concepts
- The entropy decreases when a gas dissolves in a solvent
CO2(g) → CO2(aq)
- Solute molecules go from the gas phase to the liquid phase
- Solute molecules become less random
- Increase in disorder of the solvent is small
- Hence a net decrease in disorder
ENTROPY (S)
Some General Concepts
- Entropy increases with increase in temperature
- Kinetic energy of particles increase as temperature
increases
- Disorder increases as kinetic energy (energy of motion)
increases
ENTROPY (S)
Other Concepts
- Entropy increases in a chemical reaction when ∆n is
positive
- Entropy decreases in a chemical reaction when ∆n is
negative
- Entropy of a system increases with increasing volume
- Entropy of a solution increases with dilution
- Osmosis (spontaneous process) is driven by positive ΔS
THE SECOND LAW OF THERMODYNAMICS
- For a spontaneous process there is always an increase in
the entropy of the universe
ΔSuniv = ΔSsys + ΔSsurr
ΔSuniv > 0 for a spontaneous process
If ΔSsys is negative, ΔSsurr must be large and positive to make
ΔSuniv positive
Example
- Combustion of hydrocarbons
ΔSsys is negative but ΔSuniv is positive
THE SECOND LAW OF THERMODYNAMICS
If ΔS < 0
- The reverse reaction is spontaneous
If ΔS = 0
- The system is at equilibrium
- The process is not spontaneous in either direction
- The process has no tendency to occur
THE THIRD LAW OF THERMODYNAMICS
- The entropy of any pure crystalline substance at a
temperature of 0 K is zero
- Absolute value of S can be measured
- As the temperature of a substance is increased from 0 K,
the motion of the particles increases and entropy increases
THE THIRD LAW OF THERMODYNAMICS
- Change in entropy is proportional to the added energy
- The transfer of a given amount of energy as heat has
greater impact on entropy at lower temperatures than at
higher temperatures
- ΔS is directly proportional to quantity of heat transferred
(q) and inversely proportional to temperature (T)
Units: J/K
q
ΔS 
T
THE THIRD LAW OF THERMODYNAMICS
- The entropy change of a chemical reaction
∆Srxn = ΣnSo[products] - ΣmSo[reactants]
So = standard molar entropy of a substance (at 298 K)
- n and m are the number of moles of products and reactants
- The standard molar entropy of an element in its standard
state is not zero
THE THIRD LAW OF THERMODYNAMICS
Calculate ∆So at 25 oC for the reaction
2NiS(s) + 3O2(g) → 2SO2(g) + 2NiO(s)
Obtain So values from appendix
∆So
= [(2 mol)(248.11 J/mol·K) + (2 mol)(37.99 J/mol·K)]
–
[(2 mol)(52.99 J/mol·K) + (3 mol)(205.03 J/mol·K)]
= -148.87 J/K
(∆So is negative as ∆n is negative)
GIBBS FREE ENERGY (G)
G = H – TS
- Change in Gibbs free energy at constant temperature and
pressure
∆G = ∆H – T∆S
- The absolute value of G cannot be measured but ∆G can
be measured
- ∆G is a state function
GIBBS FREE ENERGY (G)
If ∆G < 0
- Forward reaction is spontaneous
If ∆G = 0
- System is at equilibrium
If ∆G > 0
- Reverse reaction is spontaneous
GIBBS FREE ENERGY (G)
Standard Gibbs Free energy of formation (∆Gfo)
- The Gibbs free energy change during the formation of one mole
of a substance in its standard state from its constituent elements
in their standard states
∆Gfo = ∆Hfo – T∆Sfo
GIBBS FREE ENERGY (G)
Change in Gibbs free energy of a chemical reaction
∆Gorxn = Σn∆Gfo[products] - Σm∆Gfo[reactants]
- n and m are the number of moles of products and reactants
- The standard Gibbs free energy of an element in its standard
state is zero
GIBBS FREE ENERGY (G)
Influence of Temperature
∆G = ∆H - T∆S
From the equation above
- Decrease in ∆H (more negative) favor spontaneous change
- Increase in ∆S (more positive) favor spontaneous change
- ∆G is strongly influenced by temperature through the T∆S term
GIBBS FREE ENERGY (G)
Influence of Temperature
At Low Temperatures
- The sign of ∆H determines the sign of ∆G
At High Temperatures
- The sign of ∆S determines the sign of ∆G
- When ∆H and ∆S have opposite signs temperature change does
not influence the direction of spontaneity
- It is assumed that the numerical values of ∆H and ∆S are not
affected by temperature change
GIBBS FREE ENERGY (G)
Equilibrium Constant
∆G = ∆Go + RTlnQ
∆G = Gibbs free-energy change at non-standard-state conditons
∆Go = standard Gibbs free-energy change
R = ideal gas constant (8.314 J/mol·K)
T = temperature (K)
Q = reaction quotient
GIBBS FREE ENERGY (G)
Equilibrium Constant
∆G = ∆Go + RTlnQ
At equilibrium
∆G = 0 and Q = Keq
Implies
∆Go = - RTlnKeq
or
K eq  e
 ΔG o /RT
GIBBS FREE ENERGY (G)
In Summary
- Reaction is spontaneous in the forward direction if ∆G is negative
- Reaction is spontaneous in the reverse direction if ∆G is positive
- Reaction is spontaneous in the forward direction if Q < Keq
- Reaction is spontaneous in the reverse direction if Q > Keq
- At equilibrium ∆G approaches zero and Q approaches Keq
GIBBS FREE ENERGY (G)
Temperature and Equilibrium Constant
∆Go = - RTlnKeq
Combining
and
∆Go = ∆Ho - T∆So
We obtain
∆Ho - T∆So = - RTlnKeq
and
lnK eq
ΔS o
ΔH o


R
RT
GIBBS FREE ENERGY (G)
Temperature and Equilibrium Constant
- Equilibrium constant changes with temperature
- If K1 and K2 are the equilibrium constant values at temperatures T1
and T2, respectively
K1 ΔH o  1
1

ln

 

K2
R  T2 T1 
- Known as the Clausius-Clapeyron equation
GIBBS FREE ENERGY (G)
Useful Work
- For a spontaneous process at constant temperature and pressure
- The maximum useful work that can be performed equals the
change in Gibbs free energy
wmax = ∆G
- The energy that is free to perform useful work
- The equation gives the minimum amount of work required to
cause a change when ∆G is positive