Thermodynamics Essentials

THERMODYNAMICS.
Elements of Physical Chemistry. By P. Atkins
Dr. H.A. Ellis
18/1/05
Concerned with the study of
transformation of energy:
Heat ↔ work

CONSERVATION OF ENERGY – states that:
 Energy can neither be created nor destroyed in chemical
reactions. It can only be converted from one form to the other.
 UNIVERSE
 System – part of world have special interest in…
 Surroundings – where we make our observations

→ →
Open system Closed system Isolated system
Example:
↔ matter
↔ energy ↔ energy not matter matter ×
Energy
×

 If system is themally isolated called Adiabatic system eg: water
in vacuum flask.

WORK and HEAT
 Work – transfer of energy to change height of the weight in
surrounding eg: work to run a piston by a gas.
 Heat – transfer of energy is a result of temperature difference
between system and surrounding eg:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) - heat given off.
If heat released to surroundings – exothermic.
If heat absorbed by surroundings – endothermic.

Example: Gasoline, 2, 2, 4 trimethylpentane
CH3C(CH3)2CH2CH(CH3)CH3 + 25/2 O2 → 8CO2(g) +H2O(l)
5401 kJ of heat is released (exothermic)
Where does heat come from?
From internal energy, U of gasoline. Can represent chemical reaction:
Uinitial = Ufinal + energy that leaves system (exothermic)
Or
Ui = Uf – energy that enters system (endothermic)

Hence, FIRST LAW of THERMODYNAMICS (applied to a closed
system)
 The change in internal energy of a closed system is the energy
that enters or leaves the system through boundaries as heat or
work. i.e.
 ∆U = q +w
 ∆U = Uf – Ui
 q – heat applied to system
 W – work done on system
 When energy leaves the system, ∆U = -ve i.e. decrease internal
energy
 When energy enter the system, ∆U = +ve i.e. added to internal
energy

Different types of energies:
1. Kinetic energy = ½ mv2
(chemical reaction) kinetic energy
(KE) ∝ k T (thermal energy) where k = Boltzmann constant
2. Potential energy (PE) = mgh – energy stored in bonds
Now, U = KE + PE

3. Work (W)
w = force × distance moved in direction of force
i.e. w = mg × h = kg × m s-2
× m = kg m2
s-2
(m) (g) (h)
1 kg m2
s-2
= 1 Joule
- Consider work – work against an opposing force, eg: external
pressure, pex. Consider a piston

Piston
Pex
pressure (P)
pex
A = area of piston
P
h
h is distance moved

w = distance × opposing force
w = h × (pex × A) = pex × hA
Work done on system = pex × ∆V
∆V – change in volume (Vf – Vi)
∴ Work done by system = -pex × ∆V
Since U is decreased

Example:
C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l) at 298 K ∝ 1 atm
(1 atm = 101325 Pa), -2220 kJ = q
What is the work done by the system?
For an ideal gas;
pV = nRT (p = pex)
n – no. of moles
R – gas constant
T = temperature
V – volume
p = pressure

∴ V= nRT/p or Vi = niRT/pex
6 moles of gas:
Vi = (6 × 8.314 × 298)/ 101325 = 0.1467 m3
3 moles of gas:
Vf = (3 × 8.314 × 298)/ 101325 = 0.0734 m3
work done = -pex × (Vf – Vi)
= -101325 (0.0734 – 0.1467) = +7432 J

NB: work done = - pex (nfRT/pex – niRT/pex)
= (nf – ni) RT
Work done = -∆ngasRT
i.e. work done = - (3 – 6) × 8.314 × 298 = + 7432.7 J
Can also calculate ∆U
∆U = q +w q = - 2220 kJ
w = 7432.7 J = 7.43 kJ
∴ ∆U = - 2220 + 7.43 = - 2212.6 kJ

NB:
qp ≠ ∆U why?
Only equal if no work is done i.e. ∆V = 0
i.e. qv = ∆U

Example: energy diagram
C3H8 + 5 O2 (Ui)
3CO2 + 4H2O(l) (Uf)
U
U
progress of reaction
reaction path

Since work done by system = pex∆V
System at equilibrium when pex = pint (mechanical equilibrium)
Change either pressure to get reversible work i.e.
pex > pint or pint > pex at constant temperature by an
infinitesimal change in either parameter

 For an infinitesimal change in volume, dV
 Work done on system = pdV
For ideal gas, pV = nRT
∴p = nRT/ V
∴ work = p dV = nRT dV/ V
= nRT ln (Vf/Vi) because ∫dx/x = ln x
 Work done by system= -nRT ln(Vf/Vi)
∫
Vf
Vi∫
Vf
Vi

Enthalpy, H
 Most reactions take place in an open vessel at constant
pressure, pex. Volume can change during the reaction
i.e. ∆V ≠ 0 (expansion work).
Definition: H = qp i.e. heat supplied to the system at
constant pressure.

Properties of enthalpy
 Enthalpy is the sum of internal energy and the product of pV of
that substance.
i.e H = U + pV (p = pex)
 Some properties of H

Hi = Ui + pVi
Hf = Uf + pVf
∴Hf – Hi = Uf – Ui +p(Vf – Vi)
or
∆H = ∆U + p ∆V

Since work done = - pex ∆V
∆H = (- pex ∆V + q) +p∆V
(pex= p)
∴ ∆H = ( -p ∆V + q) + p ∆V = q
 ∴ ∆H = qp

suppose p and V are not constant?
∀ ∆H = ∆U + ∆( pV) expands to:
∀ ∆H = ∆U + pi ∆V + Vi ∆P + (∆P) (∆V)
• i.e. ∆H under all conditions.
• When ∆p = 0 get back
∆H = ∆U + pi ∆V ≡ ∆U + p ∆V
• When ∆V = 0:
∆H = ∆U + Vi ∆p

Enthalpy is a state function.
lattitude
longitude
A
B
path 1
path 2
- does not depend on the path taken

NB: work and heat depend on the path taken and are written as
lower case w and q. Hence, w and q are path functions. The
state functions are written with upper case.
eg: U, H, T and p (IUPAC convention).

Standard States
 By IUPAC conventions as the pure form of the substance at 1 bar
pressure (1 bar = 100,000 Pa).
What about temperature?
 By convention define temperature as 298 K but could be at any
temperature.

Example:
C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l)
at 1 bar pressure, qp = - 2220 kJmol-1
.
 Since substances are in the pure form then can write
∆Hθ
= - 2220 kJ mol-1
at 298 K
θ represents the standard state.

H2(g) → H(g) + H(g), ∆Hθ
diss = +436kJmol-1
H2O(l) → H2O(g), ∆Hθ
vap = +44.0 kJmol-1
Calculate ∆Uθ
for the following reaction:
CH4(l) + 2 O2(g) → CO2(g) + 2H2O(l), ∆Hθ
= - 881.1kJmol-1

∆H = ∆U + ∆(pV)
= ∆U + pi ∆V + Vi ∆p + ∆p ∆V
NB: p = 1 bar, i.e. ∆p = 0
∴ ∆Hθ
=∆Uθ
+ pi ∆V
Since -pi ∆V = - ∆nRT,
∆Uθ
= ∆Hθ
- ∆nRT

calculation
∴ ∆Uθ
= - 881.1 – ((1 – 2)(8.314) × 298)/ 1000
= - 881.1 – (-1)(8.314)(0.298) = - 2182.99 kJ mol-1

STANDARD ENTHALPY OF FORMATION, ∆Hf
θ
 Defined as standard enthalpy of reaction when substance is formed
from its elements in their reference state.
 Reference state is the most stable form of element at 1 bar
atmosphere at a given temperature eg.
At 298 K Carbon = Cgraphite
Hydrogen = H2(g)
Mercury = Hg(l)
Oxygen = O2(g)
Nitrogen = N2(g)

NB: ∆Hf
θ
of element = 0 in reference state
 Can apply these to thermochemical calculations
eg. Can compare thermodynamic stability of substances in
their standard state.
 From tables of ∆Hf
θ
can calculate ∆H θ
f rxn for any reaction.

Eg. C3H8 (g) + 5O2(g) → 3CO2(g) + 4H2O(l)
 Calculate ∆Hθ
rxn given that:
∆Hθ
f of C3H8(g) = - 103.9 kJ mol-1
∆Hθ
f of O2(g) = 0 (reference state)
∆Hθ
f of CO2(g) = - 393.5 kJ mol-1
∆Hθ
f of H2O(l) = - 285.8 kJ mol-1
∆Hrxn = Σn ∆Hθ
(products)- Σn ∆Hθ
(reactants)

∆Hθ
f(products) = 3 × (- 393.5) + 4 × (- 285.8)
= - 1180.5 -1143.2 = - 2323.7 kJ mol-1
∆Hθ
f(reactants) = - 103.9 + 5 × 0 = - 103.9 kJ mol-1
∆Hθ
rxn = - 2323.7 – (- 103.9) = - 2219.8 kJ mol-1
=
- 2220 kJ mol-1

 Answer same as before. Eq. is valid.
 Suppose: solid → gas (sublimation)
 Process is: solid → liquid → gas
 ∆Hsub = ∆Hmelt + ∆Hvap
 Ie. ∆H ( indirect route) = . ∆H ( direct route)

Hess’ Law
 - the standard enthalpy of a reaction is the sum of the standard
enthalpies of the reaction into which the overall reaction may be
divided. Eg.
 C (g) + ½ O2(g) → CO (g) , ∆Hθ
comb =? at 298K

– From thermochemical data:
 C (g) +O2 (g) → CO 2(g) ∆Hθ
comb =-393.5 kJmol-
1
…………………………….(1)
 CO (g) +1/2 O2 (g) →CO 2(g), ∆Hθ
comb = -283.0 kJ mol-
1……………………. (2)
 Subtract 2 from 1 to give:
 C (g) + O2 (g) – CO (g) – 1/2 O2 (g) → CO2 (g) – CO2 (g)
 ∴ C (g) + ½ O2 (g) → CO (g) , ∆Hθ
comb= -393.5 –
 (-283.0) = - 110.5 kJ mol-1

Bond Energies
 eg. C-H bond enthalpy in CH4
 CH4 (g) → C (g) + 4 H (g) , at 298K.
 Need: ∆Hf
θ
of CH 4 (g) =- 75 kJ mol-1
 ∆Hf
θ
of H (g) = 218 kJ mol-1
 ∆Hf
θ
of C (g) = 713 kJ mol-1

 ∆Hθ
diss = Σ n∆Hf
θ
(products)- Σ n∆Hf
θ
( reactants)
 = 713 + ( 4x 218) – (- 75) = 1660 kJ
 mol-1
 Since have 4 bonds : C-H = 1660/4 = 415
 kJ mol-1

Variation of ∆Hθ
with temperature
Suppose do reaction at 400 K, need to know
∆Hθ
f at 298 K for comparison with literature value. How?
 As temp.î ∆Hθ
mÎ ie. ∆Hθ
m α T
 ∴ ∆Hθ
m = Cp,m ∆ T where Cp,m is the molar
 heat capacity at constant pressure.

 Cp,m = ∆Hθ
m/ ∆T = J mol-1
/ K
 = J K-1
mol-1
 ∴ ∆ HT2 = ∆ HT1 + ∆ Cp ( T2 - T1)
 Kirchoff’s equation.
 and
 ∆ Cp = Σn Cp(products)- ΣnCp(reactants)
 For a wide temperature range: ∆Cp ∫ dT between T1 and T2.
 Hence : qp = Cp( T2- T1) or ∆H = Cp ∆T and.


 qv = Cv ( T2 – T1) or Cv ∆T = ∆U
 ie. Cp = ∆H / ∆T ; Cv =∆U /∆T
 For small changes:

 Cp = dH / dT ; Cv = du / dT
 For an ideal gas: H = U + p V
 For I mol: dH/dT = dU/dT + R
 ∴ Cp = Cv + R
 Cp / Cv = γ ( Greek gamma)

Work done along isothermal paths
 Reversible and Irreversible paths
 ie ∆T =0 ( isothermal)
 pV = nRT= constant
 Boyle’s Law : piVi =pf Vf
 Can be shown on plot:

pV diagram
P
V
P
i
v
i
P
f
v
f
pV
= nRT = constant

 Work done = -( nRT)∫ dV/V
= - nRT ln (Vf/Vi)
 Equation is valid only if : piVi=pfVf and therefore: Vf/Vi = pi/pf and
 Work done = -( nRT) ln (pi/pf) and follows the path shown.

pV diagram
 An irreversible path can be
followed: Look at pV
diagram again.
V
P
Isothermal reversible process
(ie. at equlb. at every stage
of the process)
Irreversible reaction
PiVi
PfVf

An Ideal or Perfect Gas
 NB
For an ideal gas, ∆u = 0
Because: ∆U α KE + PE
α k T + PE (stored in bonds)
Ideal gas has no interaction between molecules (no bonds broken
or formed)

Therefore ∆u = 0 at ∆T = 0
Also ∆H = 0 since ∆(pV) = 0 ie no work done
This applies only for an ideal gas and NOT a chemical reaction.

Calculation
 eg. A system consisting of 1mole of perfect gas at 2 atm and
298 K is made to expand isothermally by suddenly reducing the
pressure to 1 atm. Calculate the work done and the heat that
flows in or out of the system.

 w = -pex ∆V = pex(Vf -Vi)
Vi = nRT/pi = 1 x 8.314 x (298)/202650
= 1.223 x 10-2
m3
Vf = 1 x 8.314 x 298/101325 = 2.445 x 10-2
m3
therefore, w = -pex (Vf- Vi)
= -101325(2.445-1.223) x 10-2
= -1239 J
∆U = q + w; for a perfect gas ∆U = 0
therefore q = -w and
q = -(-1239) = +1239 J

Work done along adiabatic path
 ie q = 0 , no heat enters or leaves the system.
 Since ∆U = q + w and q =0
 ∆U = w
 When a gas expands adiabatically, it cools.
 Can show that: pVγ
= constant, where ( Cp/Cv =γ )
 and: piVi
γ
= pfVf
γ
and since:
 -p dV = Cv dT

 Work done for adiabatic path = Cv (Tf- Ti)
 For n mol of gas: w = n Cv (Tf –Ti)
 Since piVi
γ
= pfVf
γ
 piVi
γ
/Ti = pfVf
γ
/ Tf
 ∴
Tf = Ti(Vi/Vf)γ-1
 ∴ w =n Cv{ ( Ti( Vi/ Vf)γ-1
– Ti}
 An adiabatic pathway is much steeper than pV = constant
pathway.

Summary
 piVi = pfVf for both reversible and irreversible
 Isothermal processes.
 For ideal gas: For ∆T =0, ∆U = 0, and ∆H=0
 For reversible adiabatic ideal gas processes:
 q=0 , pVγ
= constant and
 Work done = n Cv{ ( Ti( Vi/ Vf)γ-1
– Ti}
 piVi
γ
= pfVf
γ
for both reversible and irreversible adiabatic ideal
gas.

2nd
Law of Thermodynamics
 Introduce entropy, S (state function) to explain spontaneous
 change ie have a natural tendency to occur- the apparent
 driving force of spontaneous change is the tendency of energy
 and matter to become disordered. That is, S increases on
disordering.
 2nd
law – the entropy of the universe tends to increase.

Entropy
 ∆S = qrev /T ( J K-1
) at equilibrium
 ∆Sisolated system > 0 spontaneous change
 ∆Sisolated system < 0 non-spontaneous change
 ∆Sisolated system = 0 equilibrium

Properties of S
 If a perfect gas expands isothermally from
 Vi to Vf then since ∆U = q + w = 0
 ∴ q = -w ie
 qrev = -wrev and
 wrev = - nRT ln ( Vf/Vi)
 At eqlb., ∆S =qrev/T = - qrev/T = nRln (Vf/Vi)
 ie ∆S = n R ln (Vf/Vi)
 Implies that ∆S ≠ 0 ( strange!)
 Must consider the surroundings.

Surroundings
 ∆Stotal = ∆Ssystem + ∆Ssurroundings
 At constant temperature surroundings give heat to the system
to maintain temperature.
 ∴ surroundings is equal in magnitude to heat gained or loss but
of opposite sign to make

 ∆S = 0 as required at eqlb.

 Rem: dq = Cv dT and
 dS = dqrev / T
 ∴ dS = Cv dT/ T and
 ∆S = Cv ∫ dT /T between Ti and Tf
 ∆S = Cv ln ( Tf/ Ti )
 When Tf/ Ti > 1 , ∆S is +ve
 eg. L → G , ∆S is +ve
 S → L , ∆S is +ve and since qp = ∆H
 ∆Smelt = ∆Hmelt / Tmelt and
 ∆Svap = ∆Hvap / Tvap

Third Law of Thermodynamics
 eg. Standard molar entropy, ∆Sθ
mThe entropy of a perfectly
crystalline substance is zero at T = 0
 ∆Sθ
m/ J K-1
at 298 K
 ice 45
 water 70 NB. Increasing disorder
 water vapour 189
 For Chemical Reactions:
 ∆Sθ
rxn = Σ n ∆Sθ
(products) - Σ n ∆Sθ
( reactants)
 eg. 2H2 (g) + O2( g) → 2H2O( l), ∆Hθ
= - 572 kJ mol-1

Calculation
 Ie surroundings take up + 572kJ mol-1
of heat
∆Sθ
rxn = 2∆Sθ
(H2Ol) - (2 ∆Sθ
(H2g ) + ∆Sθ
(O2g) )
 = - 327 JK-1
mol-1
( strange!! for a spontaneous reaction;
for this ∆Sθ
is + ve. ).
 Why? Must consider S of the surroundings also.
 ∆Stotal = ∆S system + ∆S surroundings
 ∆Ssurroundings = + 572kJ mol-1
/ 298K = + 1.92 x 103
JK-1
mol-1
∴ ∆S total =( - 327 JK-1
mol-1
) + 1.92 x 103
= 1.59 x 103
JK-1
mol-1
 Hence for a spontaneous change, ∆S > 0

Free Energy, G
 Is a state function. Energy to do useful work.
 Properties
 Since ∆Stotal = ∆Ssystem + ∆Ssurroundings
 ∆Stotal= ∆S - ∆H/T at const. T&p
 Multiply by -T and rearrange to give:
 -T∆Stotal = - T ∆S + ∆H and since ∆G = - T ∆Stotal
 ie. ∆G = ∆H - T ∆S
 Hence for a spontaneous change: since ∆S is + ve, ∆G = -ve.

Free energy
 ie. ∆S > 0, ∆G < 0 for spontaneous change ;
 at equilibrium, ∆G = 0.
 Can show that : (dG)T,p = dwrev ( maximum work)
 ∴ ∆G = w (maximum)

Properties of G
 G = H - T S
 dG = dH – TdS – SdT
 H = U + pV
 ∴dH = dU + pdV + Vdp
 Hence: dG = dU + pdV + Vdp – TdS – SdT
 dG = - dw + dq + pdV + Vdp – TdS – SdT

 ∴ dG = Vdp - SdT

For chemical Reactions:
 For chemical reactions
 ∆Gθ
=Σ n Gθ
(products) - Σ n Gθ
(reactants)

 and
 ∆Gθ
rxn = ∆Hθ
rxn - T ∆Sθ
rxn

Relation between ∆Gθ
rxn and position of
equilibrium
 Consider the reaction: A = B
 ∆Gθ
rxn = Gθ
B - Gθ
A
 If Gθ
A> Gθ
B , ∆Gθ
rxn is – ve ( spontaneous rxn)
 At equilibrium, ∆Gθ
rxn = 0.
 ie. Not all A is converted into B; stops at equilibrium point.

For non-spontaneous rxn. GB > GA,
∆G is + ve

Gas phase reactions
 Consider the reaction in the gas phase:
 N2(g) + 3H2(g)→ 2NH3(g)
 Q =( pNH3 / pθ
)2
/( ( pN2/ pθ
) (pH2/ pθ
)3 )
where :
 Q = rxn quotient ; p = partial pressure and pθ
= standard
pressure = 1 bar
 Q is dimensionless because units of partial pressure cancelled
by pθ
.
 At equilibrium:
 Qeqlb = K = (( pNH3 / pθ
)2
/ ( pN2/ pθ
) (pH2/ pθ
)3
)eqlb

Activity ( effective concentration)
 Define: aJ = pJ / pθ
where a = activity or effective concentration.
 For a perfect gas: aJ = pJ / pθ
 For pure liquids and solids , aJ = 1
 For solutions at low concentration: aJ = J mol dm-3
 K = a2
NH3 / aN2 a3
H2
 Generally for a reaction:
 aA + bB → cC + dD
 K = Qeqlb = ( ac
C ad
D / aa
A ab
B ) eqlb = Equilibrium constant

Relation of ∆G with K
 Can show that:
 ∆Grxn = ∆Gθ
rxn + RT ln K
 At eqlb., ∆Grxn = 0
 ∴ ∆Gθ
rxn = - RT ln K
 Hence can find K for any reaction from thermodynamic data.

 Can also show that:
 ln K = - ∆Gθ
/ RT
 ∴K = e - ∆Gθ / RT
 eg
 H2 (g) + I2 (s) = 2HI (g) , ∆Hθ
f HI = + 1.7 kJ mol-1
at 298K; ∆Hθ
f H2
=0 ; ∆Hθ
f I 29(s)= 0

calculation
 ∆Gθ
rxn = 2 x 1.7 = + 3.40 kJ mol-1
 ln K = - 3.4 x 103
J mol-1
/ 8.314J K-1
mol-1
x 298K
 = - 1.37
 ie. K = e– 1.37
= 0.25

 ie. p2
HI / pH2 pθ
=0.25 ( rem. pθ
= 1 bar; pθ 2
/ pθ
= pθ
)
 ∴ p2
HI = pH2 x 0.25 bar

Example: relation between Kp and K
Consider the reaction:
N2 (g) + 3H2 (g) = 2NH3 (g)
Kp =( pNH3 / pθ
)2
/ ( pN2/ pθ
)( pH2 / pθ
)3
and
K = [( pNH3 / pθ
)2
/ ( pN2/ pθ
)( pH2 / pθ
)3
] eqlb
∴ Kp = K (pθ
)2
in this case. ( Rem: (pθ
)2
/ (pθ
)4
= pθ -2

 For K >> 1 ie products predominate at eqlb. ~ 103
 K<< 1 ie reactants predominate at eqlb. ~ 10-3
 K ~ 1 ie products and reactants in similar amounts.

Effect of temperature on K
 Since ∴ ∆Gθ
rxn = - RT ln K = ∆Hθ
rxn - T∆Srxn
 ln K = - ∆Gθ
rxn / RT = - ∆Hθ
rxn/RT + ∆Srxn
θ
/R
 ∴ ln K1 = - ∆Gθ
rxn / RT1 = - ∆Hθ
rxn/RT1 + ∆Srxn
θ
/R
 ln K2 = - ∆Gθ
rxn / RT2 = - ∆Hθ
rxn/ RT2 + ∆Srxn
θ
/ R

 ln K1 – ln K2 = - ∆Hθ
rxn / R ( 1/ T1 - 1/ T2 ) 0r
 ln ( K1/ K2) = - ∆Hθ
rxn / R ( 1/ T1 - 1/ T2 ) van’t Hoff equation

Thermodynamics Essentials

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