Download THERMODYNAMICS. Elements of Physical Chemistry. By P. Atkins

THERMODYNAMICS.
Elements of Physical Chemistry. By P. Atkins
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
Example:
↔
↔
matter
energy
↔
→
Open system
energy not matter
→
Closed system
matter
Energy
Isolated system
×
×

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)
2.
where k = Boltzmann constant
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
Pex
h
P
pressure (P)
h is distance moved
A = area of piston
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
 ∆U = - 2220 + 7.43 = - 2212.6 kJ
q = - 2220 kJ
w = 7432.7 J = 7.43 kJ
NB:
qp  ∆U
why?
Only equal if no work is done i.e. ∆V = 0
i.e. qv = ∆U
Example: energy diagram
U
C3H8 + 5 O2 (Ui)
reaction path
U
3CO2 + 4H2O(l) (Uf)
progress of reaction
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
 work =
p = nRT/ V
p  dV = nRT  dV/ V
Vf
Vi
Vf
Vi
= nRT ln (Vf/Vi) because dx/x = ln x

Work done by system= -nRT ln(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

Some properties of H
(p = pex)
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.
path 1
longitude
B
path 2
A
lattitude
- 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
H = - 2220 kJ mol-1 at 298 K
 represents the standard state.
H2(g) → H(g) + H(g), H diss = +436kJmol-1
H2O(l) → H2O(g), H vap = +44.0 kJmol-1
Calculate U for the following reaction:
CH4(l) + 2 O2(g) → CO2(g) + 2H2O(l), H = - 881.1kJmol-1
H = U + (pV)
= U + pi V + Vi p + p V
NB: p = 1 bar, i.e. p = 0
 H = U + pi V
Since -pi V = - nRT,
U = H - nRT
calculation
 U = - 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

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 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 can calculate H f rxn for any reaction.
Eg. C3H8 (g) + 5O2(g) → 3CO2(g) + 4H2O(l)

Calculate Hrxn given that:
Hf of C3H8(g) = - 103.9 kJ mol-1
Hf of O2(g) = 0 (reference state)
Hf of CO2(g) = - 393.5 kJ mol-1
Hf of H2O(l) = - 285.8 kJ mol-1
Hrxn = n H (products)- n H(reactants)
Hf(products) = 3  (- 393.5) + 4  (- 285.8)
= - 1180.5 -1143.2 = - 2323.7 kJ mol-1
Hf(reactants) = - 103.9 + 5  0 = - 103.9 kJ mol-1
H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) , Hcomb =? at 298K
From thermochemical data:
 C (g) +O2 (g) → CO 2(g) Hcomb =-393.5 kJmol1…………………………….(1)
CO (g) +1/2 O2 (g) →CO 2(g), Hcomb = -283.0 kJ mol(2)
1…………………….
Subtract 2 from 1 to give:
C (g) + O2 (g) – CO (g) – 1/2 O2 (g) → CO2 (g) – CO2 (g)
 C (g) + ½ O 2 (g) → CO (g) , H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 of CH 4 (g) =- 75 kJ mol-1
Hf of H (g) = 218 kJ mol-1
Hf of C (g) = 713 kJ mol-1








Hdiss =  nHf (products) -  nHf ( reactants)
= 713 + ( 4x 218) – (- 75) = 1660 kJ
mol-1
Since have 4 bonds : C-H = 1660/4 = 415
kJ mol-1
Variation of H with temperature
Suppose do reaction at 400 K, need to know
Hf at 298 K for comparison with literature value. How?

As temp.î HmÎ ie. Hm  T

 Hm = Cp,m  T where Cp,m is the molar

heat capacity at constant pressure.
 Cp,m = Hm/ T = J mol-1/ K

= J K-1 mol-1
   HT2 =  HT1 +  Cp ( T2 - T1)
 Kirchoff’s equation.
 and
  Cp = n Cp(products)- nCp(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
Pv
i i
P
pV=
nRT = constant
Pv
f f
V

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
P

An irreversible path can be
followed: Look at pV
diagram again.
PiVi
Isothermal reversible process
(ie. at equlb. at every stage
of the process)
PfVf
Irreversible reaction
V
An Ideal or Perfect Gas

NB
For an ideal gas, u = 0
Because: U  KE + PE
kT+
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: piViγ = pfVfγ 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
piViγ = pfVf γ
piViγ/Ti = pfVf γ/ Tf
 T = T (V /V )γ-1
f
i
i f
 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}
piViγ = pfVfγ 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

Sisolated system > 0 spontaneous change

Sisolated system < 0 non-spontaneous change

Sisolated system = 0 equilibrium
( J K-1) at 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, SmThe entropy of a perfectly
crystalline substance is zero at T = 0


Sm/ J K-1 at 298 K
ice
45
water
70
NB. Increasing disorder
water vapour 189
For Chemical Reactions:
Srxn =  n S (products) -  n S ( reactants)

eg. 2H2 (g)




+ O2( g)
→
2H2O( l ), H = - 572 kJ mol-1
Calculation





Ie surroundings take up + 572kJ mol-1 of heat
Srxn = 2S(H2Ol) - (2 S (H2g ) + S (O2g) )
= - 327 JK-1 mol-1 ( strange!! for a spontaneous
reaction; for this S is + ve. ).
Why? Must consider S of the surroundings also.
S total = S system + S surroundings
S surroundings = + 572kJ mol-1/ 298K = + 1.92 x 103JK-1 mol-1
 S total =( - 327 JK-1mol-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 -TS
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
G = n G (products) -  n G (reactants)



and
Grxn = Hrxn - T Srxn
Relation between Grxn and position of
equilibrium

Consider the reaction:
A
= B

Grxn = GB - GA

If GA> GB , Grxn is – ve ( spontaneous rxn)

At equilibrium, Grxn = 0.

ie. Not all A is converted into B; stops at equilibrium point.
Equilibrium diagram
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 / p)2 /( ( pN2/ p) (pH2/ p)3

Q = rxn quotient ; p = partial pressure and p = standard
pressure = 1 bar
Q is dimensionless because units of partial pressure cancelled
by p .
At equilibrium:
Qeqlb = K = (( pNH3 / p)2 / ( pN2/ p) (pH2/ p)3 )eqlb




)
where :
Activity ( effective concentration)

Define: aJ = pJ / p where a = activity or effective
concentration.
For a perfect gas: aJ = pJ / p
For pure liquids and solids , aJ = 1
For solutions at low concentration: aJ = J mol dm-3
K = a2NH3 / aN2 a3H2
Generally for a reaction:
aA + bB → cC + dD

K = Qeqlb = ( acC adD / aaA abB ) eqlb = Equilibrium constant






Relation of G with K

Can show that:

Grxn = Grxn + RT ln K

At eqlb., Grxn = 0

 Grxn = - RT ln K

Hence can find K for any reaction from thermodynamic data.

Can also show that:

ln K = - G / RT

K = e - G / RT

eg
H2 (g) + I2 (s) = 2HI (g) , Hf HI = + 1.7 kJ mol-1 at 298K;
Hf H2 =0 ; Hf I 29(s)= 0

calculation

G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 p =0.25 ( rem. p = 1 bar; p 2 / p = p )

 p2 HI = pH2 x 0.25 bar
Example: relation between Kp and K
Consider the reaction:
N2 (g) + 3H2 (g) = 2NH3 (g)
Kp =( pNH3 / p)2 / ( pN2/ p )( pH2 / p)3
and
K = [( pNH3 / p)2 / ( pN2/ p )( pH2 / p)3] eqlb

Kp = K (p)2 in this case. ( Rem: (p)2 / (p)4 = p -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  Grxn = - RT ln K = Hrxn - TSrxn

ln K = - Grxn / RT = - Hrxn/RT + Srxn/R


 ln K1 = - Grxn / RT1 = - Hrxn/RT1 + Srxn/R
ln K2 = - Grxn / RT2 = - Hrxn/ RT2 + Srxn/ R



ln K1 – ln K2 = - Hrxn / R ( 1/ T1 - 1/ T2 )
ln ( K1/ K2) = - Hrxn / R ( 1/ T1 - 1/ T2 )
0r
van’t Hoff equation