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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.

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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
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
slide3
→ →

Example:

↔matter↔energy ↔ energy not mattermatter× Energy ×

Open system

Closed system

Isolated system

work and heat
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
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
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
Different types of energies:
  • Kinetic energy = ½ mv2 (chemical reaction) kinetic energy

(KE)  k T (thermal energy)where k = Boltzmann constant

  • Potential energy (PE) = mgh – energy stored in bonds

Now,U = KE + PE

slide9
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

slide11
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
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

slide13
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

slide14
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

slide15
NB:

qp ∆U why?

Only equal if no work is done i.e.∆V = 0

i.e. qv = ∆U

slide17
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

slide18
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)
enthalpy h
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
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
slide21
Hi = Ui + pVi

Hf = Uf + pVf

Hf – Hi = Uf – Ui +p(Vf – Vi)

or

H = U + p V

slide22
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
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

slide25
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
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.
example1
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.

slide28
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

slide29
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
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 h f
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)

slide32
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 c 3 h 8 g 5o 2 g 3co 2 g 4h 2 o l
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)

slide34
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

slide35
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
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
slide37
From thermochemical data:
          • C (g) +O2(g)→ CO 2(g)Hcomb=-393.5 kJmol-1…………………………….(1)
  • CO (g) +1/2 O2 (g) →CO 2(g), H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 O 2(g) → CO (g) , Hcomb= -393.5 –
  • (-283.0) = - 110.5 kJ mol-1
bond energies
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
slide39
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
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.
slide41
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.
slide42
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
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 diagram

Pivi

pV= nRT = constant

Pfvf

slide45
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.
an ideal or perfect gas
An Ideal or Perfect Gas
  • NB

For an ideal gas, u = 0

Because: U  KE + PE

 k T + PE (stored in bonds)

Ideal gas has no interaction between molecules (no bonds broken or formed)

slide48
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.

calculation1
Calculation
  • eg. A system consisting of 1mole of perfectgas 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.
slide50
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
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
slide52
Work done for adiabatic path = Cv (Tf- Ti)
  • For n mol of gas: w = n Cv (Tf –Ti)
  • Since piViγ= pfVfγ
  • piViγ/Ti= pfVfγ/ 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
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.
2 nd law of thermodynamics
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
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
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
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.
slide58
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
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
calculation2
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
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
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
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:
  • 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
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.
gas phase reactions
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 ) where :
  • 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
activity effective concentration
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
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.
slide71
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fH2 =0 ; HfI 29(s)= 0
calculation3
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. p2HI / pH2 p =0.25 ( rem. p = 1 bar; p 2 / p = p )
  •  p2HI = pH2 x 0.25 bar
example relation between k p and k
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

slide74
ForK >> 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
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 ) 0r
  • ln ( K1/ K2) = - Hrxn / R ( 1/ T1 - 1/ T2 ) van’t Hoff equation