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

→ →

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 forcew = 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 molesR – gas constantT = temperatureV – volumep = 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

ViVf

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

A

Bpath 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

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 calculationseg. 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 Hrxn given that:

Hf of C3H8(g) = - 103.9 kJ mol-1

Hf of O2(g) = 0 (reference state)

Hf of CO2(g) = - 393.5 kJ mol-1

Hf of H2O(l) = - 285.8 kJ mol-1

Hrxn = n H (products)- n H(reactants)

Hf(products) = 3 (- 393.5) + 4 (- 285.8)= - 1180.5 -1143.2 = - 2323.7 kJ mol-1

Hf(reactants) = - 103.9 + 5 0 = - 103.9 kJ mol-1

Hrxn = - 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) , Hcomb =? at 298K

– 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), Hcomb = -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

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

Hdiss = nHf (products) - nHf

( 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 knowH

f at 298 K for comparison with literature value. How?

As temp.î HmÎ ie. H

m T

\ Hm = Cp,m T where Cp,m is the molar

heat capacity at constant pressure.

Cp,m = Hm/ 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

P

V

Pivi

Pfv

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

NBFor 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)

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 = 0therefore 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 = 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 piVi

γ = pfVf γ

piViγ/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}

piViγ = 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, SmThe 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

Srxn = 2S(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: -TStotal = - 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

G = n G (products) - n G (reactants) and

Grxn = H

rxn - T Srxn

Relation between Grxn and position of

equilibrium

Consider the reaction: A = B

Grxn = G

B - GA

If GA> G

B , Grxn is – ve ( spontaneous rxn)

At equilibrium, Grxn = 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 ) 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)

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 a3

H2

Generally for a reaction: aA + bB → cC + dD

K = Qeqlb = ( acC ad

D / aaA ab

B ) eqlb = Equilibrium constant

Relation of G with K

Can show that:

Grxn = Grxn + RT ln K

At eqlb., Grxn = 0

\ Grxn = - 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 ; Hf I 29(s)= 0

calculation

Grxn = 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 \ Grxn = - RT ln K = H

rxn - TSrxn

ln K = - Grxn / RT = - H

rxn/RT + Srxn/R

\ ln K1 = - Grxn / RT1 = - H

rxn/RT1 + Srxn/R

ln K2 = - Grxn / RT2 = - H

rxn/ RT2 + Srxn/ R

ln K1 – ln K2 = - H

rxn / R ( 1/ T1 - 1/ T2 ) 0r

ln ( K1/ K2) = - Hrxn / R ( 1/ T1 - 1/ T2 ) van’t Hoff equation