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

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

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

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) - heat given off. 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: 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

w = force × distance moved in direction of force 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

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

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)

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

Hf – Hi = Uf – Ui +p(Vf – Vi) 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.

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 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 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 Pivi pV= nRT = constant Pfvf

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.

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)

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  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 = 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 ( 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, 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 - 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: 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 ) 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 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 ) 0r ln ( K1/ K2) = - Hrxn / R ( 1/ T1 - 1/ T2 ) van’t Hoff equation