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: l Energy can neither be created nor destroyed in chemical reactions. It can only be converted from one form to the other. l UNIVERSE Ø Ø 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 × ×

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

WORK and HEAT l Work – transfer of energy to change height of the weight in surrounding eg: work to run a piston by a gas. l Heat – transfer of energy is a result of temperature difference between system and surrounding eg: HCl(aq) + Na. OH(aq) → Na. Cl(aq) + H 2 O(l) - heat given off. If heat released to surroundings – exothermic. If heat absorbed by surroundings – endothermic.

Example: Gasoline, 2, 2, 4 trimethylpentane CH 3 C(CH 3)2 CH 2 CH(CH 3)CH 3 + 25/2 O 2 → 8 CO 2(g) +H 2 O(l) 5401 k. J 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) l l 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 Ø ∆U = Uf – Ui Ø q – heat applied to system Ø W – work done on system Ø l 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 mv 2 (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 m 2 s-2 (m) (g) (h) 1 kg m 2 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 × h. A Work done on system = pex × ∆V ∆V – change in volume (Vf – Vi) Work done by system = -pex × ∆V Since U is decreased

Example: C 3 H 8(g) + 5 O 2(g) → 3 CO 2(g) + 4 H 2 O(l) at 298 K 1 atm (1 atm = 101325 Pa), -2220 k. J = q What is the work done by the system? For an ideal gas; p. V = n. RT (p = pex) n – no. of moles R – gas constant T = temperature V – volume p = pressure

V= n. RT/p or Vi = ni. RT/pex 6 moles of gas: Vi = (6 × 8. 314 × 298)/ 101325 = 0. 1467 m 3 3 moles of gas: Vf = (3 × 8. 314 × 298)/ 101325 = 0. 0734 m 3 work done = -pex × (Vf – Vi) = -101325 (0. 0734 – 0. 1467) = +7432 J

NB: work done = - pex (nf. RT/pex – ni. RT/pex) = (nf – ni) RT Work done = -∆ngas. RT 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 k. J q = - 2220 k. J w = 7432. 7 J = 7. 43 k. J

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

l l For an infinitesimal change in volume, d. V Work done on system = pd. V For ideal gas, p. V = n. RT p = n. RT/ V work = p d. V = n. RT d. V/ V = n. RT ln (Vf/Vi) because dx/x = ln x l Work done by system= -n. RT ln(Vf/Vi)

Enthalpy, H l 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 l Enthalpy is the sum of internal energy and the product of p. V of that substance. i. e H = U + p. V l Some properties of H (p = pex)

Hi = Ui + p. Vi Hf = Uf + p. Vf 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 l H = qp

suppose p and V are not constant? • H = U + ( p. V) 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 l By IUPAC conventions as the pure form of the substance at 1 bar pressure (1 bar = 100, 000 Pa). What about temperature? l By convention define temperature as 298 K but could be at any temperature.

Example: C 3 H 8(g) + 5 O 2(g) → 3 CO 2(g) + 4 H 2 O(l) at 1 bar pressure, qp = - 2220 k. Jmol-1. l Since substances are in the pure form then can write H = - 2220 k. J mol-1 at 298 K represents the standard state.

H 2(g) → H(g) + H(g), H diss = +436 k. Jmol-1 H 2 O(l) → H 2 O(g), H vap = +44. 0 k. Jmol-1 Calculate U for the following reaction: CH 4(l) + 2 O 2(g) → CO 2(g) + 2 H 2 O(l), H = - 881. 1 k. Jmol-1

H = U + (p. V) = U + pi V + Vi p + p V NB: p = 1 bar, i. e. p = 0 H = U + pi V Since -pi V = - n. RT, U = H - n. RT

calculation U = - 881. 1 – ((1 – 2)(8. 314) 298)/ 1000 = - 881. 1 – (-1)(8. 314)(0. 298) = - 2182. 99 k. J mol-1

STANDARD ENTHALPY OF FORMATION, Hf l Defined as standard enthalpy of reaction when substance is formed from its elements in their reference state. l Reference state is the most stable form of element at 1 bar atmosphere at a given temperature eg. At 298 K Carbon = Cgraphite Hydrogen = H 2(g) Mercury = Hg(l) Oxygen = O 2(g) Nitrogen = N 2(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. Ø Ø From tables of Hf can calculate H f rxn for any reaction.

Eg. C 3 H 8 (g) + 5 O 2(g) → 3 CO 2(g) + 4 H 2 O(l) l Calculate H rxn given that: H f of C 3 H 8(g) = - 103. 9 k. J mol-1 H f of O 2(g) = 0 (reference state) H f of CO 2(g) = - 393. 5 k. J mol-1 H f of H 2 O(l) = - 285. 8 k. J mol-1 Hrxn = n H (products)- n H (reactants)

H f(products) = 3 (- 393. 5) + 4 (- 285. 8) = - 1180. 5 -1143. 2 = - 2323. 7 k. J mol-1 H f(reactants) = - 103. 9 + 5 0 = - 103. 9 k. J mol-1 H rxn = - 2323. 7 – (- 103. 9) = - 2219. 8 k. J mol-1 = - 2220 k. J mol-1

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

Hess’ Law l l - 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 O 2(g) → CO (g) , H comb =? at 298 K

From thermochemical data: l C (g) +O 2 (g) → CO 2(g) H comb =-393. 5 k. Jmol 1………………. (1) CO (g) +1/2 O 2 (g) →CO 2(g), H comb = -283. 0 k. J mol(2) 1…………. Subtract 2 from 1 to give: C (g) + O 2 (g) – CO (g) – 1/2 O 2 (g) → CO 2 (g) – CO 2 (g) C (g) + ½ O 2 (g) → CO (g) , H comb= -393. 5 – (-283. 0) = - 110. 5 k. J mol-1 – l l l

Bond Energies l eg. C-H bond enthalpy in CH 4 l CH 4 (g) → C (g) + 4 H (g) , at 298 K. Need: Hf of CH 4 (g) =- 75 k. J mol-1 Hf of H (g) = 218 k. J mol-1 Hf of C (g) = 713 k. J mol-1 l l l

l l l H diss = n Hf (products) - n Hf ( reactants) = 713 + ( 4 x 218) – (- 75) = 1660 k. J mol-1 Since have 4 bonds : C-H = 1660/4 = 415 k. J 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? l As temp. î H mÎ ie. H m T l H m = Cp, m T where Cp, m is the molar l heat capacity at constant pressure.

l Cp, m = H m/ T = J mol-1/ K l l l l = J K-1 mol-1 HT 2 = HT 1 + Cp ( T 2 - T 1) Kirchoff’s equation. and Cp = n Cp(products)- n. Cp(reactants) For a wide temperature range: Cp ∫ d. T between T 1 and T 2. Hence : qp = Cp( T 2 - T 1) or H = Cp T and.

l l l qv = Cv ( T 2 – T 1) or Cv T = U ie. Cp = H / T ; Cv = U / T For small changes: l l Cp = d. H / d. T ; Cv = du / d. T l For an ideal gas: H = U + p V For I mol: d. H/d. T = d. U/d. T + R Cp = C v + R Cp / Cv = γ ( Greek gamma) l l l

Work done along isothermal paths l Reversible and Irreversible paths ie T =0 ( isothermal) l p. V = n. RT= constant l Boyle’s Law : pi. Vi =pf Vf l Can be shown on plot: l

p. V diagram Pv i i p. V= n. RT = constant Pv f f

l Work done = -( n. RT)∫ d. V/V = - n. RT ln (Vf/Vi) l Equation is valid only if : pi. Vi=pf. Vf and therefore: Vf/Vi = pi/pf and l Work done = -( n. RT) ln (pi/pf) and follows the path shown.

p. V diagram l An irreversible path can be followed: Look at p. V diagram again.

An Ideal or Perfect Gas l 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 (p. V) = 0 ie no work done This applies only for an ideal gas and NOT a chemical reaction.

Calculation l eg. A system consisting of 1 mole 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.

l w = -pex V = pex(Vf -Vi) Vi = n. RT/pi = 1 x 8. 314 x (298)/202650 = 1. 223 x 10 -2 m 3 Vf = 1 x 8. 314 x 298/101325 = 2. 445 x 10 -2 m 3 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 l l l l 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: p. Vγ = constant, where ( Cp/Cv =γ ) and: pi. Viγ = pf. Vfγ and since: -p d. V = Cv d. T

l l l l Work done for adiabatic path = Cv (Tf- Ti) For n mol of gas: w = n Cv (Tf –Ti) Since pi. Viγ = pf. Vf γ pi. Viγ/Ti = pf. Vf 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 p. V = constant pathway.

Summary l l l l pi. Vi = pf. Vf 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 , p. Vγ = constant and Work done = n Cv{ ( Ti( Vi/ Vf)γ-1 – Ti} pi. Viγ = pf. Vfγ for both reversible and irreversible adiabatic ideal gas.

2 nd Law of Thermodynamics l Introduce entropy, S (state function) to explain spontaneous l 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. l l l 2 nd law – the entropy of the universe tends to increase.

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

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

Surroundings l Stotal = Ssystem + Ssurroundings l 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 l l l S = 0 as required at eqlb.

l l l l l Rem: dq = Cv d. T and d. S = dqrev / T d. S = Cv d. T/ T and S = Cv ∫ d. T /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 l eg. Standard molar entropy, S m. The entropy of a perfectly crystalline substance is zero at T = 0 l 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) l eg. 2 H 2 (g) l l l + O 2( g) → 2 H 2 O( l ), H = - 572 k. J mol-1

Calculation l l l Ie surroundings take up + 572 k. J mol-1 of heat S rxn = 2 S (H 2 Ol) - (2 S (H 2 g ) + S (O 2 g) ) = - 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 = + 572 k. J mol-1/ 298 K = + 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 l Hence for a spontaneous change, S > 0

Free Energy, G l 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 l ie. G = H - T S l Hence for a spontaneous change: since S is + ve, G = -ve. l l l

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

Properties of G l l l G =H -TS d. G = d. H – Td. S – Sd. T H = U + p. V d. H = d. U + pd. V + Vdp Hence: d. G = d. U + pd. V + Vdp – Td. S – Sd. T d. G = - dw + dq + pd. V + Vdp – Td. S – Sd. T l l d. G = Vdp - Sd. T

For chemical Reactions: l l For chemical reactions G = n G (products) - n G (reactants) l l l and G rxn = H rxn - T S rxn

Relation between G rxn and position of equilibrium l Consider the reaction: A = B l G rxn = G B - G A l If G A> G B , G rxn is – ve ( spontaneous rxn) l At equilibrium, G rxn = 0. l 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 l Consider the reaction in the gas phase: N 2(g) + 3 H 2(g)→ 2 NH 3(g) l Q =( p. NH 3 / p )2 /( ( p. N 2/ p ) (p. H 2/ p )3 l 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 = (( p. NH 3 / p )2 / ( p. N 2/ p ) (p. H 2/ p )3 )eqlb l l ) where :

Activity ( effective concentration) l Define: a. J = p. J / p where a = activity or effective concentration. For a perfect gas: a. J = p. J / p For pure liquids and solids , a. J = 1 For solutions at low concentration: a. J = J mol dm-3 K = a 2 NH 3 / a. N 2 a 3 H 2 Generally for a reaction: a. A + b. B → c. C + d. D l K = Qeqlb = ( ac. C ad. D / aa. A ab. B ) eqlb = Equilibrium constant l l l

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

l Can also show that: l ln K = - G / RT l K = e - G / RT l eg H 2 (g) + I 2 (s) = 2 HI (g) , H f HI = + 1. 7 k. J mol-1 at 298 K; H f H 2 =0 ; H f I 29(s)= 0 l

calculation l G rxn = 2 x 1. 7 = + 3. 40 k. J mol-1 l ln K = - 3. 4 x 103 J mol-1 / 8. 314 J K-1 mol-1 x 298 K = - 1. 37 ie. K = e – 1. 37 = 0. 25 l l ie. p 2 HI / p. H 2 p =0. 25 ( rem. p = 1 bar; p 2 / p = p ) l p 2 HI = p. H 2 x 0. 25 bar

Example: relation between Kp and K Consider the reaction: N 2 (g) + 3 H 2 (g) = 2 NH 3 (g) Kp =( p. NH 3 / p )2 / ( p. N 2/ p )( p. H 2 / p )3 and K = [( p. NH 3 / p )2 / ( p. N 2/ p )( p. H 2 / p )3] eqlb Kp = K (p )2 in this case. ( Rem: (p )2 / (p )4 = p -2

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

Effect of temperature on K l Since G rxn = - RT ln K = H rxn - T Srxn l ln K = - G rxn / RT = - H rxn/RT + Srxn /R l l ln K 1 = - G rxn / RT 1 = - H rxn/RT 1 + Srxn /R ln K 2 = - G rxn / RT 2 = - H rxn/ RT 2 + Srxn / R l ln K 1 – ln K 2 = - H rxn / R ( 1/ T 1 - 1/ T 2 ) ln ( K 1/ K 2) = - H rxn / R ( 1/ T 1 - 1/ T 2 ) 0 r van’t Hoff equation