Chem 340 - Lecture Notes 7 – Fall 2013 – Gibbs & Helmholtz Free Energy

With entropy we found a state function that indicates direction of a spontaneous process, however it requires consideration of the entropy change of both system and surroundings, as chemists we prefer to focus on the system, usually not isolated, i.e. in contact with surroundings, and often we work in conditions of constant pressure and sometimes also at constant temperature. Alternatively one can imagine cases where constant volume is a characteristic. Use of these ideas helps design two new new state functions that can indicate direction of a process and one of which most useful for chemistry. The Clausius inequality will be critical in making the indication work.

dS ≥ ᵭq/T  T dS ≥ ᵭq recall: only equal for reversible process, dqrev recall 1st law: dU = ᵭq + ᵭw rearrange, substitute dq: TdS ≥ dU - ᵭw or TdS + ᵭw - dU ≥ 0 now there are two kinds of work, external or expansion, and non-expansion work, split up:

TdS - PextdV + ᵭwnonexp - dU ≥ 0 Spontaneity now a function of state functions: U, V, S, T, and path dependent ones  w

Consider isothermal process: (const T)

Define new state function, Helmholtz free energy: A = U - TS

or equivalently:

A is a state function that calculates maximum work from system on surround, i.e. A = wrev

Now if add constraint: constant V and only expansion work possible then dA ≤ 0

Chemists prefer const P conditions, at const T,P: d(PV) = PdV, d(TS) = TdS, back to inequ.

dU - TdS + PdV ≤ 0  d(U +PV –TS) = d(H - TS) ≤ ᵭwnonexp let new state function be Gibbs Free energy: G = H – TS,

dG ≤ ᵭwnonexp or if ᵭwnonexp = 0, then dG ≤ 0 indicates spont, const T,P only PV work so now have two criterion for spontaneity, each works best under different conditions, dA for constant V, dG for constant P. Go back to Clausius

dS - ᵭq/T ≥ 0 recall dSsur = - ᵭq/T so dS - dSsur ≥ 0 or the spontaneity condition

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but for G and A - criteria for spontaneity - not depend on surroundings, also from Clausius. G = H - TS and G < 0  spontaneous isothermal, constant P process

See both enthalpy and entropy contribute to spontaneity, Entropy greater impact at higher temperatures Chemical reaction spontaneous if H < 0 (exothermic) and S > 0 Chemical reaction never spontaneous if H > 0 (endothermic) and S < 0 Balance of H and S control reaction, if H < 0 & S < 0 or H > 0 & S > 0

Similar for constant V change  A = U - TS and A < 0  spontaneous isothermal, constant V process

Examples: -from Engel,on web site, but - sign/magnitude problems in text!! Calculating A, U, S and w for reversibly contracting collagen from l = 0.20.1 m with -1 absorption of qrev = 0.15 J at const T = 310 K, let tension  = - l,  = 10 Nm

For this, Uw = -q, A = wrev = -0.15 J  same mechanical stretch w = -∫dl = ∫ldl -4 -1 -1. . 2 S = qrev/T = 0.15 J / 310 K = 4.83x10 JK 10Nm ½ (-0.03 m ) = -0.15 J

If dU = TdS – PdV - dl determine (G/l)P,T and maximum work obtainable by contract dG = dH – d(TS) = d(U +PV) – d(TS) = TdS – PdV - dl + PdV + VdP – TdS – SdT = - dl + VdP – SdT sign issue! l for contraction

Constant P,T: dT = dP = 0  (G/l)P,T = - l, dG =ldl

Integrate and get G =∫ldl = -0.15 J = wrev and see same as A = wrev

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Example of RNA stretching. If hook opposite ends of strand to two polystyrene beads, can pull apart and measure force required a function of distance. Need very small scale and to grab spheres (~1m)  laser tweezer can hold an object and measure force exerted See (middle, black stretch, red refold) at 10 pNs-1, the path not reversible, but at 1 pNs-1 (right) black and red overlap, approximately reverses. See at ~13 pN an abrupt increase in length, this is unfolding, and length difference (~20 nm) is length of the RNA. From this work, determined: G = 174 kJ/mol, compare to melting (phase trans) at ~80 C

Fundamental equations - Differential forms of U, H, A, G Start from definitions and re-express as differentials in terms of thermodynamic variables: H = U + PV method use dU = dq + dw but recognize state function

A = U – TS dU = dqrev + dwrev = TdS - PdV G = H – TS assume only P,V work Take differentials: dH = dU + d(PV) = TdS – PdV + PdV + VdP = TdS + VdP dA = dU – d(TS) = TdS – PdV – TdS – SdT = -SdT – PdV dG = dH – d(TS) = TdS + VdP – TdS – SdT = -SdT + VdP Used a lot, called the fundamental equations: dU = TdS – PdV dH = TdS + VdP dA = -SdT – PdV dG = -SdT + VdP - text calls this fundamental equation of chemical thermodynamics these are used to develop a set of ”natural variables”; total differential two variables, U(S,V)

 Coefficients of dS and dV must be same in both equations for dU

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Same for the other equations (left): right use: (/V(U/S)V)S = (/S(U/V)S)V

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Left, define variances of thermodynamic right, Maxwell relationships, shows inter- state functions in terms of variables relationships of variables (T,P,V,S) for example H inc. with S,P (T,V = +) note first two have derivative at const S but G dec. w/ T and inc. w/ P (S,V = +) since dS=dqrev/T, const S  dq=0 adiabatic these relations look obscure but allow express state fct. in measurable things : T,P,V,

Example from text: in Notes 4 introduced internal pressure (U/V)T = T = T(P/T)V – P

From complete differential: dU = (U/S)VdS + (U/V)SdV divide through by dV, at const T

(U/V)T = (U/S)V(S/V)T + (U/V)S

Now use (U/S)V = T and (U/V)V = - P  T = (U/V)T = T(S/V)T – P and then use Maxwell relation: ( S/ V) = ( P/ T) = T( P/ T) – P (result)   T   V  T   V

Dependence of Free energies on P, V, T Just described how free energy depends on “natural variables:

So since S and P always positive, see A always decreases with increasing T or V Similarly for G; S and V always positive

So G always decreases with T and increases with P

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If at const T, and want change in free energy from standard state, can integrate over P Liquid and solid, V~ const, so simplify:

But in gases volume strongly depends on P, so substitute in for V in integral:

See Gm  - ∞ as P 0, since for gas that means V∞ and S = nR ln(Vf/Vi)  large +, See text, fig 3.19, 3.20 p. 125, show high T,P dependence of G for gas, more S,V

Standard molar Gibbs Free Energy o Normally only use G, so absolute values not needed, but for standard state can get a G m o If we take H m = 0 in its standard state, for a pure element, then o o o o o o G m = H m - TS m = - TS m (since for elements in std. state: H mH f = 0) o o o Pure compounds are formed from elemental react – formation: G f = G m prod +iG m, react i o o o Rearrange: G m = G f + iTS m, react i - reactants in formation i - only elements o so only TS m, react i term and i all negative, since reactant o o o note for molar enthalpy, equation simpler, misses elements, no S m contribut.: H m = H f

o Example. Calculate G m (a) for Ar(g) and (b) for H2O(l) at T=298 K, P=1 bar o o o o -1 -1 o (a) G m = H m - TS m = - TS m = -298Kx155 JK mol = - 4.62 kJ/mol (S m from table) o o o o o o (b) G m = G f + iTS m =G f (H2O) - TS m(H2) -1/2 TS m(O2) (- for reactants) = -237kJ/mol – 298K(131 + 205/2) J/Kmol = -307 kJ/mol

o o o Can just extend this to a Hess’s law formulation as well, G f = ProdG f - ReacG f o o o using standard free energy of formation: G f = H f - TS f [- for reactant] o o o So problems solved by just looking up G f values or alternatively H f and S f in tables

+ For ionic solutions, need to have both cations and anions, so define H aq for reference as 0: o + G f (H aq) = 0

Text example: HCl formation in solution: + - o ½ H2 + ½ Cl2  H aq +Cl aq G f = -131 kJ/mol o o + o - o - o - But G f =G f (H aq) +G f (Cl aq) = G f (Cl aq) or G f (Cl aq ) = -131 kJ/mol + - o Now consider: Ag(s) + ½ Cl2  Ag aq +Cl aq G f = -54 kJ/mol o + This implies G f (Ag aq) = 77 kJ/mol (i.e. -54 –(-131) = 77)

Remember that the process of forming an aqueous ion involves going from the element to the ion and then dissolving ion, so need to construct a cycle to model it (text p.120)

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Gibbs-Helmholtz equations For temperature dependence, one normally looks at the variation of G/T because equilibrium constants depend on G/T. This results in the Gibbs-Helmholtz equations:

This is also written, using; d(1/T)/dT = -1/T2 as:

(look ahead, Keq will depend on G/T and if get variation of it with 1/T gives solution for H)

For a reaction at constant P need to substitute G and H in equation above, and next “multiply” both sides by d(1/T), then if integrate both sides get:

Where in the last step we assume H is constant over the T range This equation lets you determine G at some temperature, if know Go and Ho at 298K

o Example: Calculate Gf for glycine, H2N(CH2)COOH, at 350K if G f = -378 kJ/mol and o o H f = -537 kJ/mol at T=298 K. Assume H f constant over T range.

Gf (T2) = T2[Gf (T1)/T1 + Hf (T1)x(1/T2 – 1/T1)] = 350K [-378/398 – 537(1/350 -1/298)]kJ/mol K = -350 kJ/mol (accident that [ ]≈1)

Gm varies with T and P as

Slopes of surface give –S and V

Gas has large S, steep dG/dT Solid least slope

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Example. Hydrolysis is the reaction to break up protein peptide bonds to form amino acids. As a test of protein stability, determine the standard free energy for hydrolysis of Gly-Gly, at T=298 K, p = 1 bar. + - + - Reaction: H3NCH2CONHCH2COO + H2O  2 H3NCH2COO Look up std free energy values, note only solid values available, but G small for Gly  solliq o o o o G = 2G (Gly) – G (GlyGly) - G (H2O) Go = [2(-378) – (-491) – (-237)]kJ/mol = - 28 kJ/mol So this process appears to be spontaneous – luckily it is slow unless one has proteolytic enzymes around, such as in digestion or in cell regulation pathways or cell death

But our body functions at ~37 C, is it still spontaneous?

Need Gibbs Helmholtz for this, which requires H, or use (G/T)P = -S  G = S(T) 1st determine Ho and So using Hess’s law/standard state methods (like for Go)  Ho = [2(-537)-(-745)-(286)]kJ/mol = -43 kJ/mol So = [2(104)-190-70]J/molK = -53 J/molK Use So  Go(310) = Go(298) -So(298)x(310-298) = [-28-12x(-0.53)]kJ/mol = -27 kJ/mol o o o o Use H  G /T2 = G /T1 +H (1/ T2 -1/T1) = [-28/298 + (-43)(1/310 -1/298)]kJ/molK Go(310) /310K = -0.087 kJ/molK Go(310) = - 27 kJ/mol Both give same results, but need to assume Ho and So constant, which is OK for T=12o

Pressure dependence: (G/P)T = V  ∫dG = ∫VdP = G(P2) – G(P1)

For gas, if ideal: G(P2) – G(P1) = nRT∫dP/P = nRT ln(P2/P1) For solid, liquid: G(P2) – G(P1) ≈ V(P2 – P1)

For solution reaction: G(P2) – G(P1) ≈ V(P2 – P1) where V=V(prod)-V(react) So if volume of products greater than reactants, increased P increases G,

if a gas, then ignore sol-liq volumes : G(P2) – G(P1) ≈ nRT ln(P2/P1)

Example. Can you make diamond form graphite at T=298, P=1 bar? C(s,graphite)  C(s,diamond) has Go = +2.9 kJ/mol So no, it does not work, sorry!

How much pressure does it take? Use G(P2) – G(P1) ≈ V(P2 – P1) want G(P2)=0 From densities: (g) = 2.25 g/cm3 and (d) = 3.51 g/cm3 3 3 Get molar volumes: Vm (g) = 5.33 cm /mol and Vm (d) = 3.42 cm /mol  G(P) = G(1) + V(P – 1) = [2.84 + (3.42-5.33)(P-1)] kJ/mol = 0 P = 1 + 2.84/1.94x10-4 = 15k bar = P (eq) Problem, P so large that V will not be constant, so actual P is higher, but does work Increasing T will not help because S = -3.4 J/molK so adds to G

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