Free Energy and Equilibrium.

If a reaction is at equilibrium, then DG for those conditions (which will not be standard conditions) must be zero:

This leads to:

DG_T = 0 = DGº_T + RT ln Q

but, at equilibrium, Q = K

so

DG_T = 0 = DGº_T + RT ln K

or

DGº_T = - RT ln K

This is an important equation!! Do not forget the sign.

There is a direct relationship between the equilibrium constant in terms of activities, and DGº_T.

ln K = - DGº_T / RT

Do not forget the sign!!

Calculate the equilibrium constant at 400 K for the process:

	N2(g)	+ 3 H2(g)		2 NH3(g)
DHºf				-46.1 (kJ/mol)
Sº298	192	131		192 (J/K.mol)

DHº = 2 (-46.1) - 1 (0) - 3 (0) = -92.2 kJ

DSº = 2 (192) - 1 (192) - 3 (131) = -201 J/K

DGº₄₀₀ = DHº - 400 DSº = -92200 - 400 (-201)

= -11800 J

(Note unfavourable entropy, favourable enthalpy)

ln K = - DGº_T / RT

ln K = - (-11800 J) / (8.314 J * 400)

ln K = 3.55

K = 34.8

K is > 1 showing that the equilibrium favours products.

K has no units: it is terms of activities:

K = a(NH₃)² /a(N₂) * a(H₂)³

From the thermodynamic tables in your text,

1. calculate the normal boiling point of bromine, Br₂

2. calculate the vapour pressure of Br₂ at 25ºC.

The process is: Br₂(l) Br₂(g)

1. At normal boiling point, the equilibrium is under standard conditions, so DGº_T = 0.

DGº_T = DHº - TDSº = 0

so

T = DHº / DSº = 30.9 kJ / 93.3 J.K-1 (watch UNITS)

T = 331 K (58ºC)

2. At equilibrium at 298 K (25ºC)

ln K = - DGº_T / RT

ln K = -DHº / RT + DSº / R

ln K = -30900 / 8.314*298 + 93.3 / 8.314

(watch the UNITS)

ln K = - 1.25

K = 0.287

K = a(Br_2(g))/a(Br_2(l)) = {P(Br₂) kPa / 100 kPa} / 1

P(Br₂) = 28.7 kPa at 25ºC

DG_T = DGº_T + RT ln Q

but, DGº_T = -RT ln K

so DG_T = -RT ln K + RT ln Q

= RT ln {Q/K}

Q is a non-equilibrium quantity. As the reaction proceeds, Q will change, and the direction of that change will be towards the value of K.

Thus:

if Q > K, then Q must decrease, reaction must go left.

if Q < K, then Q must increase, reaction must go right.

N₂(g) + 3 H₂(g) 2 NH₃(g)

In an experiment, the partial pressures were observed to be:

P(N₂) = 150 kPa, P(H₂) = 100 kPa, and P(NH₃) = 200 kPa.

Predict the way the equilibrium will move and calculate DG₄₀₀ for these conditions.

1. Obtain a value for Q which can then be compared to K:

Q = P a(i)^m(i)

= a(NH₃)² * a (N₂)^-1 * a(H₂)^-3

= {200kPa/100kPa}²*{150 kPa/100 kPa}^-1*{100 kPa/100kPa}^-3

= 2² / {1.5 * 1³} = 2.67

2. Compare Q to K: Q < K

so: Q must increase, reaction must go to right (to products).

3. DG₄₀₀ = RT ln {Q/K} = 8.314 * 400 ln {2.67/34.8}

= 8.314 J/K * 400 K * (-2.57)

= -8540 J

(negative DG means process is spontaneous towards products)

DGº_T = -RT ln K = DHº - TDSº

so:

ln K = DGº_T/RT = DSº/R - DHº/RT

Note that this is the equation which was paralleled by Arrhenius when he proposed the variation of k (rate constant with T):

ln k = ln A - E_a/RT

The equation:

ln K = DSº/R - DHº/RT

shows how K, the equilibrium constant varies with T.

A graphical plot of ln K against 1/T should give a straight line of slope -DHº/R and intercept DSº/R.

Thus DHº and DSº can be obtained graphically from K measured at different temperatures.

In the simplest case, two values of K at different temperatures can be used to determine DHº and DSº.

Example:

For the reaction NiO(s) + CO(g) Ni(s) + CO₂(g)

K₉₃₆ = 4540 and K₁₁₂₅ = 1580.

Calculate DHº and DSº for this process, assuming that both are temperature independent.

ln K₉₃₆ = DSº/R - DHº/R * 936 = ln 4540 = 8.42

ln K₁₁₂₅ = DSº/R - DHº/R * 1125 = ln 1580 = 7.37

Subtract:

8.42 - 7.37 = {DHº/R}*{1/1125 - 1/936}

1.05 = {DHº/R}*(-0.0001795)

DHº = -48 600 J (same units as R)

substitute:

8.42 = DSº/8.324 - {(-48 600)/8.314*936)}

DSº = 18.0 J/K.

{Take care to differentiate between DH (non-unit activity) and DHº (unit activity)}

DS(univ) > 0 for a spontaneous process

DS(univ) = 0 at equilibrium.

DS(univ) = DS(system) + DS(surrounds)

DE = q + w = q + PDV

DE = q_v (measurement of DE in bomb calorimeter)

DH = DE - PDV = q_p

(measurement of DH in open calorimeter)
{note for gases only: PDV = RTDn.}

DS(univ) = DS - DH/T

DS(univ) = -DG/T

DHº₂₉₈ = åm_iDHº_f(i)

DSº₂₉₈ = åm_iSº₂₉₈(i)

DG_T = DH - TDS

DGº₂₉₈ = åm_iDGº_f(i)

DG_T = DGº_T + RT ln Q (Q = Pa(i)^m(i))

At equilibrium: DG_T = 0 and Q = K

DGº_T = - RTln K

DG_T = RT ln {Q/K}

ln K_T = DSº/R - DHº/RT = - DGº_T/RT

Some ways to obtain DH:

by direct measurement: DH = q_p

from tables: DHº₂₉₈ = åm_iDHº_f(i)

from K: ln {K₁/K₂} = {DHº/R}{1/T₂ - 1/T₁}

Some ways to obtain DS:

from DH at equilibrium: DS = DH/T

from C_p on heating: DS = C_p ln {T₂/T₁}

from tables: DSº₂₉₈ = åm_iSº₂₉₈(i)

from DG and DH: DG = DH - TDS

from a graph of ln K against 1/T (intercept)

Some ways to obtain DG:

from DH and DS: : DG = DH - TDS

from tables: DGº₂₉₈ = åm_iDGº_f(i)

from Q (Q = Pa(i)^m(i)): DG_T = DGº_T + RT ln Q

from K (= Q at equilibrium): DGº_T = -RT ln K_T

from K and Q: DG_T = RT ln {Q/K}

A note on phase changes:
Phase changes are equilibria.

Normal boiling points and melting points are equilibria under unit activity. (DGº_T = 0.)

Vapour pressure is an equilibrium under non-unit activity conditions. (DG_T = 0.)

From thermodynamic tables, calculate the solubility of silver chloride in water at 75ºC.

The equilibrium involved is:

	AgCl(s)		Ag+(aq)	+ Cl-(aq)
DHºf (kJ/mol)	-127.1		105.6	-167.6
Sº298 (J/K.mol)	96.2		72.68	56.5

The solubility of a substance is the number of moles/L that will dissolve. (Note solubilty may also be reported in g/100mL of solution)

If s moles/L of AgCl dissolve, then the [Ag⁺] = [Cl^-] = s M

K = a(Ag⁺)*a(Cl^-)*a(AgCl)^-1

K = {[Ag⁺]/1 M}*{[Cl^-]/1M}*1

K = s²

Also ln K_T = DSº/R - DHº/RT

Calculate DHº₂₉₈

= åm_iDHº_f(i) = {1*105.6 + 1*(-167.6) - 1*(-127.1)} kJ

= 65.1 kJ

Calculate DSº₂₉₈

= åm_iSº₂₉₈(i) = {1*72.68 + 1*56.5 - 1*96.2}J/K

= 32.98 J/K

Calculate K₃₄₈ (watch units!!)

ln K₃₄₈ = 32.98/8.314 - 65100/(8.314*348)

= -18.53

so K = 8.93 x 10^-9

and K = s²

so s = 9.45 x 10^-5.

The solubility of AgCl in water at 75ºC is predicted to be:

9.45 x 10^-5 moles/L.

From thermodynamic tables, estimate the pH of a 1M solution of HF in water at 25ºC

The equilibrium involved is:

	HF(aq)	+ H2O		H3O+(aq)	+ F-(aq)
DGºf (kJ/mol)	-296.8	-237.1		-237.1	-278.8

Known relationships:

pH = - log₁₀ a(H₃O⁺)

K = a(H₃O⁺) * a(F^-) * a(HF)^-1

ln K = - DGº₂₉₈/R*298

Calculation:

DGº₂₉₈ = {1*(-237.1) + 1*(-278.8) - 1*(-296.8) - 1*(-237.1)}kJ

= 18.0 kJ

ln K = - 18000/(8.314*298) UNITS!!

= - 7.265

so K = 0.000700

K = a(H₃O⁺)² (since [H₃O⁺] = [F^-])

a(H₃O⁺) = 0.0265

so pH = 1.6

1. For a discussion of activity (though it is not called this) see your text, p634, a box: Entropy and Concentration.

2. K_p = P P(i)^m(i)

3. K_c = P [i]^m(i)

4. K = P a(i)^m(i)

K(for gases) = P {P(i) / Pº} ^m(i)

K = {P P(i)^m(i)}{ Pº}^-m(i)

K = K_p * {Pº}^-m(i)

If P is measured in atmospheres, Pº = 1 atm, or

if P is measured in bars, where Pº = 1 bar (= 100kPa),

K(for gases) = K_p (without the units)

K(for solutions) = P {[i] / 1 M}^m(i)

K = K_c (without the units)

• Since P(i) = {n/V}RT = [i]RT



• K_p = P P(i)^m(i)
• K_p = P {[i]RT}^m(i)
• K_p = {P [i]^m(i)}{RT}^Sm(i)
• K_p = K_c * {RT}^Dn
  • (since Sm(i) = Sn(products) - Sn(reactants) = Dn.)