Free Energy and Equilibrium.

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

DGT = 0 = DT + RT ln Q

but, at equilibrium, Q = K

so

DGT = 0 = DT + RT ln K

or

DT = - 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 DT.

ln K = - DT / RT

Do not forget the sign!!

Example.

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

D400 = DHº - 400 DSº = -92200 - 400 (-201)

= -11800 J

(Note unfavourable entropy, favourable enthalpy)

ln K = - DT / 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

Example:

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 DT = 0.

DT = 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 = - DT / 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(Br2(g))/a(Br2(l)) = {P(Br2) kPa / 100 kPa} / 1

P(Br2) = 28.7 kPa at 25ºC

DT is a measure of how far the reaction will go to reach equilibrium:

 DGºT K300 K500 K1000 100 3.87 x 10-18 3.57 x 10-11 0.00000598 50 1.97 x 10-9 0.00000598 0.00244 25 0.0000444 0.00244 0.0495 10 0.0181 0.0902 0.300 5 0.135 0.300 0.548 2 0.448 0.618 0.786 1 0.670 0.786 0.887 0 1 1 1 -1 1.49 1.27 1.13 -2 2.23 1.62 1.27 -5 7.41 3.33 1.82 -10 55.2 11.1 3.33 -25 22 500 410 20.2 -50 508 000 000 16 700 410

K, Q, and DGT:

DGT = DT + RT ln Q

but, DT = -RT ln K

so DGT = -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.

Example, at 400 K, K for the ammonia synthesis was 34.8:

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)

Determination of Thermodynamic Parameters from Equilibrium Measurements.

DT = -RT ln K = DHº - TD

so:

ln K = DT/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 - Ea/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.

ln K = DSº/R - DHº/RT

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.

Thermodynamics: Equation Summary.

{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 = qv (measurement of DE in bomb calorimeter)

DH = DE - PDV = qp

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

DS(univ) = DS - DH/T

DS(univ) = -DG/T

D298 = åmiDf(i)

D298 = åmi298(i)

DGT = DH - TDS

D298 = åmiDf(i)

DGT = DT + RT ln Q (Q = Pa(i)m(i))

At equilibrium: DGT = 0 and Q = K

DT = - RTln K

DGT = RT ln {Q/K}

ln KT = DSº/R - DHº/RT = - DT/RT

Calculations in Thermodynamics - Summary.

Some ways to obtain DH:

by direct measurement: DH = qp

from tables: D298 = åmiDf(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 Cp on heating: DS = Cp ln {T2/T1}

from tables: D298 = åmi298(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: D298 = åmiDf(i)

from Q (Q = Pa(i)m(i)): DGT = DT + RT ln Q

from K (= Q at equilibrium): DT = -RT ln KT

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

A note on phase changes:
Phase changes are equilibria.

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

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

Example 1.

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 KT = DSº/R - DHº/RT

Calculate D298

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

= 65.1 kJ

Calculate D298

= åmi298(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.

Example 2:

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 = - D298/R*298

Calculation:

D298 = {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

Relationship between the various Equilibrium Constants

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

2. Kp = P P(i)m(i)

3. Kc = 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 = Kp * {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) = Kp (without the units)

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

K = Kc (without the units)

• Since P(i) = {n/V}RT = [i]RT

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