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:



DGT = 0 = DGT + RT ln Q



but, at equilibrium, Q = K



so



DGT = 0 = DGT + RT ln K



or

DGT = - 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 DGT.



ln K = - DGT / 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)
DHf-46.1 (kJ/mol)
S298192131192 (J/K.mol)




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



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



DG400 = DH - 400 DS = -92200 - 400 (-201)

= -11800 J



(Note unfavourable entropy, favourable enthalpy)



ln K = - DGT / 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 25C.





The process is: Br2(l) Br2(g)





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



DGT = DH - TDS = 0

so

T = DH / DS = 30.9 kJ / 93.3 J.K-1 (watch UNITS)

T = 331 K (58C)



2. At equilibrium at 298 K (25C)



ln K = - DGT / RT

ln K = -DH / RT + DS / 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 25C












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



DGT 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 = DGT + RT ln Q



but, DGT = -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.




DGT = -RT ln K = DH - TDS



so:

ln K = DGT/RT = DS/R - DH/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 = DS/R - DH/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 -DH/R and intercept DS/R.





Thus DH and DS can be obtained graphically from K measured at different temperatures.














ln K = DS/R - DH/RT




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





Example:

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



K936 = 4540 and K1125 = 1580.



Calculate DH and DS for this process, assuming that both are temperature independent.





ln K936 = DS/R - DH/R * 936 = ln 4540 = 8.42



ln K1125 = DS/R - DH/R * 1125 = ln 1580 = 7.37



Subtract:

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

1.05 = {DH/R}*(-0.0001795)

DH = -48 600 J (same units as R)



substitute:

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



DS = 18.0 J/K.












Thermodynamics: Equation Summary.

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



DH298 = miDHf(i)

DS298 = miS298(i)



DGT = DH - TDS

DG298 = miDGf(i)

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



At equilibrium: DGT = 0 and Q = K

DGT = - RTln K

DGT = RT ln {Q/K}



ln KT = DS/R - DH/RT = - DGT/RT












Calculations in Thermodynamics - Summary.


Some ways to obtain DH:

by direct measurement: DH = qp

from tables: DH298 = miDHf(i)

from K: ln {K1/K2} = {DH/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: DS298 = miS298(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: DG298 = miDGf(i)

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

from K (= Q at equilibrium): DGT = -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. (DGT = 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 75C.



The equilibrium involved is:



AgCl(s) Ag+(aq) + Cl-(aq)
DHf (kJ/mol) -127.1 105.6 -167.6
S298 (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 = DS/R - DH/RT





Calculate DH298

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

= 65.1 kJ



Calculate DS298

= miS298(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 75C 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 25C



The equilibrium involved is:



HF(aq) + H2O H3O+(aq) + F-(aq)
DGf (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 = - DG298/R*298





Calculation:



DG298 = {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) / P} m(i)

K = {P P(i)m(i)}{ P}-m(i)

K = Kp * {P}-m(i)



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

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

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





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

K = Kc (without the units)