Chemical Equilibria:
More on Connection to Thermodynamics
and other matters
CHEM 102!
T. Hughbanks!
Reconsider ∆G
  In
thermodynamics, when we consider a
reaction αA + βB ⇌ γC + δD, we calculate
∆G° for the forward reaction as follows:!
∆G°rxn = ∑ni∆Gf°products – ∑nj∆Gf°reactants!
= γ∆Gf°(C) + δ∆Gf°(D) – α∆Gf°(A) – β∆Gf°(B)!
  Each
of the reactants and products are
defined to be in standard states in this eqn.!
  Look more carefully at standard states &
consider ∆G when conditions aren’t standard.!
Standard States α A + β B  γ C + δ D
  298
K, 1 atm. Each gas phase reactant and
product considered to be at 1 atm. pressure
under standard conditions. (eg., if A & C are
gases, PA = PC = 1atm)!
  Example:
!NH3(g) + HCl(g) ⇌ NH4Cl(s)!
!from data in Appendix E (all in kJ):!
∆G°rxn= -202.87 - (-16.45) - (-95.30) = -91.12 kJ!
!This is ∆G for converting 1 mol NH3 and 1 mol
HCl at 1.0 atm. pressure into solid NH4Cl at
298 K.!
∆G = ∆G° + RT lnQ
Q =
PCγ PDδ
PAα
PBβ
Q =
γ
[C] [D]δ
[A]α [B]β
  Q
is variable (determined by the amount of
reactants and products at any time.)!
  Check: standard conditions, Q = 1, ∆G = ∆G˚!
  At equilibrium, Q = K. But we know that
when we are at equilibrium, ∆G = 0.!
!0 = ∆G˚ + RTlnK !⇒ ! ∆G˚ = -RTlnK!
∆G° = - RT lnK , Example
  Given
the following data, calculate K for the
reaction:
!I2(g) + H2(g) ⇌ 2 HI(g)!
From Appendix:!
∆Hf°(I2(g)) = 62.44;
∆Hf°(HI(g)) = 26.48
both in kJ/mol!
S°(I2(g)) = 260.69;
S°(H2(g)) = 130.68!
! !
!S°(HI(g)) = 206.59 ! all in J/mol K!
I 2 (g) + H 2 (g)  2HI(g)
= – RT lnK !⇒ K = exp{-∆G°/RT}!
  We need ∆G° in order to get K, so....!
∆Hr° = 2[∆Hf°(HI(g))] – ∆Hf°(I2(g)) – ∆Hf°(H2(g))!
! != 2[26.48] – 62.44 – 0 = – 9.48 kJ!
∆Sr° = 2[S°(HI(g))] – S°(I2(g)) – S°(H2(g))!
!= 2[206.59] – 260.69 – 130.68 = + 21.81 J/K!
∆G° = ∆H° – T∆S° = – 9.48 – (298)(.0218) kJ!
∆G° = –15.98 kJ!
  ∆G°
I 2 (g) + H 2 (g)  2HI(g)
∆G° = – RT lnK K = exp{-∆G°/RT}!
We found: ! ∆G° = –15.98 kJ!
⇒ K = exp{-15980 J/(8.314 J/mol•K)(298 K)}!
K = 1.58 × 10-3 (no units)!
I 2 (g) + H 2 (g)  2HI(g)
= 1.58 × 10-3!
  But, what is this equal to?!
The equilibrium constant must be
expressed using pressures for gases when
obtained directly from thermodynamic data.!
  K
KP = 1.58 × 10-3 = (PHI)2/(PI2)(PH2)!
I 2 (g) + H 2 (g)  2HI(g)
KP = 1.58 × 10-3 = (PHI)2/(PI2)(PH2)!
convert to concs. (KC)? !
use P = (n/V)RT = [concentration]RT!
(PHI)2/(PI2)(PH2) = [HI]2/[I2][H2] × (RT)2/(RT)(RT)!
!in this case, KP and KC are numerically equal!
This is not true generally (when ∆(nRT) ≠ 0)!
Equilibria Involving Solutions
(more detail coming soon)
  Many
equilibrium reactions occur in
solutions. Weak acids & bases, etc.!
  Example:
solution of ammonia
NH3(aq) + H2O ⇌ NH4+(aq) + OH–(aq)!
  Write
an equilibrium constant for this.!
Equilibria Involving Solutions
NH3(aq) + H2O ⇌ NH4+(aq) + OH–(aq)!
K eq =
[NH +4 ][OH – ]
[NH 3 ][H 2 O]
  But
“[H2O]” is constant, so we can
incorporate it into the constant itself. !
Equilibria Involving Solutions
NH3(aq) + H2O ⇌ NH4+(aq) + OH–(aq)!
[NH +4 ][OH – ]
K eq =
= Kb
[NH 3 ]
  By
convention, tabulated K’s are written like
this, without including water term.!
  NH3 produces OH– in water and is therefore
a base. Keq for a base is called Kb. !
  For NH3, Kb = 1.8 × 10-5 . What % of NH3 is
converted to NH4+ in a 1.0 M solution? !
Acids - Definition of Ka
HA(aq) + H2O ⇌ H3O+(aq) + A–(aq)!
K eq =
[H 3 O + ][A – ]
[HA]
= Ka
  Once
again, the water term is omitted in
tabulating the equilibrium constants of acids.!
  An acid, say HA, produces H3O+ in water.
Keq for an acid is called Ka. !
  For CH3COOH (acetic acid), Ka = 1.8 × 10-5 .
What is [H3O+] in a 0.1 M solution? !
Equilibria Involving Solutions
Water self-dissociates, even in the absence!
of added acids or bases:!
!H2O + H2O ⇌ H3O+(aq) + OH–(aq)!
K eq = [H 3O + ][OH – ] = 1.0 × 10 −14 = K w
  Again,
“[H2O]” is constant, so Kw does
not include it. !
  In pure water, what are [H3O+] & [OH–]?!
Equilibria Involving Solutions
In all aqueous solutions, equilibrium between!
H3O+ & OH– is continually re-established:!
!H2O + H2O ⇌ H3O+(aq) + OH–(aq)!
K eq = [H 3O + ][OH – ] = 1.0 × 10 −14 = K w
  This
is true even with added acids or bases.!
  In a 1.0 M NH3 solution, what is [H3O+]?!
Solubility Equilibrium
The process is in
equilibrium
when rate of ions
(or molecules)
leaving the solid
= rate returning
to the solid.
That is, when
dissolution rate =
precipitation rate
Solubility Products
MpXq(s) ⇌ p Mq+(aq) + q Xp-(aq)!
Ksp = [M q + ]p [X p – ]q
  Note
that no “concentration” of the MpXq
solid appears in the equilibrium constant.!
  This when the solid salt and dissolved salt
are in equilibrium, the solution is “saturated”.
When the solution is saturated, the amount
of “extra” salt does not affect the amount of
dissolved MpXq.!
Solubility Products - Example
  In
a saturated aqueous solution of MgF2,
the concentration of Mg2+ ions is 1.14 ×
10-3 M. Write the equilibrium constant
expression, and compute Ksp for MgF2.!