Equilibrium Law
1) Equilibrium Constant, Keq:
If, aA + bB cC + dD
then beq[B] a
eq[A]
deq[D] c
eq[C] eqK unitless
temperature dependent!
eg. For A2 + B2 2AB
then]2[B ]2[A
2[AB] eqK
Note: all “eq”
2) Significance of Keq magnitude:
• If Keq is very large then ][reactants
[products]
is large and the reaction nears completion.
If Keq is very small then there is no reaction.
3) Keq and the Balanced Chemical Eq’n any action performed on the chemical rxn,
the Keq expression is raised to that action.
eg. H2 + I2 2HI ]
2[I ]
2[H
2[HI] eqK
Flip the rxn or x by -1:eg. 2HI H2 + I2
2[HI]
]2
][I2
[H '
eqK
eqK1
1- eqK '
eqK
]2
[I ]2
[H
2[HI] eqK
by 2 or x 1/2:
eg. 1/2 H2 + 1/2 I2 HI
1/2]2
[I 2/1]2
[H
[HI] 'eqK eqK
1/2eqK '
eqK
4) Keq and Reaction Kinetics:
If A + B C + D
and both the forward and reverse reactions are elementary steps, then
Ratef = kf [A]1[B]1 and Rater = kr [C]1[D]1
At equilibrium:
kf [A]1[B]1 = kr [C]1[D]1
Ratef = Rater
and
then
eqK [B] [A][D] [C]
rkf
k
5) Keq and the Effect of Temperature:
usually Eaf ≠ Ear.
an in T won’t affect the forward and reverse reactions equally, so,
Ratef ≠ Rater
and the equilibrium will change
][reactants
[products] '
eqK
If exothermic:At higher temperature, Rater creates more reactants than at lower T, and
If endothermic then the opposite occurs:
][reactants
[products] '
eqK
6) Heterogeneous Equilibria:
reactants and products in different physical states (s, l ,g, aq)
Pure liquids (not aq) and solids have constant densities
as a result, pure solids and liquids are not written in the Keq expression.
Vm D
M x Vn D
constant a is C
thenVn C and
Mn x m
M x C D MD Cor
VMn x D
eg. 2H2O(l) 2H2 (g) + O2 (g)
then ]2[O2]2[H eqK
eg. Zn(s) + Cu2+(aq) Cu (s) + Zn2+
(aq)
]2[Cu
]2[Zn eqK
Equilibrium and Spontaneity
∆Gº = -RT ln Keq R = 8.314 J/mol•K
Calculate ∆Gº from Keq
eg. PCl3 + Cl2 PCl5 ; Keq = 0.18
∆Gº = -RT ln Keq
= (-8.314 J/ mol•K x 298K) x ln (0.18)= 4.2 kJ/mol
Calculate Keq from ∆Gº
∆Gº = - RT ln Keq
ln Keq = - (∆Gº) / (RT)
Keq = e-(∆Gº)/(RT) * Note : ∆Gº in J/mol
eq. 2CO(g) + O2(g) 2CO2(g)
and ∆Gº = -514.5 kJKeq =
e-(∆Gº)/(RT)= e-(-514.5 x 103 J)/(8.314 J/mol•K x 298 K)
= 1.54 x 1090
the products are highly favoured!