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!