Chem 1310: Introduction to physical chemistry

Part 3: Equilibria

Peter H.M. Budzelaar

Kinetics and Equilibria

Kinetics:how fast does a reaction go (initially)

Equilibrium:what will be the final compositionof the reaction mixture?

What is "equilibrium" ?

A system is at (dynamic) equilibrium when its composition doesn't change any more over time.

There are equal numbers of molecules going to the left as to the right (and this number is not zero!).

A mixture with a composition that does not appear to change over time is not necessarily at equilibrium. The reaction could just be very slow!

How do you know that equilibrium has been established?

• The reaction started, but then did not proceed any further.This is not an infallible criterion!

• The equilibrium was approached from both the product and the reactant side (in two separate experiments), and both resulted in the same composition.Molecules have no memory, so the equilibrium

composition does not depend on how you got there.

Approaching equilibrium from different sides

pure H2 + pure I2(1:1)

pure HI

pure H2 + pure I2 + pure HI(1:1:1)

equilibrium mixture ofH2, I2, HI

How do you know that equilibrium has been established?

Heating accelerates approach to equilibrium, but also changes the equilibrium composition.

On cooling, the reaction might slow down so much that the new equilibrium will not be established.

To make sure you are in a dynamic equilibrium, you could also add one pure component and see how the system responds. If nothing happens, you are probably not at equilibrium.

For a reaction A B ⇋ at equilibrium:

For a reaction A + B C + ⇋ 2 D at equilibrium:

KC is constant for a given temperature, independent of pressure, volume, etc.

]B][A[

]D][C[ 2

CK

The equilibrium constant

]A[

]B[CK

Units and equilibrium constantsMSJ says (p678) that we should simply omit units from the equilibrium constant, because they are always mol/L.

That is not the real reason. We do it because the equilibrium constant is actually something like

where [A]° etc are the concentrations at an agreed reference state (which is 1 mol/L for solutes). Because of this division, units cancel and KC is truly dimensionless. Because the reference state is 1 mol/L, we can ignore it when we use mol/L as units.

)]B/[]B)([]A/[]A([

)]D/[]D)([]C/[]C([ 2

CK

Kinetics vs equilibria

Rate laws can have a complicated dependence on concentrations, unrelated to the reaction stoichiometry. They might depend e.g. on concentrations of catalysts and poisons.

Equilibrium constant expressions can be deduced directly from the reaction equation and do not depend on the reaction mechanism, nor on catalysts etc.

Writing an equilibrium constant

KC: equilibrium constant expressed in concentrations. Preferably in a single phase, usually gas or solution.

NH4+ + CH3COO- NH⇋ 3 + CH3COOH (in water)

Adding NaOH or HCl may change the values of some of the concentrations involved, but KC will indeed remain constant.

]COOCH][NH[

]COOHCH][NH[

34

33CK

Writing an equilibrium constant

NH3 + ¾ O2 ½ N⇋ 2 + 1½ H2O

4 NH3 + 3 O2 2 N⇋ 2 + 6 H2O

KC(b) = (KC(a))4

An equilibrium constant belongs toa chemical equation as written.

75.023

5.12

5.02

]O][NH[

]OH[]N[)( aKC

32

43

62

22

]O[]NH[

]OH[]N[)( bKC

Reactions involvingsolids and liquids

Pure solids and liquids are not included in KC.

Nearly-pure solvents are also left out.

If you have a vapour or solute in equilibrium with the pure solid or liquid, you also leave out that vapour or solute.

So if you have ⅛ S8(s) + O2 SO⇋ 2, it doesn't matter whether you also have some S8(g) present, or whether that is involved in the reaction. You don't need to include S8(g) as long as S8(s) remains present.

Why drop pure solids and liquids?

This is not simply because their concentrations are constant (MSJ 677). The correct KC would contain not [S8(s)] but [S8(s)]/[S8(s)]°, where [S8(s)]° is the concentration in the reference state, which is the pure solid/liquid by convention.

Since solids and liquids are not compressible, [S8(s)]/[S8(s)]° will always be very close to 1, and there is no reason to include it. We are not just dropping a constant, we are dropping a factor 1.

Reactions involvingsolids and liquids

CH3COOH + H2O CH⇋ 3COO- + H3O+

Hg(l) + Cl2 HgCl⇋ 2(s)

[Cl2](eq) = 1/KC

Hg(l) + ⅛ S8(s) HgS⇋ (s)

No equilibrium constant! If product-favoured, reaction will proceed until one or both reactants consumed.

]OH][COOHCH[

]OH][COOCH[

23

33

CK

]Cl][Hg[

]HgCl[

2(l)

(s)2CK

Using equilibrium constants

• Calculate KC, given equilibrium concentrations.

Just plug the concentrations in the formula.

• Calculate whether a reaction will go forward or backward, given KC and initial concentrations.

– Calculate QC (same formula as KC, but now for non-equilibrium concentrations).

– If QC < KC : reaction will go forward

– If QC > KC : reaction will go backward

– If QC = KC : reaction is at equilibrium

Using equilibrium constants

• Calculate concentrations at equilibrium, given KC and initial conditions.This may involve serious calculations; use the "x method"

of the book (MSJ p682-684), e.g.

A B C

initial 1 1 0

change -x -x +x

equilibrium 1-x 1-x x

Dissociation of HIH2 + I2 2 HI⇋

We begin with 1 mol/L of HI, and let this reach equilibrium with H2 and I2 at 25°C (KC = 25, see MSJ p685). What will be the final concentrations?

H2 I2 HI

initial 0 0 1

change + ½ x + ½ x -x

equilibrium ½ x ½ x 1-x

Dissociation of HI (2)Plug into KC equation:

Final concentrations: [HI] = 0.71, [H2] = [I2] = 0.14.

Put into KC expression for final check.

29.015.35.215½

1

25)(½

)1(

))(½½(

)1(

]I][H[

]HI[2

22

22

2

xxxxx

x

x

x

xx

xKC

Dissociation of N2O4

N2O4 2 NO⇋ 2

We begin with 1 mol/L of N2O4, and let this reach equilibrium with NO2 at 500K (KC = 46, see MSJ p702). What will be the final concentrations?

N2O4 NO2

initial 1 0

change -x + 2 x

equilibrium 1-x 2 x

Dissociation of N2O4 (2)

Plug into KC equation:

Final concentrations: [N2O4] = 0.07, [NO2] = 1.85.

Put into KC expression for final check.

93.08

4.5346

8

46*4*44646

04646446464

461

)2(

]ON[

]NO[

2

22

2

42

22

x

xxxx

x

xKC

Quadratic equations

Concentration or pressure dependence

Some equilibria are concentration- or pressure-dependent: if all concentrations/pressures change by the same factor (by compressing a gas mixture, or diluting a solution), they are no longer at equilibrium.

This happens if the sums of exponents in numerator and in denominator of KC differ. For the case of N2O4 dissociation: numerator 2, denominator: 1. The HI dissociation equilibrium is not pressure dependent (both exponents 2).

Dissociation of N2O4 (3)

What will happen if we compress the previous equilibrium mixture ([N2O4] = 0.07, [NO2] = 1.85) to 1/10th of its original volume?

Immediately after compression: [N2O4] = 0.7,[NO2] = 18.5, new QC = 489 (should have been 460, but there are rounding errors). So QC > KC, the equilibrium will shift towards the reactant N2O4.

Dissociation of N2O4 (4)

Change table:

N2O4 NO2

initial 0.7 18.5

change +x - 2 x

equilibrium 0.7+x 18.5-2 x

Dissociation of N2O4 (5)

Plug into KC equation:

(the other solution, x = 27.1, would make [NO2] negative and is unphysical)

Final concentrations: [N2O4] = 3.56, [NO2] = 12.8.

Put into KC expression for final check.

86.28

2.97120

8

1.310*4*4120120

01.31012042.32463.342744

467.0

)25.18(

]ON[

]NO[

2

22

2

42

22

x

xxxxx

x

xKC

Le Chatelier's principle

The system reacts to counteract an imposed change:

A + B X + Y⇋Add A (reactant): drive reaction towards X, Y.

Add X (product): drive reaction towards A, B.

Add something non-reacting (within same volume):no change in concentrations, QC stays equal to KC, no change.

Compress: all concentrations change, but QC does not (no change in #particles): no change.

Le Chatelier's principle (2)

A X + Y⇋Add A (reactant): drive reaction towards X, Y.

Add X (product): drive reaction towards A.

Add something non-reacting (within same volume):no change in concentrations, QC stays equal to KC, no change.

Compress: QC increases, so reaction is driven towards A (fewer particles), reducing pressure.

Le Chatelier's principle (3)

Effect of temperature:• heating drives an exothermic reaction (H < 0)

towards reactants,an endothermic reaction (H > 0) towards products.

(but not always to completion; the figure on p687 is incorrect!)

• cooling has the opposite effect.

KP and KC

For gases, instead of KC one usually uses KP:

where the pressures p are in bar (atm).

For any component X, with the ideal-gas lawpV = nRT, one has

For the above example:

]B][A[

]D][C[ 2

CKB

DCP pp

ppK

A

2

RTRTV

np X

X ]X[

CB

DCP KRT

RTRT

RTRT

pp

ppK

]B[]A[

)(]D[]C[ 22

A

2

KP and KC

Whenever the system is pressure-dependent(sums of exponents in numerator and denominator differ) you will also find RT terms in the conversion between KP and KC.

If gases are involved, always make clear whether you are using KC/QC or KP/QP !!!!!

Ammonia synthesis