Chemical Equilibrium

Chemistry 100

The concept

A condition of balance between opposing physical forces

A state in which the influences or processes to which a thing is subject cancel one another and produce no overall change

Oxford English Dictionary

Static and Dynamic

A book sitting on a desk is in static equilibrium; The book remains at rest; its position is

constant. The moon circles the earth.

There is movement but the (average) distance between the two is unaltered. This is dynamic equilibrium.

Equilibrium

The molecules of A are able to turn into molecules of B

The rate at which this happens is proportional to [A].

Ratefor = kfor[A]

Likewise, if B can turn into A, then

Raterev = krev[B]

The Equilibrium Condition

Start with pure A. [A] decreases and [B] increases as A

turns into B What happens to the rate at which A

turns into B, and the rate at which B turns into A? The rate of A B decreases, while BA

increases

What eventually happens? Rate of A B = Rate B A

Ratefor = Raterev kfor[A] = krev[B]

Time

Concentration A

B

Time

Rate

A

B

The Equilibrium Condition #2

And then what?

eq

rev

for Kkk

AB

We have as an equilibrium condition

kfor[A] = krev[B]

Keq the thermodynamic equilibrium constant

The Meaning of PB/PA = K

K is a constant number such as 2.3, 0.65, etc

What the equilibrium expression means is:

No matter how much A or B we start with, when the system reaches equilibrium

A

Beq K

P

P

Reversible reactions

If these two reactions are possible A B and B A,

we have a reversible reaction A ⇌ B Here is a real reversible reaction

N2(g) +3H2(g) ⇌ 2NH3(g)

Equilibrium can be reached from either side

At start PH2

= 3; PN2 = 1; PNH3

=

0

At start PH2

= 0; PN2 = 0; PNH3

=2

Law of Mass ActionExpression for K

riumat equilib pressures partial the are

.. ,P,P where QP

For the reactionaA (g) + bB (g) ⇌ pP (g) + qQ

(g)

bB

aA

qQ

pP

eqPP

PPK

Examples of Keq

22 ClrB

2BrCl

eq PPP

K

3HN

2NH

eq

22

3

PP

PK

N2(g) +3H2(g) ⇌ 2NH3(g)

Br2(g) +Cl2(g) ⇌ 2BrCl(g)

21

OSO

SOeq

22

3

PP

PK SO2(g) +½O2(g) ⇌ SO3(g)

Magnitude of Keq

2 HI(g) ⇌ H2(g) + I2(g) Keq = 0.016

The magnitude (size) of Keq provides

information K >> 1 the products are favoured K << 1 the reactants are favoured

CO(g) + Cl2(g) ⇌ COCl2(g) Keq= 4.57109 Equilibrium lies far to the right - there is

very little CO and Cl2 in the equilibrium mixture.

Heterogeneous Equilibrium

When the substances in the reaction are in the same phase (e.g., all gases), reactions are termed homogeneous equilibria.

When different phases are present, we speak of heterogeneous equilibrium.

We will look at reactions involving gases and solids, and gases and liquids

Solids do not appear in Keq

Examine the reaction CaCO3(s) ⇌ CaO(s) + CO2(g)

][CaCO

[CaO]PK

3

COeq

2

For a pure solid X (or liquid X) [X] = density/molar mass.

Note molar mass and density are intensive properties!! [X] = constant

2CO'eqeq

2eq PK

1constant 2constant

Kor 2][constant

]1][CO[constant K

Heterogeneous Equilibrium

At a given temperature, the equilibrium between CaCO3(s), CaO(s), and CO2(g) yields the same concentration (same partial pressure) of CO2(g).

True as long as all three components are present.

2COeq P K Note that it does not matter how much of the

two solids are present; we just need some.

More heterogeneous equilibria

CO2(g) + H2(g) ⇄CO(g) + H2O(l)

SnO2(s) + 2CO(g) ⇄ Sn(s) + 2CO2(g)

22 HCO

COeq PP

P K

2CO

2CO

eq P

P K 2

Keq values for forward and reverse reactions

For the reaction

2 HI(g) ⇌ H2(g) + I2(g), Keq = 0.016 What is Keq for H2(g) + I2(g) ⇌ 2HI(g)

? Call the first reaction (F) and the

second (R)

22

22

IH

2HIR

eq2HI

IHFeq PP

PK and

P

PPK

Forward and Reverse (II)

62.5 0.016

1K so

K1

Kor 1K K

Req

Feq

Req

Feq

Req

An aside

For the equilibrium H2O(l) ⇌ H2O(g), write down the expression for Keq.

OHeq 2PK

When liquid water and water vapour are in equilibrium the vapour has a fixed pressure

at a given temperature!

Applications

Obtaining the equilibrium constant from the measured equilibrium concentrations

Calculating the composition of the equilibrium system have the concentration of all but one

component at equilibrium and the value of Keq

given initial amounts of reactants and the equilibrium constant

Applications (II)

For a given reaction, Keq has a set value for a given temperature

eq

bB

aA

qQ

pP

eqPP

PPK

PP

PPQ

bB

aA

qQ

pP

Q depends on the experimental conditions

Q = Keq at equilibrium

Summary of the Q and Keq story

When Q > Keq

reaction shifts left When Q = Keq

equilibrium When Q < Keq

reaction shifts right

Applications (III)

Equilibrium is approachable from either side of the reaction.

Le Châtelier’s Principle

Perturb a system at equilibrium Change in temperature, pressure, or the

concentration of a component The system will shift its equilibrium

position so as to counteract the disturbance.

The effect of the last two disturbances can also be be predicted by the Law of Mass Action

Changing concentration (I)

Examine the system2 NO2Cl(g) ⇌ 2 NO2(g) + Cl2(g)

Introduce a small amount of substance X that reacts with Cl2 to make XCl. The value of PCl2

has been decreased

Le Châtelier’s Principle predicts that more NO2Cl will react to increase PCl2

Changing concentration (II)

At equilibrium

2.11(0.0212)

)108.0((0.216)

P

PP K

2

2

2ClNO

Cl2NO

eq

2

22

Remove Cl2 - the new value of PCl2 is 0.05

atm2.5

(0.0212))05.0((0.216)

P

PP Q 2

2

2ClNO

Cl2NO

2

22

Q is now smaller than Keq. The reaction moves to the right to increase Qeq. Same

prediction!

Changing concentration (III)

CaCO3(s) ⇌ CaO(s) + CO2(g)If we have this system in equilibrium and add either CaCO3(s) or CaO(s), there will be no effect on the equilibrium.

Changing the concentration

Changing pressure (I)

2 NO2Cl(g) ⇌ 2 NO2(g) + Cl2(g) Increase the pressure in the system

by making the vessel smaller. Note that there are a total of 3

moles on the right of the reaction and 2 on the left. The left “takes up less space”

Le Châtelier’s Principle predicts that the species on the left will react to form more NO2Cl.

Changing pressure (II)

2 NO2Cl (g) 2 NO2 (g) + Cl2 (g) System is initially at equilibrium. Increase the pressure by making the

vessel smaller. We could use the Keq expression to

predict what happens but Le Châtelier’s Principle is much easier to use!

Cautionary Note

Le Châtelier’s Principle predicts what occurs when we change the partial pressure of one or more of the species in the reaction

Change the total pressure by adding/removing an inert gas (not involved in the reaction)

NO EFFECT ON THE EQUILIBRIUM

More pressure changes

Predict what happens N2(g) + 3H2(g) ⇌ 2NH3(g); total

pressure is decreasedReaction shifts to left; more moles

on left H2(g) + I2(g) ⇌ 2HI(g); total

pressure is increasedNo effect; 2 moles on each side

What is Heat - not a substance!

Some textbooks Heat is treated as a chemical reagent

when applying Le Châtelier’s Principle to change of temperature problems.

Treating heat as a substance can lead to confusion. There is a better way!

Changing the temperature

An exothermic reaction causes an increase in temperature. The reverse causes cooling.

Warm up an exothermic reaction Le Châtelier’s Principle predicts the

system will move in the direction that will bring the temperature back down The direction that cools. So the reverse

reaction () occurs

Changing Temperature

N2O4(g) 2 NO2(g) H = 58.0 kJ The forward reaction is endothermic

Absorbs heat. Decrease the temperature - reaction

shifts to the left Brings T back up.

If we increase the temperature, opposite effect Reaction takes in heat and lower the

temperature

Temperature and Keq

Endothermic reactions – increasing temperature increases the value of the equilibrium constant!

Exothermic reactions – increasing temperature decreases the value of the equilibrium constant!

Temperature changes are the only stresses on the systems that change the numerical

values of Keq

Temperature and Keq (II)

Co(H2O)62+(aq)+ 4 Cl- (aq) ⇌ CoCl42- (aq) + 6 H2O (l)

∆H > 0

Catalyst does NOT change K

A catalyst speeds up a reaction by providing and alternate reaction pathway with a lower Ea.

Reversible reaction the forward and backward reactions

have their Ea’s changed by the same amount.

Keq is not altered.

A catalyst cannot alter K!! Otherwise we would be able to build a perpetual

motion machine!!