Chemical Equilibrium

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Chemical EquilibriumChemical Equilibrium

Chapter 17Chapter 17

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EquilibriumEquilibriumWe’ve already used the phrase “equilibrium”

when talking about reactions.

In principle, every chemical reaction is reversible ... capable of moving in the forward or backward direction.

2 H2 + O2 ⇌ 2 H2O

Some reactions are easily reversible ...Some not so easy ...

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Equilibrium: the extent of a reactionEquilibrium: the extent of a reactionIn stoichiometry we talk about theoretical yields,

and the many reasons actual yields may be lower.

Another critical reason actual yields may be lower is the reversibility of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce.

Equilibrium looks at the extent of a chemical

reaction.

Chemical Equilibrium AnalogyChemical Equilibrium Analogy

Let us introduce the idea of chemical equilibrium by an analogy, seemingly far-fetched at first sight, but actually mathematically correct. Imagine that a crabapple tree sits on the dividing line between two homes, one inhabited by a crochety old man, and the other by a father who has told his young son to go out and rid the back yard of crabapples. The boy quickly realizes that the easiest way to dispose of the crabapples is to throw them into the neighbouring yard. He does so, arousing the ire of the old man.

The boy and the man start throwing crabapples back and forth across the fence as fast as they can. Who will win?

The battle is outlined in five phases, as shown on the following pages.


Assuming that the boy is more energetic and agile than the old man, you might think at first that the conflict would end with all of the apples on the old man's side (Phases I and II).

It is true that with equal numbers of crabapples on either side, the boy will throw apples across the fence faster than the old man can return them. But this only means that apples will become more plentiful on the old man's side, and easier to reach.


They will become scarcer on the boy's side, and require more running around to locate. Eventually a standoff, or equilibrium, will be reached, in which the number of apples crossing the fence is the same in both directions.

The old man will throw less quickly but will have less trouble finding apples (Phase III); the boy will throw more rapidly but will waste time scurrying around hunting for the relatively few crabapples on his side (Phase IV). The ratio of apples on the two sides of the fence ultimately will be determined by the relative agility of the two combatants, but all of the apples will not end up on one side (Phase V).

Chemical Equilibrium AnalogyChemical Equilibrium Analogy We can express the rate at which the old man throws apples by

RateM = kMCM

The rate is measured in apples per second across the fence, and CM is the concentration of apples on the man's side of the fence in apples per square foot of ground. The rate constant, kM , has units of square feet per second:

The value of kM expresses the agility of the old man, and his speed in covering the territory on his side of the fence.

The rate at which the boy throws apples back across the fence is given by

RateB = kBCB

in which CB is the concentration of apples in the boy's yard, and kB is the rate constant, or agility constant, which tells how fast the boy gets around on his side of the fence, in square feet per second. Since we have assumed that the boy is livelier than the man, kB is greater than kM .

Chemical Equilibrium AnalogyChemical Equilibrium Analogy If the boy had cleaned up his yard completely before the old man came

out, then as the battle began,RateM would be greater than RateB , and there would be a net flow of apples to the boy's side. His agility would do him no good if there were no apples on his side to pick up.

Conversely, if the battle had begun with equal concentrations of apples on each side, then RateB would have been greater than RateM because the agility constant kB is greater than kM .

With the same number of apples at their disposal, the boy always can do better than the old man because he gets around faster.

In either case, a neutral observer would have found to his surprise that the battle eventually settled down into a stalemate, or equilibrium in which RateM = RateB , at a point where the extra apples on the old man's side just compensated for the extra agility of the boy.

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The Concept of EquilibriumThe Concept of Equilibrium• Consider colorless frozen N2O4. At room temperature, it

decomposes to brown NO2:

N2O4(g) 2NO2(g).

• At some time, the color stops changing and we have a mixture of N2O4 and NO2.

• Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant.

• Using the collision model:

– as the amount of NO2 builds up, there is a chance that two NO2 molecules will collide to form N2O4.

– At the beginning of the reaction, there is no NO2 so the reverse reaction (2NO2(g) N2O4(g)) does not occur.

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The Concept of EquilibriumThe Concept of Equilibrium• As the substance warms it begins to decompose:

N2O4(g) 2NO2(g)

• When enough NO2 is formed, it can react to form N2O4:

2NO2(g) N2O4(g).

• At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4

• The double arrow implies the process is dynamic.N2O4(g) 2NO2(g)

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The Concept of EquilibriumThe Concept of EquilibriumAs the reaction progresses

– [A] decreases to a constant,

– [B] increases from zero to a constant.

– When [A] and [B] are constant, equilibrium is achieved.

A B

Equilibrium: Q and KEquilibrium: Q and K

• If you mix A and B, A will react to become B, and B will react to become A

– This equation also states that one molecule of A reacts to give one molecule of B. This will be an important part of the concept. A is just turning into B, and vice versa. No other substances are involved. Associated with this system are two quantities, Q, the reaction quotient, and K, the equilibrium constant.

• It is important to understand the distinction between Q and K.Q is a quantity that changes as a reaction system approaches equilibrium. K is the numerical value of Q at the "end" of the reaction, when equilibrium is reached.

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If Q < Keq, shift to right (toward product)

If Q > Keq, shift to left (toward reactant)

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The Equilibrium ConstantThe Equilibrium Constant• No matter the starting composition of reactants and

products, the same ratio of concentrations is achieved at equilibrium.

• For a general reaction

the equilibrium constant expression is

where Kc is the equilibrium constant.

aA + bB(g) pP + qQ

ba

qp

cKBA

QP

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The Equilibrium ConstantThe Equilibrium Constant• Kc is based on the molarities of reactants and

products at equilibrium.

• We generally omit the units of the equilibrium constant.

• Note that the equilibrium constant expression has products over reactants.

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The Equilibrium ExpressionThe Equilibrium Expression• Write the equilibrium expression for the

following reaction:

N2(g) + 3H2(g) 2NH3(g)

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The Equilibrium ConstantThe Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• The equilibrium constant, K, is the ratio of products

to reactants.• Therefore, the larger K the more products are present

at equilibrium.• Conversely, the smaller K the more reactants are

present at equilibrium.• If K >> 1, then products dominate at equilibrium and

equilibrium lies to the right.• If K << 1, then reactants dominate at equilibrium and

the equilibrium lies to the left.

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The Equilibrium ConstantThe Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• An equilibrium can be approached from any

direction.

Example:N2O4(g) 2NO2(g)

212.0

ONNO

42

22 cK

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The Equilibrium ConstantThe Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• However,

• The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

2NO2(g) N2O4(g)

72.4212.01

NO

ON2

2

42 cK

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The Equilibrium ConstantThe Equilibrium ConstantHeterogeneous EquilibriaHeterogeneous Equilibria• When all reactants and products are in one phase, the

equilibrium is homogeneous.• If one or more reactants or products are in a different

phase, the equilibrium is heterogeneous.• Consider:

– experimentally, the amount of CO2 does not seem to depend on the amounts of CaO and CaCO3. Why?

CaCO3(s) CaO(s) + CO2(g)

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The Equilibrium ConstantThe Equilibrium ConstantHeterogeneous EquilibriaHeterogeneous Equilibria

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The Equilibrium ConstantThe Equilibrium ConstantHeterogeneous EquilibriaHeterogeneous Equilibria• Neither density nor molar mass is a variable, the

concentrations of pure solids and pure liquids are constant. (You can’t find the concentration of something that isn’t a solution!)

• We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions.

• The amount of CO2 formed will not depend greatly on the amounts of CaO and CaCO3 present.

Kc = [CO2]

CaCO3(s) CaO(s) + CO2(g)

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Calculating Equilibrium ConstantsCalculating Equilibrium Constants• Steps to Solving Problems:

1. Write an equilibrium expression for the balanced reaction.

2. Write an ICE table. Fill in the given amounts.

3. Use stoichiometry (mole ratios) on the change in concentration line.

4. Deduce the equilibrium concentrations of all species.

• Usually, the initial concentration of products is zero. (This is not always the case.)

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Applications of Equilibrium ConstantsApplications of Equilibrium ConstantsPredicting the Direction of ReactionPredicting the Direction of Reaction• We define Q, the reaction quotient, for a reaction at

conditions NOT at equilibrium

as

where [A], [B], [P], and [Q] are molarities at any time.• Q = K only at equilibrium.

aA + bB(g) pP + qQ

ba

qpQ

BA

QP

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Applications of Equilibrium ConstantsApplications of Equilibrium ConstantsPredicting the Direction of ReactionPredicting the Direction of Reaction

• If Q > K then the reverse reaction must occur to reach equilibrium (go left)

• If Q < K then the forward reaction must occur to reach equilibrium (go right)

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Example Problem: Calculate ConcentrationExample Problem: Calculate Concentration

Note the moles into a 10.32 L vessel stuff ... calculate molarity.Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M

2 HI H2 + I2

222

][

]][[

HI

IHKeq

Initial:Change:Equil:

0.242 M 0 0 -2x +x +x0.242-2x x x

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2

21026.1

]2242.0[]2242.0[

]][[

xx

x

x

xxKeq

What we are asked for here is the equilibrium concentration of H2 ... ... otherwise known as x. So, we need to solve this beast for x.

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Example Problem: Calculate ConcentrationExample Problem: Calculate Concentration

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2

1026.1]2242.0[

xx

x

232 ]2242.0[1026.1 xxx

]4968.00586.0[1026.1 23 xxx

2335 1004.51022.11038.7 xxxxx

01038.71022.1995.0 532 xxxx

And yes, it’s a quadratic equation. Doing a bit of rearranging:

a

acbbx

2

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x = 0.00802 or –0.00925Since we are using this to model a real, physical system,we reject the negative root.The [H2] at equil. is 0.00802 M.

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Example Problem: Calculate KeqExample Problem: Calculate Keq

This type of problem is typically tackled using the “three line” approach:2 NO + O2 2 NO2

Initial:

Change:

Equilibrium:

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ApproximatingApproximating

If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants.

0.20 – x is just about 0.20 if x is really dinky.

If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic formula.

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ExampleExample

Initial Concentration of I2: 0.50 mol/2.5L = 0.20 MI2 2 I

Initialchangeequil:

0.20 0-x +2x0.20-x 2x 10

2

10

2

2

1094.2]20.0[

]2[

1094.2][

][

xx

x

xI

IKeq

With an equilibrium constant that small, whatever x is, it’s nearzero, and 0.20 minus almost zero is 0.20 (like a million dollars minus a nickel is still a million dollars).

0.20 – x is the same as 0.20

102

1094.220.0

]2[ xx

x = 3.83 x 10-6 M

More than 3orders of mag.between thesenumbers. The simplification willwork here.

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ExampleExample

Initial Concentration of I2: 0.50 mol/2.5L = 0.20 MI2 2 I

Initialchangeequil:

0.20 0-x +2x0.20-x 2x 209.0

]20.0[

]2[

209.0][

][

2

2

2

x

x

I

IKeq These are too close to

each other ... 0.20-x will not betrivially close to 0.20here.

Looks like this one has to proceed through the quadratic ...

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Le Chatelier’s Principle:Le Chatelier’s Principle: if you disturb an if you disturb an equilibrium, it will shift to undo the equilibrium, it will shift to undo the disturbance.disturbance.

Remember, in a system at equilibrium, come what Remember, in a system at equilibrium, come what may, the concentrations will always arrange may, the concentrations will always arrange themselves to multiply and divide in the Keq themselves to multiply and divide in the Keq equation to give the same number (at constant equation to give the same number (at constant temperature).temperature).

Le Châtelier’s PrincipleLe Châtelier’s Principle

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Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• Adding a reactant or product shifts the equilibrium

away from the increase.• Removing a reactant or product shifts the equilibrium

towards the decrease.• To optimize the amount of product at equilibrium, we

need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier).

• We illustrate the concept with the industrial preparation of ammonia

N2(g) + 3H2(g) 2NH3(g)

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Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• Consider the Haber process

• If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2 (by Le Châtelier).

• That is, the system must consume the H2 and produce products until a new equilibrium is established.

• Therefore, [H2] and [N2] will decrease and [NH3] increases.

N2(g) + 3H2(g) 2NH3(g)

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Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations• The unreacted nitrogen and hydrogen are recycled

with the new N2 and H2 feed gas.

• The equilibrium amount of ammonia is optimized because the product (NH3) is continually removed and the reactants (N2 and H2) are continually being added.

Effects of Volume and PressureEffects of Volume and Pressure• As volume is decreased pressure increases.• Le Châtelier’s Principle: if pressure is increased the

system will shift to counteract the increase.

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Le Châtelier’s PrincipleLe Châtelier’s Principle• Consider the production of ammonia

• As the pressure increases, the amount of ammonia present at equilibrium increases.

• As the temperature decreases, the amount of ammonia at equilibrium increases.

• Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

N2(g) + 3H2(g) 2NH3(g)

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Le Châtelier’s PrincipleLe Châtelier’s PrincipleChange in Reactant or Product ConcentrationsChange in Reactant or Product Concentrations

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ExampleExample

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Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffects of Volume and PressureEffects of Volume and Pressure• The system shifts to remove gases and decrease

pressure.• An increase in pressure favors the direction that has

fewer moles of gas.• In a reaction with the same number of product and

reactant moles of gas, pressure has no effect.• Consider

N2O4(g) 2NO2(g)

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Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffects of Volume and PressureEffects of Volume and Pressure• An increase in pressure (by decreasing the volume)

favors the formation of colorless N2O4.

• The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased.

• The system moves to reduce the number moles of gas (i.e. the forward reaction is favored).

• A new equilibrium is established in which the mixture is lighter because colorless N2O4 is favored.

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The Equilibrium ConstantThe Equilibrium ConstantThe Equilibrium Constant in Terms of PressureThe Equilibrium Constant in Terms of Pressure• If KP is the equilibrium constant for reactions

involving gases, we can write:

• KP is based on partial pressures measured in atmospheres.

ba

qp

PPP

PPK

BA

QP

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Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffect of Temperature ChangesEffect of Temperature Changes• The equilibrium constant is temperature dependent.• For an endothermic reaction, H > 0 and heat can be

considered as a reactant.• For an exothermic reaction, H < 0 and heat can be

considered as a product.• Adding heat (i.e. heating the vessel) favors away from

the increase:– if H > 0, adding heat favors the forward reaction,

– if H < 0, adding heat favors the reverse reaction.

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Le Châtelier’s PrincipleLe Châtelier’s PrincipleEffect of Temperature ChangesEffect of Temperature Changes• Removing heat (i.e. cooling the vessel), favors towards

the decrease:– if H > 0, cooling favors the reverse reaction,

– if H < 0, cooling favors the forward reaction.

• Consider

for which H > 0.– Co(H2O)6

2+ is pale pink and CoCl42- is blue.

Cr(H2O)6(aq) + 4Cl-(aq) CoCl42-(aq) + 6H2O(l)

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Solubility Product PrincipleSolubility Product Principle

• Another equilibrium situation is slightly soluble products

• Ksp is the solubility product constant

• Ksp can be found on a chart at a specific temperature

• Since the product is solid on the left side, only the products (ions) are involved in the Ksp expression

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• Example: Find the concentration of ions present in calcium fluoride (in water) and the molar solubility.

CaF2(s) --> Ca+2 + 2 F-

Ksp = [Ca+2] [F-]2 = 2 X 10 -10

If x = [Ca+2 ], then [F-] = 2x (the x and 2x come from coefficients)

[x] [2x]2 = 2 X 10 -10

4x3 = 2 X 10 -10

x3 = 5 X 10 -11

x = 3.68 X 10 -4

[Ca+2 ] = x = 3.68 X 10 -4 [F-] = 2x = 7.37 X 10 -4

Molar solubility = the lowest of the ion concentrationsSolubility of CaF2 = 3.68 X 10 -4 (same as Ca+)

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• Example: Find the molar solubility of silver chloride (in water).

AgCl (s) --> Ag+ + Cl -

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• Example: The Ksp for PbS(s) is 1x10-27 at 25°C. Find the solubility of PbS(s) in water at 25°C.

Chemical Equilibrium

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