Chemical EquilibriumChapter 14

Equilibrium: 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.

Equilibrium is a state in which there are no observable changes as time goes by.

Chemical equilibrium is achieved when:

• the rates of the forward and reverse reactions are equal and

• the concentrations of the reactants and products remain constant

Physical equilibrium

H2O (l)

Chemical equilibrium

N2O4 (g)

14.1

H2O (g)

2NO2 (g)

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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.

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)

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

N2O4 (g) 2NO2 (g)

Start with NO2 Start with N2O4 Start with NO2 & N2O4

equilibrium

equilibrium

equilibrium

14.1

14.1

constant

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

N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]bLaw of Mass Action

K >> 1

K << 1

Lie to the right Favor products

Lie to the left Favor reactants

Equilibrium Will

14.1

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.

N2O4 (g) 2NO2 (g)

Kc = [NO2]2

[N2O4]Kp =

NO2P2

N2O4P

In most cases

Kc Kp

aA (g) + bB (g) cC (g) + dD (g)

14.2

Kp = Kc(RT)Δn

Δn = moles of gaseous products – moles of gaseous reactants

= (c + d) – (a + b)

Homogeneous Equilibrium

CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)

Kc =‘[CH3COO-][H3O+][CH3COOH][H2O]

[H2O] = constant

Kc = [CH3COO-][H3O+]

[CH3COOH]= Kc [H2O]‘

General practice not to include units for the equilibrium constant.

14.2

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

following reaction:

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

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)

Kc = [COCl2]

[CO][Cl2]=

0.140.012 x 0.054

= 220

Kp = Kc(RT)Δn

Δn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K

Kp = 220 x (0.0821 x 347)-1 = 7.7

14.2

The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2

2NO2 (g) 2NO (g) + O2 (g)

14.2

Kp = 2PNO PO

2

PNO2

2

PO2 = Kp

PNO2

2

PNO2

PO2 = 158 x (0.400)2/(0.270)2 = 347 atm

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

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

Kc =‘[CaO][CO2]

[CaCO3][CaCO3] = constant[CaO] = constant

Kc = [CO2] = Kc x‘[CaCO3][CaO]

Kp = PCO2

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

14.2

PCO 2= Kp

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

PCO 2 does not depend on the amount of CaCO3 or CaO

14.2

Writing Equilibrium Constant Expressions

• The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.

• The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.

• The equilibrium constant is a dimensionless quantity.

• In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.

• If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

14.2

Calculating Equilibrium Concentrations

1. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.

2. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.

3. Having solved for x, calculate the equilibrium concentrations of all species.

14.4

At 12800C the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)

Br2 (g) 2Br (g)

Let x be the change in concentration of Br2

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

[Br]2

[Br2]Kc = Kc =

(0.012 + 2x)2

0.063 - x= 1.1 x 10-3 Solve for x

14.4

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

x = -0.00178x = -0.0105

At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M

At equilibrium, [Br2] = 0.062 – x = 0.0648 M14.4

Example 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

32

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.

Example Problem: Calculate Concentration

32

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

42

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.

Example Problem: Calculate Keq

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

Initial:

Change:

Equilibrium:

Approximating

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 is 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.

Example

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 neardink, and 0.20 minus dink 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.

Example

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 ...

A + B C + D

C + D E + F

A + B E + F

Kc =‘[C][D][A][B]

Kc =‘‘[E][F][C][D]

[E][F][A][B]

Kc =

Kc ‘Kc ‘‘Kc

Kc = Kc ‘‘Kc ‘ x

If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

14.2

N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

2NO2 (g) N2O4 (g)

K = [N2O4]

[NO2]2‘ =

1K

= 216

When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

14.2

The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.

IF

• Qc > Kc system proceeds from right to left to reach equilibrium

• Qc = Kc the system is at equilibrium

• Qc < Kc system proceeds from left to right to reach equilibrium

14.4

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s Principle

• Changes in Concentration

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

AddNH3

Equilibrium shifts left to offset stress

14.5

Le Châtelier’s Principle

• Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left14.5

aA + bB cC + dD

AddAddRemove Remove

Le Châtelier’s Principle

• Changes in Volume and Pressure

A (g) + B (g) C (g)

Change Shifts the Equilibrium

Increase pressure Side with fewest moles of gas

Decrease pressure Side with most moles of gas

Decrease volume

Increase volume Side with most moles of gas

Side with fewest moles of gas

14.5

Le Châtelier’s Principle

• Changes in Temperature

Change Exothermic Rx

Increase temperature K decreases

Decrease temperature K increases

Endothermic Rx

K increases

K decreases

14.5colder hotter

uncatalyzed catalyzed

14.5

Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner

Le Châtelier’s Principle

Example

Chemistry In Action

Life at High Altitudes and Hemoglobin Production

Kc = [HbO2]

[Hb][O2]

Hb (aq) + O2 (aq) HbO2 (aq)

Chemistry In Action: The Haber Process

N2 (g) + 3H2 (g) 2NH3 (g) H0 = -92.6 kJ/mol

Le Châtelier’s Principle

Change Shift EquilibriumChange Equilibrium

Constant

Concentration yes no

Pressure yes no

Volume yes no

Temperature yes yes

Catalyst no no

14.5