Equilibrium

UNIT 12

Overview Concept of

Equilibrium Equilibrium constant Equilibrium

expression Heterogeneous vs

homogeneous equilibrium

Solving for equilibrium constant Solving for K Manipulating

equilibrium constant Using gases

Equilibrium Calculations ICE method ICE Stoichiometry

Reaction Quotient (Q) Relationship K and Q

Le Chatlier’s Principle Pressure Volume Concentration Temperature Catalysts

The Concept of Equilibrium We’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 ...

The Concept of Equilibrium Chemical equilibrium occurs when a

reaction and its reverse reaction proceed at the same rate.

The Concept of Equilibrium

As a system approaches equilibrium, both the forward and reverse reactions are occurring.

At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium

Once equilibrium is achieved, the amount of each reactant and product remains constant.

A System at EquilibriumRates become equal Concentrations become constant

Depicting Equilibrium In a system at equilibrium, both the

forward and reverse reactions are running simultaneously. We write the chemical equation with a double arrow:

Equilibrium Constant

Forward reaction: Reverse reaction:

Rate LawRate law

Equilibrium ConstantAt equilibrium

Rearranging gives:

r

Equilibrium Constant The ratio of the rate constants is a constant

(as long as T is constant).

The expression becomes

Equilibrium Expression (Law of Mass Action)To generalize, the reaction:

Has the equilibrium expression:

This expression is true even if you don’t know the elementary reaction mechanism.

Equilibrium Expression

[A], [B], [C], [D] = molar concentrations or partial pressures at equilibrium

Products = numerator and reactants = denominator

Coefficients in balanced equation = exponents

Equilibrium Expression Equilibrium constant can have different

subscripts

Kc = concentration Moles, liters, or molarity

Kp = partial pressure Atmospheres

Ksp = solubility product Reactant is a solid so only products are in

expression

Equilibrium Expression

Does not depend on initial concentration of reactants and products

Solids and pure liquids are not included in equilibrium expression (only gases and solutions)

Keq does not have units (describes activity of reaction)

Depends only on the particular reaction and the temperature

Equilibrium Expression The Concentrations of Solids and Liquids

Are Essentially Constant Therefore, the concentrations of solids and

liquids do not appear in the equilibrium expression

Kc = [Pb2+] [Cl−]2

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Sample Exercise 1 Write the equilibrium expression for Kc for

the following reactions:

Sample Exercise 1 – Answers

Homogeneous vs Heterogeneous Homogeneous equilibrium

Reactants and products are all in the same phase Example: 2NOBr(g) ↔ 2NO(g) + Br2(g)

Heterogeneous equilibrium Reactants and products are occur in different

phases Example: PbCl2 (s) ↔ Pb2+(aq) + 2Cl-(aq)

Equilibrium - Summary At equilibrium…

Concentrations of reactants and products no longer change with time

Neither reactants nor products can escape from the system

The equilibrium constant comes from a ratio of concentration

Solving for Keq

Kc, the final ratio of [NO2]2 to [N2O4], reaches a constant no matter what the initial concentrations of NO2 and N2O4 are.

Solving for Keq

= = 0.211

[0.0172]2

[0.00140]

Solving for Keq

This graph shows data from the last two trials from the table.

What does the value of K mean? If K >> 1, the

reaction is product-favored; contains mostly products at equilibrium

• If K << 1, the reaction is reactant-favored; contains mostly reactants at equilibrium.

Manipulating Equilibrium Constants The equilibrium constant of a reaction in

the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

Manipulating Equilibrium Constants The equilibrium constant of a reaction that

has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

Manipulating Equilibrium Constants If two reactions can be added together to

create a third reaction, then the Keq for the two reactions can be multiplied together to get the Keq for the third reaction.

If A + B ↔ C Keq = K1

and C ↔ D + E Keq = K2

then A + B ↔ D + E Keq = K1K2

Manipulating Equilibrium Constants (Summary) The equilibrium constant in the reverse

direction is the inverse of the equilibrium constant in the forward direction.

The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number.

The equilibrium constant for a net reaction made up of multiple steps is the product of the equilibrium constants for individual steps.

Equilibrium Constant and Gases Because pressure is proportional to

concentration for gases, the equilibrium expression can also be written in terms of partial pressures (instead of concentration):

Mixed versions are also used sometimes:

Relationship between Kc and Kp

From the ideal gas law we know that

= Pressure in terms of concentration

Relationship between Kc and Kp

Substituting P=[A]RT into the expression for Kp for each substance, the relationship between Kc and Kp becomes

Where:Kp = Kc (RT)n

n = (moles of gaseous product) – (moles of gaseous reactant)R = 0.0821 L∙atm/mol∙K

GIVEN TO YOU ON AP CHEAT SHEET!!

Relationship between Kc and Kp

In the synthesis of ammonia from nitrogen and hydrogen, N2(g) + 3H2(g) ↔ 2NH3(g). Given Kc of 9.60 at 300°C, calculate Kp.n = (moles gaseous products – moles

gaseous reactants) = 2-4 = -2

Kp = Kc (RT)n

Kp = 9.60(0.0821×573K)-2= 4.34×10-3

TheICEMethod

ICE

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)

Comparing initial concentrations to equilibrium

concentrations

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 – 0.0011x4x2 + 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 Mor 0.00844 MAt equilibrium, [Br2] = 0.062 – x = 0.0648 M

14.4

37

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

][]][[

HIIHKeq

Initial:Change:Equil:

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

32

2

2 1026.1]2242.0[]2242.0[

]][[

xx

xx

xxKeq

38

32

2

1026.1]2242.0[

xx

x

01038.71022.1995.0 532 xxxx

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

aacbbx

242

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.

Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 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

xIIKeq

40

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. (Example: A million dollars minus a nickel is still

a million dollars)

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.

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

Initialchangeequil:

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

102

10

2

2

1094.2]20.0[

]2[

1094.2][

][

xx

x

xIIKeq

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

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

Initialchangeequil:

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

102

10

2

2

1094.2]20.0[

]2[

1094.2][

][

xx

x

xIIKeq

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

102

1094.220.0]2[ xx x = 3.83 x 10-6 M

Equilibrium Stoichiometry

Given initial concentrations of reactantsAND

Equilibrium concentration of products

(or vice versa)

Equilibrium StoichiometryA closed system initially containing1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448C for the reaction:

H2 (g) + I2 (g) 2 HI (g)

Equilibrium Stoichiometry

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At Equilibrium

1.87 x 10-3

ICE method:

Write what we know.

Equilibrium Stoichiometry

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At equilibrium

1.87 x 10-3

Calculate the change in [HI]. [HI] Increases by 1.87 x 10-3 M

Equilibrium Stoichiometry

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

1.87 x 10-3

Use stoichiometry to solve for the change in concentration of [H2] and [I2] from the given change in [HI].

Equilibrium Stoichiometry

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

Now calculate concentrations at equilibrium

…and, therefore, the equilibrium constant

Kc =[HI]2

[H2] [I2]

= 51

= (1.87 x 10-3)2

(6.5 x 10-5)(1.065 x 10-3)

The Reaction Quotient (Q) To calculate Q, one substitutes the initial

concentrations of reactants and products into the equilibrium expression.

Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

Q [A], [B], [C], [D] = initial molar

concentrations or partial pressures

If Q = K,

The system is at equilibrium.

If Q > K,

There is too much product

and the reaction proceeds

backwards (equilibrium

shifts to the left).

If Q < K,

There is too much reactant, and the reaction

proceeds forward

(equilibrium shifts to the

right).

Summary Q uses same as equilibrium expression but

with initial concentrations or pressures

Q = K, reaction is at equilibrium Q > K, reaction goes backwards (shifts left) Q < K, reaction goes forward (shifts right)

Le Châtelier’s Principle“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

Le Chatlier’s Principle1. System starts at equilibrium.2. A change/stress is then made to system

at equilibrium. Change in concentration Change in volume Change in pressure Change in Temperature Add Catalyst

3. System responds by shifting to reactant or product side to restore equilibrium.

Le Chatlier’s Principle – basic terms When you take something away from a system at

equilibrium, the system shifts in such a way as to replace some what you’ve taken away.

When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

Le Chatlier’s Principle - Concentration

A system shifts to offset the change in concentration of reactants or products

Le Chatlier’s Principle - Concentration Adding reactants or products causes a shift in

the other direction to offset the added reactants or products

aA + bB cC + dD

Add

aA + bB cC + dD

Add

Shifts right Shifts left

Le Chatlier’s Principle - Concentration Removing reactants or products causes a

shift toward the removed reactants or products to offset the change

aA + bB cC + dD aA + bB cC + dD

Remove Remove

Shifts rightShifts left

Le Chatlier’s Principle – Volume & Pressure

Only matters when dealing with gases

Count the total moles of gas on each side of the equation to determine shift

Le Chatlier’s Principle – Volume & 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.

62

Le Chatlier’s Principle – Volume & Pressure

CHANGE SHIFT

Increase pressure Toward side with less moles of gas Decrease volume Toward side with less moles of gas Decrease pressure Toward side with more moles of gas Increase volume Toward side with more moles of gas

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

Le Chatlier’s Principle – Temperature Endothermic: heat can be considered as a

reactant A + heat B + C

Exothermic: heat can be considered as a product A + B C + heat

Adding heat (i.e. heating the vessel) favors away from the increase

Removing heat (i.e. cooling the vessel), favors towards the decrease

Le Chatlier’s Principle – Temperature Equilibrium constant, K, is temperature

dependant Temperature is only factor that can change

the value of K

Change ExothermicIncrease temperature K decreases

Decrease temperature K increases

EndothermicK increases

K decreases

Le Chatlier’s Principle – Catalysts Does not change K Does not shift the position of an

equilibrium system System will reach equilibrium sooner

Catalysts increase the rate of both the forward and reverse reactions.

Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

Le Chatelier Example #1A closed container of ice and water is at equilibrium. Then, the temperature is raised.

Ice + Energy Water

The system temporarily shifts to the _______ to restore equilibrium.

right

Le Chatelier Example #2A closed container of N2O4 and NO2 is at equilibrium. NO2 is added to the container.

N2O4 (g) + Energy 2 NO2 (g)

The system temporarily shifts to the _______ to restore equilibrium.

left

Le Chatelier Example #3A closed container of water and its vapor is at equilibrium. Vapor is removed from the system.

water + Energy vapor

The system temporarily shifts to the _______ to restore equilibrium.

right

Le Chatelier Example #4A closed container of N2O4 and NO2 is at equilibrium. The pressure is increased.

N2O4 (g) + Energy 2 NO2 (g)

The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation.

left