Chemical Equilibrium

Chemical EquilibriumChapter 14

Equilibrium: the extent of a ________In stoichiometry we talk about ______yields,

and the many reasons actual yields may be _________.

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

Equilibrium looks at the ______ of a chemical

reaction.

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

_______equilibrium is achieved when:

• the rates of the ______ and _____ reactions are ______ and

• the _______ of the reactants and products remain _______

Physical equilibrium

H2O (l)

Chemical equilibrium

N2O4 (g)

14.1

H2O (g)

2NO2 (g)

Rate of ______ of Rate of ______ of cookies cookies

==Rate of ________ Rate of ________

cookiescookies

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

_________ to brown NO2:

N2O4(g) 2NO2(g).

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

• ________equilibrium is the point at which the rate of the ______ reaction is equal to the rate of the _______ reaction. At that point, the concentrations of all species are ________.

• Using the collision model:

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

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

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

N2O4(g) 2NO2(g)

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

2NO2(g) N2O4(g).

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

• The _____ _______ implies the process is dynamic.

N2O4(g) 2NO2(g)

The Concept of EquilibriumThe Concept of EquilibriumAs the reaction progresses

– [A] __________ to a constant,

– [B] __________ from zero to a constant.

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

A B

N2O4 (g) 2NO2 (g)

Start with NO2 Start with N2O4 Start with NO2 & N2O4

_________

__________

__________

14.1

14.1

________

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

products, the ______ ______ of ________ is achieved at equilibrium.

• For a general reaction

the equilibrium constant expression is

where ____ is the equilibrium constant.

aA + bB(g) pP + qQ

N2O4 (g) 2NO2 (g)

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

[N2O4]

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]b

K >> 1

K << 1

Lie to the ______ ______ products

Lie to the ______ ______reactants

Equilibrium Will

14.1

_________________applies to reactions in which all reacting species are in the _________ ______.

N2O4 (g) 2NO2 (g)

In most cases

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

14.2

n = moles of _________ products – moles of ______ reactants

= (c + d) – (a + b)

_______________ Equilibrium

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

Kc =‘ [H2O] = constant

Kc = = Kc [H2O]‘

General practice _____ 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 =

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

_________equilibrium applies to reactions in which reactants and products are in ________ phases.

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

The concentration of ______ and _______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 ____. In the gaseous phase, the concentrations can be expressed in ___ or in ___.

• The concentrations of pure ____, pure ____ and solvents do ____ appear in the equilibrium constant expressions.

• The equilibrium constant is a ________ quantity.

• In quoting a value for the equilibrium constant, you must specify the ______________ and the __________.

• If a reaction can be expressed as a _________ of two or more reactions, the equilibrium constant for the overall reaction is given by the _____ of the equilibrium _______ of the individual __________.

14.2

Calculating Equilibrium Concentrations

1. Express the equilibrium concentrations of all species in terms of the ______ concentrations and a single unknown ___, which represents the ______ in concentration.

2. Write the equilibrium constant expression in terms of the equilibrium ______. 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)

14.4

14.4

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

Initial:Change:Equil:

0.242 M 0 0

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 ______ very far, meaning the equilibrium concentrations will be nearly the _____ as the _____ concentrations of your ______.

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

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

Example

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

Initialchangeequil:

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

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 ___orders 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

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 ____________is calculated by substituting the ______ 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 _____ stress is applied to a system at equilibrium, the system _____ in such a way that the stress is partially _____as the system reaches a new _______ position.

__________’s Principle

• Changes in _______________

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)

Decrease concentration of product(s)

Decrease concentration of reactant(s)

Increase concentration of reactant(s)

14.5

aA + bB cC + dD

Le Châtelier’s Principle

• Changes in Volume and Pressure

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

Change Shifts the Equilibrium

_______ pressure Side with ______ moles of gas

________ pressure Side with _____ moles of gas

________ volume

________ volume Side with _____ moles of gas

Side with _____ moles of gas

14.5

Le Châtelier’s Principle

• Changes in Temperature

Change Exothermic Rx

_______ temperature K _______

_______temperature K ________

Endothermic Rx

K _______

K ________

14.5colder hotter

uncatalyzed catalyzed

14.5

Catalyst lowers Ea for ______ forward and reverse reactions.

Catalyst ________ change equilibrium constant or shift equilibrium.

• Adding a _______• does not change ___• does not shift the _______of an equilibrium system• system will ______ equilibrium sooner

Le Châtelier’s Principle

Example

Le Châtelier’s Principle

Change Shift EquilibriumChange Equilibrium

Constant

Concentration

Pressure

Volume

Temperature

Catalyst

14.5