1

1

Static Equilibrium

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Dynamic Equilibrium

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2

Chemical (Dynamic) Equilibrium1. How do we recognize it?How do we recognize it?

Constant Macroscopic PropertiesTemperatureSweetnessColor

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2.

2. How do we explain it?How do we explain it?

Molecular behavior

Equilibrium with George Pimmentel 1987.wmv Begin 3:40

Properties of Equilibrium

• While not all reactions start out in equilibrium, many reactions tend towards equilibrium.

• The reaction displays no macroscopic evidence of change.

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g

• It is reached by a spontaneous process and minimizes energy of the system in question.

• It shows a dynamic balance between the forward and reverse direction.

• It is the same regardless of direction of approach.

Chemical Equilibrium• Consider colorless frozen N2O4. At room

temperature, it decomposes to brown NO2:

N2O4(g) 2NO2(g).

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2 4(g) 2(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 concentrations of all species no longer change.

3

Consider frozen N2O4: only white solid is present. On the microscopic level, only N2O4 molecules are present.

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The point at which the rate of decomposition:

N2O4(g) 2NO2(g)

equals the rate of dimerization:2NO2(g) N2O4(g).

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2(g) 2 4(g)

is dynamic equilibrium.The equilibrium is dynamic becausethe reaction has not stopped: theopposing rates are equal.

N2O4 (g) 2NO2 (g)

equilibrium

equilibrium equilibrium

Start with NO2 Start with N2O4 Start with NO2 & N2O4

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4

Connection between Equilibrium and Kinetics

• ConsiderForward reaction: A B Rate = kf[A]Reverse reaction: B A Rate = kr[B]

• At equilibrium kf[A] =kr[B].

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• For an equilibrium we write A B

• As the reaction progresses– [A] decreases to a constant,– [B] increases from zero to a constant.– When [A] and [B] are constant, equilibrium is achieved.

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

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Physical equilibrium

H2O (l)

Chemical equilibrium

N2O4 (g)

H2O (g)

2NO2 (g)

NO2

•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

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

•Chemical equilibrium is the point at which the concentrations of all species no longer change.

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The Concept of EquilibriumThe Concept of Equilibrium

The decomposition of N2O4 is a reversible process, meaning the products of the reaction can react to form reactants.

The system is in equilibrium when the rates of the forward reaction and the reverse reaction are the same.

N2O4(g) ⇌ 2NO2(g)

rate forward = kf[N2O4] and rate reverse = kr[NO2]2

Graph View

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Equilibrium Constant Kc

• No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium.

• For a general reaction

aA + bB cC + dD

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aA + bB cC + dD

The equilibrium expression is given by the

Law of Mass ActionLaw of Mass Action

K = [C] [D][A] [B]c

c d

a b

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Example

• How does one write the equilibrium expression (or the law of mass action for the following

reactions in terms of concentration:

• H2O(g) + CO(g) H2(g) + CO2(g)

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H2O(g) CO(g) H2(g) CO2(g)

• 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

• 2NO2(g) N2O4(g)

QUESTION

•One of the environmentally important reactions involved in acid rain production has the following equilibrium expression.

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• K = [SO3] / ([SO2][O2]1/2)

•From the expression, what would be the balanced chemical reaction?

Note: all components are in the gas phase.

ANSWER

SO2 + ½ O2 SO3

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Note: This slide was corrected. The equation was reversed during lecture

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The Equilibrium Constant

• Kc is based on the molarities of reactants and products at equilibrium.

• Kp is based on the partial pressures of reactants d d t t ilib i

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and products at equilibrium.

• We generally omit the units of the equilibriumconstant.

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

QUESTION

Starting with the initial concentrations:

[NH3] = 2.00 M; [N2] = 2.00 M; [H2] = 2.00 M,

what would you calculate as the equilibrium ratio K once the equilibrium position is reached for the ammonia synthesis reaction?

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reaction?

N2 + 3H2 2 NH3

A. 1.00B. 0.250C. 4.00D. Wait, I don’t think this can be done.

ANSWER

Choice D is the correct response. The concentrations given are for the INITIAL position. In order to calculate the K ratio of products to reactants the equilibrium position concentrations

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must be observed.

8

2 Cu2+(aq) + Sn2+(aq) 2 Cu+(aq) + Sn4+(aq)

K =

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(0.0640)2 (0.0320)K =

(0.0360)2 (0.0680)= 1.49

Equilibrium Constant Kp

• If all of the species in the rxn are gaseous then

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

And dc PP

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And

Where all the species are represented by partial pressures.

bB

aA

dD

c

p P P

P P = K

C

QUESTION

One of the primary components in the aroma of rotten eggs is H2S. At a certain temperature, it will decompose via the following reaction.

2H2S (g) 2 H2 (g) + S2 (g)

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If an equilibrium mixture of the gases contained the following pressures of the components, what would be the value of Kp?

PH2S = 1.19 atm; PH2= 0.25 atm; PS2

= 0.25 atm

A. 0.052B. 91C. 1.1 10–2

D. 0.013

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ANSWER

Choice C provides the correct Kp value. Whether the equilibrium constant is based on pressure or molarity, the ratio

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q p yis always set to show products divided by reactants. In addition, the pressure must be raised to the power that corresponds to the coefficients in the balanced equation.

Relationship between Kp and Kc

• Since PV = nRT, the molar concentration, n/V, is given by P/RT

where R = 0.0821 L* atm mol-1 K-1

d

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KRT

= (P )RT

(P )RT

(P )RT

(P )RT

= [C] [D][A] [B]

= KpC

cD

d

Aa

Bb

c d

a b c

Relationship between Kp and Kc Continued

• Relating Kp and Kc

K = K (RT) n

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• Where n is the number of moles of gas and

n = Σn(products) – Σn(reactants)

K = K (RT)p c

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Example

Write the equilibrium expressions Kc and Kp

for:

4 O( ) 6 O( ) 4 ( ) O ( )

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4NO(g) + 6H2O(g) 4NH3(g) + 5O2(g)

QUESTION

The balanced equation shown here has a Kp value of 0.011. What would be the value for Kc ?(at approximately 1100°C)

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2H2S (g) 2 H2 (g) + S2 (g)

A. 0.000 098B. 0.011C. 0.99D. 1.2

ANSWER

Choice A is obtained from Kp= Kc (RT)n when T is expressed in K and the expression is solved for Kc

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Equilibrium Positions

• There is only 1 equilibrium constant at a given temperature, but there are an infinite number on equilibrium positions at a give temperature

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temperature.

• What does that mean?

Equilibrium Positions

• Consider

• If we start with a mixture of nitrogen and hydrogen (in any proportion), the reaction will reach equilibrium with a

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

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any proportion), the reaction will reach equilibrium with a constant concentration of nitrogen, hydrogen and ammonia.

• However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N2 and H2 will be produced until equilibrium is achieved. There are an infinite number of equilibrium positions.

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Consider the following reaction:CH3COOH + CH3CH2OH � CH3COOCH2CH3 + H2O

The concentrations below are equilibrium concentrations:

[CH3COOH] [CH3CH2OH] [CH3COOCH2CH3] [H2O] Kc

7.29 0.10 1.71 1.71 4.0

7 07 0 32 2 93 2 93 3 8

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7.07 0.32 2.93 2.93 3.8

5.86 0.75 4.14 4.14 3.9

3.33 3.33 6.67 6.67 4.0

1.42 13.8 8.58 8.58 3.8

New topic: How to Manipulate K

Given:

N2O4(g) 2NO2(g)

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What happens when we reverse the reaction?

212.0

ONNO

42

22 cK

What happens when we reverse the reaction?

Consider the reverse reaction. Kc for this reaction is the reciprocal of the forward reaction on the last slide.

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2NO2(g) N2O4(g)

72.4212.01

NO

ON2

2

42 cK

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Three new manipulations to remember…

1. Reverse a reaction, the equilbrium constant is 1/K.For a chemical reaction in equilibrium, aA � bB

Kf = [B]b / [A]a

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For the reverse reaction, bB � aA, the equilibrium constant Kr = 1/Kf i.e., Kr = [A]a / B]b

2. If you add two reactions in equilibrium, their equilibrium constants multiply.

3. If you multiply a reaction by a number, its equilibrium constant is raised to the power of that number.

Question: Suppose you wish to find the equilibrium constant for the reaction

I. 2IBr(g) + Cl2 (g) I2 (g) + 2BrCl(g) K=?

and you are given these two equations:

2BrCl(g) Cl2(g) + Br2 (g) K1 = 0.45

Br2 (g) + I2 (g) 2IBr(g) K2 = 0.051

What is K for the reaction in I ?

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2IBr(g) + Cl2 (g) I2 (g) + 2BrCl(g)

2BrCl(g) Cl2(g) + Br2 (g) K1 = 0.45

Br2 (g) + I2 (g) 2IBr(g) K2 = 0.051

add eqns:1. 2BrCl(g) Cl2 (g) + Br2 (g) K2 = 0.452. Br2 (g) + I2 (g) 2IBr(g) K2 = 0.0513. 2BrCl(g) + I2 (g) Cl2 (g) + 2IBr(g) K2 = ?

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(g) 2 (g) 2 (g) (g) 2

K3 = K1 x K2 = 0.45 x 0.051 = 0.023

Reaction of interest is the reverse of the above reaction, take the inverse of K3:

Cl2(g) + 2IBr(g) 2BrCl(g) + I2 (g) K = 1/K2

K= 43.6 = 44

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The Size of K

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About the Equilibrium Constant K …

• Remember, K, is the ratio of products to reactants.

• Therefore, the larger K the more products are present at equilibrium.

C l h ll K h

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• 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.(products are favored)

• If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left (reactants are favored)

Heterogeneous 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:CaCO (s) CaO(s) + CO (g)

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– 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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• Sample size does not enter into the equilibrium expression, concentration and partial pressure does.

• density of CaO = 3.3 g/cm3

• MW of CaO = 56 g/mol

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MW of CaO 56 g/mol

• Molarity of CaO = 59 M, regardless of size of sample

• Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant at constant temperature.

CaCO3 (s) � CaO (s) + CO2 (g)

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PCO2

does not depend on the amount of CaCO3 or CaO

PCO2

= Kp

Heterogeneous Equilibria

• For the decomposition of CaCO3:

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.COconstantCOCaCO

CaO

cK

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

• Bottom line: Don’t include any species in a rxn in the equilibrium constant K if it has a physical state of (l) or (s).

2COconstant cc KK

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QUESTION

The liquid metal mercury can be obtained from its ore cinnabar via the following reaction:

HgS(s) + O2 (g) Hg (l) + SO2 (g)

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Which of the following shows the proper expression for Kc?

A. Kc = [Hg][SO2]/[HgS][O2]B. Kc = [SO2]/[O2]C. Kc = [Hg][SO2]/[O2]D. Kc = [O2]/[SO2]

ANSWER

Choice B correctly presents the product to reactant ratio.

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Recall that pure liquids and solids are not shown in the equilibrium constant expression.

Reaction Quotient Q

• Suppose a reaction approaches equilibrium, but is not in equilibrium at some given time.

• We define Q, the reaction quotient, for a general reaction

A + bB C + dD

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aA + bB cC + dD

as

where [A], [B], [C], and [D] are molarities at any time other than equilibrium.

bo

ao

do

co

c[B] [A]

[D] [C] = Q

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Reaction Quotient Q• For reactions that tend towards equilibrium:

If the reaction is not in equilibrium, it approachesequilibrium by adjusting the concentrations ofthe reactants and products.

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p

• If Q > K then the reverse reaction must occur toreach equilibrium (i.e., products are consumed,reactants are formed, the numerator in theequilibrium constant expression decreases andQ decreases until it equals K).

• If Q < K then the forward reaction must occur to reach equilibrium.

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

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The Reaction Quotient Qc

Q K l t ilib i

bo

ao

do

co

c[B] [A]

[D] [C] = Q

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Q = K only at equilibrium.

So what does it mean whenQ>K and Q<K? How does the reaction approach

equilibrium? Does the rxn form more products orreactants? Is the rxn driven to the left or theright?

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QUESTION

The equilibrium constant for the synthesis of ammonia is 0.060 (at 500°C). Given the following concentrations of N2 and NH3, what would the concentration of H2 have to be in order for the reaction to shift toward the product NH3?

N2 (g)+ 3H2 (g) 2NH3 (g);N2 (g)+ 3H2 (g) 2NH3 (g); Initial [N2] = 0.20 M;

[NH3] = 0.050 M

A. [H2] < 2.1 MB. [H2] > 4.2 MC. [H2] ≥ 0.59 MD. [H2] < 0.59 M

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ANSWER

Choice C correctly indicates that the H2 concentration must be greater than 0.59 M in order for Q to approach K. Example: Suppose [H2] = 0.5. Then Q = 0.1 and Q>K. For Q to

h K [H ] b l if hi l i approach K, [H2] must be larger if everything else is held constant. The expression of K should be set up so that the unknown [H2] could be solved.

[H2] = {[NH3]2/([N2]K])}1/3

Or, Q = K = 0.060 if (0.050)2

(0.20)(H2)3

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How to Determine Equilibrium ConcentrationsHow to Determine Equilibrium Concentrations

• Suppose you know Q for a reaction and you also know the equilibrium constant K at the given temperature.

• Can you determine the concentration of all

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• Can you determine the concentration of allmajor species at equilibrium?

• The answer is yes, and this involves using what is known as ICEICE tables.

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Equilibrium and Concentration Calculations Equilibrium and Concentration Calculations Using an ICE TableUsing an ICE Table

1. Using Q, decide what direction the rxn will proceed.

2. Write the equilibrium rxn.

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3. Create an ICE table which includes Initial, Change in (x values), and Equilibrium Concentrations of all species in the equilibrium expression.

4. Insert the equilibrium concentrations in the equilibrium expression and solve for x.

Equilibrium and Conc. Calculations Equilibrium and Conc. Calculations Using an ICE Table, ContinuedUsing an ICE Table, Continued

5. Once you know how much the concentration changed in the process of moving to equilibrium, you can determine the equilibrium concentration of each species in the equilibrium expression.

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Note: You might need to use the quadratic equation to solve for x, or factor or use the method of successive approximations.

Example

• One 2.00 mols of dihydrogen gas and 2.00 moles of diiodine vapor react in a 2.00-L flask to form hydrogen iodide gas. The equilibrium constant is 55.3 at a given temperature.

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• What are the concentrations of each component at equilibrium in this reaction.

• This one is easy because you can take the square root of a perfect square.

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At 1280oC 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

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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

The change in Br2 is +ive while the change in Br is –ive.

Why?

What does the magnitude of Kc tell you about Q and thedi ti th t ilib i i f d?

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direction that equilibrium is favored?

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

Br2 (g) 2Br (g)

x = 0.00178x = 0.0105

2-b b - 4ac

2a

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Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

+x -2x

0.063 +- x 0.012 - 2x

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

At equilibrium, [Br2] = 0.062 + x = 0.0648 M Always think about and test the reasonableness of the values of the roots

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Another example

• Suppose that for the synthesis of HF, 16.4 atm of H2 and 32.8 atm of F2 are mixed together in a container. The equilibrium constant, Kp, is 115 at some temperature.

• Calculate the partial pressure for each component and the total pressure.

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H2 F2 2HF

I 16.4 32.8 -

C -X -X +2X

E 16.4-X 32.8-X +2X

4x2 = 6.186 x 104 - 5.658 x103 + 115 x2 111 x2 - 5.658 x103 + 6.186 x 104 = 0

PT = 31.76 + +16.92 +0.52 = 48.7 atm

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2-b b - 4ac

2a

=X =

X = 15.88, 35.04

Le Chatelier’s Principle

63

1884

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Le Chatelier’s Principle

• Consider the production of ammoniaN2 + 3H2 2NH3

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

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

• Can this be predicted?• Le Châtelier’s Principle: if a system at

equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

Le Chatelier’s Process

• Consider the Haber process (formation of ammonia

• If H2 is added while the system is at equilibrium the system must respond to

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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.

Le Chatelier’s Principle

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23

Change in Product or Reactant Concentration

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

• Removing a reactant or product shifts the equilibrium towards the decrease.

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q

• To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier).

Change in Pressure or Volume• As volume is decreased pressure increases.

• Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase.

• That is, the system shifts to remove gases and decrease pressure

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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.

Le Châtelier’s Principle

Changes in Volume and Pressure

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

Change Shifts the Equilibrium

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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

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Change in Pressure or Volume

• Consider

• An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4.Th i t t th i th t i t t

N2O4(g) 2NO2(g)

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• 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. reverse reaction is favored).

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

Change in Temperature

• 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

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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.

• What happens when you add an inert gas?

It depends on whether the container has a fixed volume or not. There is no equilbrium change as a result of the inert gas but…

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Fixed Volume – no change

Piston and cylinder – volume changes.

25

Change in Temperarture

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

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

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

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

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Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

QUESTION

The decomposition of sodium bicarbonate (shown here) is an endothermic process often used in cooking. Of the following changes, which one would cause the production of CO2 in a container to decrease?

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NaHCO3(s) CO2(g) + H2O(g) + NaCO3(s)

A. Addition of more NaHCO3

B. Decreasing the size of the containerC. Removing some, but not all, NaCO3

D. Increasing the temperature

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ANSWER

Choice B properly applies Le Châtelier’s principle to this system. As long as the solids remain present, they do not alter the concentration of gases. By decreasing the volume of the container the gas pressure is increased. To relieve this stress l h ld b h f CO ld h

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less gas should be present. Therefore, CO2 would have to decrease.

QUESTIONConsider the following reaction (assume an ideal gas mixture):

2NOBr(g) 2NO(g) + Br2(g) A 1.0-liter vessel was initially filled with pure NOBr, at a pressure of 4.0 atm, at 300 K. After equilibrium was reached, the volume was increased to 2.0 liters, while the temperature was kept at 300 K. This

ill lt i

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will result in:

A) an increase in Kp. B) a decrease in Kp. C) a shift in the equilibrium position to the right. D) a shift in the equilibrium position to the left. E) none of these

ANSWER

C) a shift in the equilibrium position to the right.

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The driving force here is entropy. Systems tend to the largest amount of randomness possible. With an increase in volume, an increase in the number of molecules is favorable.

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QUESTION

Which of the following is true for a system whose equilibrium constant is relatively small? A) It ill t k h t ti t h ilib i

79

A) It will take a short time to reach equilibrium.B) It will take a long time to reach equilibrium. C) The equilibrium lies to the left. D) The equilibrium lies to the right. E) Two of these.

ANSWER

C) The equilibrium lies to the left.

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Kk

The K value gives us information about how a reaction will progress. A large indicates that products are favored, while a small indicates that reactants are favored.

QUESTIONThe reaction quotient for a system is 7.2 10

2. If the

equilibrium constant for the system is 36, what will happen as equilibrium is approached? A) There will be a net gain in product. B) There will be a net gain in reactant.

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C) There will be a net gain in both product and reactant.

D) There will be no net gain in either product or reactant.

E) The equilibrium constant will decrease until it equals the reaction quotient.

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ANSWER

B) There will be a net gain in reactant.

Wh Q K thi th i hi h

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> When Q K, this means there is a higher concentration of products than found at equilibrium.

The reverse reaction will predominate until enough reactants are formed to reach K

.

Le Châtelier’s Principle - Summary

Change Shift EquilibriumChange Equilibrium

Constant

Concentration yes no

Pressure yes* no

Volume yes* no

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y

Temperature yes yes

Catalyst no no

*Dependent on relative moles of gaseous reactants and products

Example: The concentrations of an equilibrium mixture of O2 , CO, and CO2 were 0.18 M, 0.35 M, and 0.029 M respectively. Enough CO was added to the flask containing the equilibrium mixture to momentarily raise its concentration to 0.60 M. What will be the concentration of each species in the flask once equilibrium has been re-established after the additional carbon monoxide was added?

2 CO2(g) 2 CO(g) + O2(g)

After the addition of more CO the system is no longer in equilibrium. The once equilibrium quantities of the other three substances are now initial quantities. Let "x" represent the change in the amount of O2 gas. Adding the CO will force the reaction to proceed in the reverse direction (K < Q) causing the amount of O2 to decrease. (Note: The equilibrium quantities given in the problem, before the addition of more CO, are also used to calculate the value of the equilibrium constant, Kc.)

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Example: The concentrations of an equilibrium mixture of O2 , CO, and CO2 were 0.18 M, 0.35 M, and 0.029 M respectively. Enough CO was added to the flask containing the equilibrium mixture to momentarily raise its concentration to 0.60 M. What will be the concentration of each species in the flask once equilibrium has been re-established after the additional carbon monoxide was added?

2CO2(g) 2CO(g) O2(g)

Initial0 029 0 60 0 18

Initial Concentration (M)

0.029 0.60 0.18

Change in Concentration (M)

+ 2 x - 2 x - x

Equilibrium Concentration (M)

0.029 + 2 x 0.60 - 2 x 0.18 - x

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Example: Cl2 gas undergoes homolytic cleavage into chlorine atoms at 1100oC. Kp at 1100oC for this process is 1.13 x 10-4.

If a sample with an initial Cl2 gas pressure of 0.500 atm was allowed to reach equilibrium, what is the total pressure in the flask?

Cl2(g) 2 Cl(g)

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Example: Cl2 gas undergoes homolytic cleavage into chlorine atoms at 1100oC. Kpat 1100oC for this process is 1.13 x 10-4.

If a sample with an initial Cl2 gas pressure of 0.500 atm was allowed to reach equilibrium, what is the total pressure in the flask?

Cl2(g) 2 Cl(g)

Let "x" represent the change in the pressure of the Cl2 gas. Since the reaction will proceed forwards to establish equilibrium the pressure of the Cl2 gas will decrease. The total pressure at equilibrium will equal

Cl2(g) Cl(g)

Initial Pressure (atm) 0.500 0

Change in Pressure (atm) - x + 2 x

Pressure at Equilibrium (atm)

0.500 - x 0 + 2 x

decrease. The total pressure at equilibrium will equal the sum of the partial pressures of each gas at equilibrium.

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30

Another example

• Suppose that for the synthesis of HF, 16.4 atm of H2 and 32.8 atm of F2 are mixed together in a container. The equilibrium constant is 115 at some temperatureconstant is 115 at some temperature. Calculate the partial pressure for each component and the total pressure.

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H2 F2 2HF

I 16.4 32.8 -

C -X -X +2X

E 16.4-X 32.8-X +2X

4x2 = 6.186 x 104 - 5.658 x103 + 115 x2 111 x2 - 5.658 x103 + 6.186 x 104 = 0

X = 15.88, 35.04 PT = 31.76 + +16.92 +0.52 = 48.7 atm

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Chemistry In Action: The Haber Process

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

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31

• Equilibrium

• % Dissociation

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• Partial Pressure and Total Pressure at Equilibrium

Examples

For the rxn H2(g) + I2(s) 2HI(g) at 25 °C Kp = 0.345.If the PH2

= 1.00 atm, what is the equilibrium PHI at the above temperature?

An excess of I2 is present and the container is filled with H2

to a pressure of 4.0 atm at the same temperature. Calculate the equilbrium pressures of H2(g) and HI(g).What is the total pressure at equilibrium?

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Chemistry In Action

Life at High Altitudes and Hemoglobin Production

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

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Kc = [HbO2]

[Hb][O2]