Chapter 15 - Chemical Equilibrium

USING THE EQUILIBRIUM CONSTANT

1. Which is favored - Reactants or products?

2. Predict direction of a reactionQ reaction quotient

3. Obtaining equilibrium concentrations of reactants and products.

4. Predict effect of changing conditions - Le Chatelier’s Principle

Concept of Equilbrium

In chemical equilibria, forward and reverse reactions occur at equal rates.

A BAt equilibrium, forward rate = backward rate

Forward reaction A B Rate = kf [A]

Backward reaction B A Rate = kb [B]

At equilibrium, kf [A] = kb [B]

• Dynamic balance

• Reaching equilibrium may be slow!

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kfkb

=[B]

[A]= K eq

Haber process: N2(g) + 3H2(g) 2NH3(g)

Initial State: reactants only Initial state: products only

Same equilibrium achieved

Time Time

Final State: ratio of products to reactants is the same for both!

The relationship between the concentrations of products and reactants at “equilibrium” will be the same regardless of starting conditions.

Catalysts do not effect equilibrium concentrations

H2 H2

NH3 NH3

N2 N2

Equilibrium Constant

Equilibrium point of any reaction is characterized by a single number.

Example: 2A B (2NO N2O4)

For this reaction: the ratio of concentrations at equilibrium will be constant.

Keq is a NUMBER.

Keq (the number) DOES NOT depend on concentrationIt’s a function of temperature only.

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B[ ]

A[ ]2 =constan t ≡ K eq =

N2O4[ ]

NO[ ]2

constant

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

What is the equilibrium constant expression for the Haber process?

1.

2.

3.

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K eq =N2[ ] H 2[ ]

NH 3[ ]2

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K eq =N2[ ] H 2[ ]

3

NH 3[ ]2

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K eq =NH 3[ ]

2

N2[ ] H 2[ ]3

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K eq =NH3[ ]

2

N2[ ]2H2[ ]

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K eq =NH 3[ ]

N2[ ] H 2[ ]

4.

5.

Heterogeneous Equilibria - Reactions in more than one phase

3Fe(s) + 4H2O(g) Fe3O4(s)+4H2(g)

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K eq =Fe 3O4[ ] H 2[ ]

4

Fe[ ]3H 2O[ ]

4

What is [Fe]? [Fe 3O4]?

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K c =Fe[ ]

3

Fe 3O4[ ]K eq =

H 2[ ]4

H 2O[ ]4

⇑ constant Leave solids and pure liquids out of the Keq expre .ssion

C6H6(g) + 3H2(g) C6H12(g) CaCO3(s) CaO(s) + CO2(g)

Predicting the direction of a reaction

aA + bB cC + dD

Reaction quotient Q

Note: the concentrations used are NOT equilibrium concentrations.

When Q = Kc system IS at equilibrium

When Q < Kc reaction moves to right (produces product)

When Q > Kc reaction moves to left

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Q =C[ ]

cD[ ]

d

A[ ]aB[ ]

b

General Approach to Equilibrium Constant Problems

1) Write the balanced reaction.

1) Write the general form for Keq.

1) Set up a data table:(may need algebraic unknowns)initial conditionschanges in concentrationsequilibrium concentrations

4) Substitute equilibrium concentrations into the expression for Keq and solve.

2 IBr(g) Br2(g) + I2(g)

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K c =Br2[ ] I2[ ]

IBr[ ]2 = 2.5×10−3

Initially [IBr] = [I2] = [Br2] = 0.05M

What is the value of Q?

Which way does reaction go?

What are the final concentrations of reactants and products?

Example:

A 1.0 L container holds 224 g of Fe and 5.00 mole of H2O(l). It is heated to 1000K and reaches equilibrium. 56 g of Fe are left unreacted. What is Kc at 1000K?

3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

initial

change

final

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K c =H 2[ ]

4

H 2O[ ]4

Le Chatelier’s Principle

If a system at equilibrium is disturbed by changing

• Concentration of one of the components• Temperature• Pressure

the concentrations will shift to counteract the disturbance.

Fe3+(aq) + SCN-(aq) [Fe(SCN)]2+(aq)yellow colorless red

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

If the system is at equilibrium and we add 0.1 mole of HI, what will happen?

1. reaction shifts to right 2. reaction shifts to left 3. no change occurs

If the volume of the container decreases, what will happen?

1. reaction shifts to right 2. reaction shifts to left 3. no change occurs

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

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K c =H 2[ ] I2[ ]

HI[ ]2 = 1.25×10−3

Summary:

changing concentration (or V so that [ ] changes) puts a stress on the system.

Stresses do not change Keq!

Q changes; system shifts to re-establish equilibrium

QK

WHAT IF TEMPERATURE CHANGES? Keq changes

change depends on whether the reaction is exothermic or endothermic.

H +H

Nitrogen fixation

Must break N-N triple bond (D = 946 kJ)

Important in biological systems (proteins, nucleic acids) & industrially (fertilizer, polymers, explosives, …)

Beans, bacteria, etc: nitrogenase enzyme reduces N2 to NH3 at room temp, 1 atm pressure

Fritz Haber (a German…) developed the process for fixing N2 in 1912.

World War I: Germany imported nitrates from Chile to make explosives.Allied blockade prevented these compounds from reaching Germany.

Haber process was used to maintain soluble nitrogen supplies.

Haber process - Industrial process used to make ammonia

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

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

Do we want high or low temperature?

Do we want high or low pressure?

Liquefy ammonia as process proceeds. WHY?

Problem: rate of reaction increases as T increases, BUT equilibrium constant decreases at higher T.

CAN’T change the equilibrium constant so doing the reaction at very high temperature would never work.

Solution: need to find a way to speed up the reaction at lower temperature: Need an appropriate catalyst.

Haber process:uses Fe/Al2O3 catalystworks a 400-500oC, at pressures of 200-600 atm

At T < 400oC, lower pressures could be used to get same equilibrium conversion of N2 to NH3

However, the rate falls exponentially with decreasing temperature

Still an enormous research problem:FertilizerH2 storage (NH3 is 17.6% hydrogen)Polymer chemistry

ROLE OF CATALYST

A catalyst increases the rate at which equilibrium is achieved, but does not change the composition the equilibrium mixture.

It increases the rate by lowering the activation barrier between reactants and products

Ea is lowered the same amount for BOTH forward and backward reactions SO the rate for BOTH reactions is increased.

The value of the equilibrium constant is NOT affected by the presence of a catalyst

Catalytic converter

N2(g) + O2(g) 2NO(g) H = +180.8kJ Keq at 300K = 10-15

At high T (combustion T), NO is favored.As gases cool, equilibrium favors NO conversion to N2 and O2, BUT low T means rate is slow: need a catalyst: Pt, Rh, Pd

In catalytic converter, NO is reduced fast (~100-400x in 10-3 s)

What happens if T is increased? At 2400K, Keq = 0.05