Chapter 9: Chemical Equilibrium The forward and reverse reaction are both taking place at the same...

Chapter 9: Chemical Equilibrium

The forward and reverse reaction are both taking place at the same rate

Production and Decomposition of Ammonia

Forward Reaction: N2 (g) + 3H2 (g) 2NH3 (g)

Reverse Reaction: 2NH3 (g) N2 (g) + 3H2 (g)

Equilibrium Reaction: N2 (g) + 3H2 (g) 2NH3 (g)

Note the double headed arrow!

The ammonia is decomposing as fast as it is being made at equilibrium

Equilibrium and the Law of Mass Action

2SO2 (g) + O2 (g) 2SO3 (g)

5 mixtures of different initial compositions of gases were made and allowed to reach equilibrium at 1000K

At first, you don’t see a trend in the data…

Equilibrium and the Law of Mass Action

No trends, but if you calculate:

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K =

PSO3

P°

⎛

⎝ ⎜

⎞

⎠ ⎟2

PSO2

P°

⎛

⎝ ⎜

⎞

⎠ ⎟2

PO2

P°

⎛

⎝ ⎜

⎞

⎠ ⎟

or K =PSO3( )

2

PSO2( )2

PO2( )

You get the same value, regardless of initial concentration

Note: K is unitless!

The Equilibrium Constant

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K = Product Concentration( )

n

Reactant Concentration( )n

K is the equilibrium constant for the reaction

The Equilibrium Constant

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K = Product Concentration( )

n

Reactant Concentration( )n

At equilibrium, the composition of the reaction mixture can be expressed in

terms of an equilibrium constant where:

For ideal gases, the concentrations are the partial pressures of the individual gases

For solutions, the concentrations are the molar values of the individual atoms/ions/molecules

Examples of K setup

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

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K =PC( )

cPD( )

d

Pa( )a

PB( )b

Units and Equilibrium Constants

When working with equilibrium Constants, we’ll use the following unit conventions:

1. Gases: Units are bar

2. Aqueous Solutions: Unit is Molarity

3. Solids: The number 1

Solids have a single value (1) because the concentration of a solid doesn’t change.

Thermodynamic Origin of Equilibrium Constants

The Free Energy changes as the composition of the reaction mixture changes

• All reactions will proceed towards equilibrium (by either forward or reverse reaction)

Gº is the free energy difference b/w the pure products and pure reactants

Thermodynamic Origins of Equilibrium Constants

• We can calculate the Free Energy change at any point along the reaction coordinate with the equation

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

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G r = ΔG r° + RTln

[C]c[D]d

[A]a[B]b

⎛

⎝ ⎜

⎞

⎠ ⎟

or

ΔG r = ΔG r° + RTlnQ

Gr° is the textbook Free Energy of reaction

Gr is the Free Energy of value when the reactants and products are at particular concentrations

Example:

The standard free energy of reaction for:

2SO2 (g) + O2 (g) 2SO3 (g)

Is Gr°= -141.74 kJ/mole at 25°C. What is the Gibbs Free Energy of reaction when the partial pressure of each gas is 100.0 bar?

Example:

The Standard Gibbs Free Energy of Reaction for

N2O4 (g) --> 2NO2 (g)

Is Gr° = +4.73 kJ/mole at 298K. What is the value of Gr when the partial pressures are PN2O4 = 0.8 bar and PNO2 = 2.10 bar?

Free Energy of a Reaction at Equilibrium

• Q=K at equilibrium• At equilibrium, G=___• Therefore,

G = Grº + RTlnK

Grº = -RTlnK (only at equilibrium)

• We can use this to compute equilibrium constants from Grº values

K and the Extent of Reactions

• When K is very large, the reaction favors the products

• When K is very small, the reaction favors the reactants

• When K=1, the reaction is neither reactant nor product favored (Equilibrium)

The Direction of ReactionHow can we tell if a reaction will

continue towards the products or back towards the reactants at a given point along the reaction coordinate?

When Q<K, G is negative (product favored)

When Q=K, G = 0

When Q>K, G is positive (reactant favored)

Q = Reaction quotient used at any point in the coordinate

K = Equilibrium constant

Equilibrium Calculations

Toolbox 9.1: Know it. Love it. Use it.

Example:

Under certain conditions, nitrogen and oxygen react to form dinitrogen oxide, N2O. Suppose that 0.482 moles of N2 and 0.933 moles of O2 are transferred to a reaction vessel of volume 10.0L and allowed to form N2O @ 800K. At this temperature, K=3.2x10-28 for the reaction:

2N2 (g) + O2 (g) 2N2O (g)

What are the partial pressures of the gases at equilibrium?

Example:

Chlorine and fluorine react at 2500K to produce ClF and reach the equilibrium:

Cl2 + F2 2ClF

With an equilibrium constant value of 20. If a gaseous mixture of 0.2 bar Cl2, 0.1 bar F2 and 0.1 bar ClF is allowed to reach equilibrium, what is the partial pressure of ClF in the mixture?

LeChatelier’s Principle

When the equilibrium composition is perturbed by adding or removing a reactant of product, the reaction tends to proceed in the direction that brings Q closer to that of K.

Consider the Equilibrium Reaction:

4NH3 (g) + 3O2 (g) 2N2 (g) + 6H2O (g)

What would result from the:

a) Addition of N2

b) Removal of NH3

c) Removal of H2O

Effects of the Environment on Equilibria

• Compressing a Gas Phase Reaction– The reaction shifts so as to decrease the pressure

• Decrease the number of gas molecules

• Changing the Temperature of a Reaction– For exothermic reactions, lowering the

temperature causes a shift towards the products– For endothermic reactions, increasing the

temperature causes a shift towards the products