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Chemical

Equilibrium

(2)

Chemical Equilibrium

Chemical equilibrium occurs in

chemical reactions that are

reversible. In a reaction such as:

CH

4

(g) + H

2

O(g)

CO(g) + 3H

2

(g)

The reaction can proceed in both

directions

(3)

An Equilibrium System

CH

4

(g) + H

2

O(g)

CO(g) + 3H

2

(g)

After some of the products are created the products begin to react to form the reactants

At equilibrium there is no net change in the concentrations of the reactants and products

The concentrations do not change but

(4)

Chemical Equilibrium

CH

4(g) + H2O(g) <=> CO(g) + 3H2 (g)

At the beginning of the reaction, the

rate that the forward direction is

higher

As the reactants decrease the rate in

the forward direction slows

As the products form, the rate in the

reverse direction increases

When the two rates are the same

(5)

Dynamic Equilibrium

An equilibrium is Dynamic

CH

4

(g) + H

2

O(g)

CO(g) + 3H

2

(g)

The amount of products and the

reactants are constant.

(6)

The Equilibrium Constant

aA + bB

cC + dD

The upper case letters are the molar

concentrations of the reactants and

products. The lower case letters

(7)

The Equilibrium Constant

(8)

Equilibrium Constant

Calculations

Example

N

2

(g) + 3H

2

(g)

2NH

3

(g)

At equilibrium, a one-liter container has

1.60 moles NH

3

, .800 moles N

2

, and

1.20 moles of H

2

. What is the

(9)

Equilibrium Constant

Calculations

At equilibrium, a one-liter container has

1.60 moles NH

3

, .800 moles N

2

, and 1.20

moles of H

2

. What is the equilibrium

(10)

Reaction quotient

The equilibrium constant is a constant

ratio only when the system is in

equilibrium.

If the system it not at equilibrium the

ratio is known as a Reaction Quotient

If the reaction quotient is equal to the

(11)

Using Equilibrium Constants

for other calculations

If a solution is not at equilibrium the ratio of the right side over the left is called a reaction

quotient

.

(12)

Equilibrium Constants and

calculations

If Q > Keq , the product side is too high and the equilibrium will shift to the left to restore

equilibrium

(13)

Equilibrium Calculations –

Using I. C. E. Models

Equilibrium constants and

concentrations can often be

deduced by carefully examining

data about initial and equilibrium

concentrations

(14)

Equilibrium Calculations

ICE Model problem 1

Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction:

H2 + I2 2HI

Suppose that 1.5 mole of H2 and 1.2 mole of I2 are placed in a 1.0 dm3 container. At equilibrium it was

(15)

Equilibrium Calculations

ICE Model Problem 1 Solution

Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction:

H2 + I2 2HI

Suppose that 1.5 mole of H2 and 1.2 mole of I2 are placed in a 1.0 dm3 container. At equilibrium it was found that there were 0.4

mole of HI. Calculate the equilibrium concentrations of [H2] and [I2] and the equilibrium constant.

I C E Since 2x = 0.4, x = 0.2

[H2 ] 1.5 -x 1.5- x [H2 ] = 1.5 – 0.2 = 1.3

[ I2 ] 1.2 -x 1.2 –x [I2 ] = 1.2 – 0.2 = 1.0

[HI ] 0 +2x 0.4

Keq = [HI]2 . = (0.4)2 = 0.123

(16)

Equilibrium Calculations

ICE Model Problem 2

Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction:

2 SO2 + O2 2SO3

Suppose that 1.4 mole of SO2 and 0.8 mole of SO3 are placed in a 1.0 dm3 container. At equilibrium it was

found that there were 0.6 dm3 of SO

3. Calculate the

(17)

Equilibrium Calculations

ICE Model Problem 2 Solution

Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction:

2 SO2 + O2 2SO3

Suppose that 1.4 mole of SO2 and 0.8 mole of SO3 are placed in a 1.0 dm3 container. At equilibrium it was found that there were 0.6

dm3 of SO

3. Calculate the equilibrium concentrations of [SO2]

and [O2] and the equilibrium constant.

I C E Since 2x = 0.6, x = 0.3

[SO2 ] 1.4 -2x 1.4-2x [SO2 ] =1.4 – 2( 0.3) = 0.8

[O2 ] 0.8 -x 0.8 –x [O2 ] = 0.8 – 0.3 = 0.5

[SO3] 0 +2x 0.6

Keq = [SO3]2 . = (0.6)2 = 0.281

[SO2 ]2 [O

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Le Chatelier’s Principle

Le Chatelier's Principle

states: When a

system in chemical equilibrium is disturbed

by a change of temperature, pressure, or a

concentration, the system shifts in

equilibrium composition in a way that tends

to counteract this change of variable.

A change imposed on an equilibrium

system is called a

stress

The equilibrium always responds in such a

(19)

Le Chatelier’s Principle

Types of stresses

Change in concentration of one or

more reactants or products

Change in temperature

Change in pressure

(20)

Effect of a Change in

Temperature

An increase in the temperature causes the

equilibrium to shift in the direction of the

endothermic reaction

N

2

(g) + 3 H

2

(g)

2NH

3

(g)

H =-92 kJ mol-1

Since

H is negative the endothermic

reaction is the reverse direction. An

increase in temperature causes the reaction

to shift to the left, resulting in an increase in

(21)

Effect of a Change in

Pressure

Pressure affects only gases in an equilibrium

PV = nRT

An increase in pressure causes the equilibrium to

shift in the direction that has the fewer number of moles

N

2

(g) + 3 H

2

(g)

2NH

3

(g)

H =-92 kJ mol-1

An increase in pressure results in a an decrease in

(22)

Effect of a Change in one of

the reactants or products

The equilibrium responds in such a way so

as to diminish the increase

Substances on the same side of the arrow

respond in opposite directions.

Substances on the opposite side of the

arrow move in the same direction

N

2

(g) + 3 H

2

(g)

2NH

3

(g)

An increase in [N

2

] results in a decrease in

(23)

Effect of a Catalyst

Catalysts affect both the forward and

reverse directions equally

A catalyst does not change the

References

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