THE POSITION OF EQUILIBRIUM

•Section 7.2

THE EQUILIBRIUM CONSTANT

The position of equilibrium for a particular reversible reaction can be defined by a constant

This constant has a numerical value found by relating the equilibrium concentrations of the products to those of the reactants

Denoted by Kc the equilibrium constant

DEDUCING THE EQUILIBRIUM CONSTANT

Consider this standard reaction: aA + bB cC + dD Where a, b, c and d represent

the stoichiometry of each substance, then:

c d

a bc

C D

A BK

CONTINUED

The concentrations must be those that occur at equilibrium, not the starting values

Kc is constant for a given temperature

Change the temperature, you change Kc

EQUILIBRIUM LAW

Equilibrium law: in any reversible reaction at a state of equilibrium, the rate of the forward reaction equals the rate of the reverse reaction

•

PROBLEM #1

Consider the following reaction: N2(g) + 3H2(g) 2NH3(g)

How is the Kc calculated?

2

33

2 2

c

NH

N HK

PROBLEM #1 CONTINUED

An experiment has this data at equilibrium at 500°C:

[N2] = 0.922 mol dm-3

[H2] = 0.763 mol dm-3

[NH3] = 0.157 mol dm-3

Calculate the Kc for this experiment

WORKING OUT

0.0602 A note about units: the

concentrations calculated would give you varying units, Kc is unitless because you are using the “activity” which is close to concentration

2

3

0 .157

0 .922 0 .763cK

ANOTHER CONSIDERATION

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

Kc for the forward reaction:

For the reverse reaction:

2

2 2c

HI

H IK

2 2

2c

H I

HIK

CONTINUED

The value of the equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction

HOW FAR WILL A REACTION GO? Consider the following reactions and Kc

H2(g) + I2(g) 2HI(g) Kc = 2 at 277°C

H2(g) + Cl2(g) 2HCl(g) Kc = 1018 at 277°C

The large difference illustrates how stable HCl is at this temperature compared to HI

Virtually all of the reaction between H2 and Cl2 has produced HCl

CONTINUED

The magnitude of Kc is a good indicator of how far a reaction goes toward completion (under given conditions)

If Kc>>1, then the reaction has gone virtually to completion

If Kc<<1, then the reaction has hardly taken place at all

CONTINUED

For Kc = 1, there are equal amounts of reactants and products

The Kc for a given reversible reaction at equilibrium only changes if the temperature changes

OTHER CONSIDERATIONS

The concentrations of certain substances remain constant, so they are omitted from the equilibrium constant expression

All solids have a fixed density, so these are omitted

The concentration of pure liquid water is also constant (important to consider when solutions are aqueous, water omitted)

CONTINUED

If water is used as a liquid with other liquids ( not aqueous solutions), then it is included

Water must be included if it is in the gas phase

LE CHÂTELIER'S PRINCIPLE

This principle is used to predict the qualitative effects of changes on the position of equilibrium

When a system at equilibrium is disturbed, the equilibrium position will shift in the direction which tends to minimize, or counteract, the effect of the disturbance

MORE Possible changes in conditions that

we need to consider are: Changes in the concentrations of

either the reactants or the products Changes in pressure for gas phase

reactions Changes in temperature The presence of a catalyst

CHANGES IN CONCENTRATION

Increasing the concentration of a reactant will move the position of equilibrium to the right

This favors the forward reaction and will increase the equilibrium concentrations of the products

MORE The addition of more product to an

equilibrium mixture would shift the position of equilibrium to the left

This would favor the reverse reaction When new equilibrium

concentrations are substituted into the equilibrium expression, the value of Kc remains unchanged

EXAMPLE

Fe(H2O)63+

(aq) + SCN-(aq) [Fe(H2O)5SCN]2+ + H2O(l)

yellow-brown colorless blood-red

If the concentration of either the thiocyanate ion, or the iron (III) ion is increased, then the intensity of the color increases

The shift in equilibrium to the right causes the concentration of the added reactant to fall again

CONTINUED If the concentration of iron(III) ions is decreased by

adding fluoride ions, which form a stable complex with the iron (III) ions, then the intensity of the coloration decreases

The shift of the equilibrium to the left produces more aqueous iron (III) ions to counteract the reduction caused by the F-

CHANGES IN PRESSURE

Changing the pressure only affects reactions that involve gases

Gas phase reactions will only be affected by a change of pressure if the reaction involves a change in the number of moles on the two sides of the equation

INCREASE IN PRESSURE

Increase the pressure and the equilibrium shifts to the side with the least moles of gas

2SO2(g) + O2(g) 2SO3(g) Increase P: equilibrium → , to side with

fewer moles of gas

C(s) + H2O(g) CO(g) + H2(g)

Increase P: equilibrium ←, to side with fewer moles of gas (

DECREASE IN PRESSURE

Decrease pressure and the equilibrium shifts towards the side with the most moles of gas

2SO2(g) + O2(g) 2SO3(g) Decrease P: equilibrium ←, to side with

more moles of gas

C(s) + H2O(g) CO(g) + H2(g) Decrease P: equilibrium →, to side with

more moles of gas

MORE ABOUT PRESSURE

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

Changing pressure has no effect as 2 moles of gas are converted to 2 moles of gas

The changes in concentrations that result from the changes in pressure are such that the value of Kc remains unchanged

TEMPERATURE CHANGE

If the temperature of a system is increased, then the equilibrium shifts in the direction of the endothermic change, this will tend to lower the temperature

N2(g) + O2(g) 2NO(g) ∆H = +180 kJ mol-1

Increased T: Kc increases, equilibrium →

Decreased T: Kc decreases, equilibrium ←

TEMPERATURE CHANGE

If the temperature of a system is decreased the equilibrium shifts in the direction of the exothermic change, this is to generate heat and raise the temperature

2SO2(g) + O2(g) 2SO3(g) ∆H = -197 kJ mol-1

Decreased T, Kc increases, equilibrium →

Increased T, Kc decreases, equilibrium ←

MORE ABOUT TEMPERATURE

Changes in temperature affect the rate constants of the forward and reverse reactions to different extents

This means that the actual value of Kc changes also

PRESENCE OF A CATALYST

The presence of a catalyst reduces the activation energy of both the forward and reverse reactions by the same amount

Both reactions are sped up by the same factor

Equilibrium is established more rapidly, but the position of equilibrium and the value of Kc are not affected

SUMMARY

INDUSTRIAL PROCESSES

Many industrial processes involve equilibria

The goal is to produce the desired product efficiently and rapidly, but with the minimum amount of waste and the minimum input of energy

Kinetics and equilibria have to be considered

HABER PROCESS

Produces ammonia from nitrogen and hydrogen gases

A heated iron catalyst is used N2(g)+ 3H2(g) 2NH3(g) ∆H = -92 kJ mol-

1 4 moles of gas are converted to 2

moles of gas A high pressure will favor the product

CONTINUED The forward reaction is exothermic,

so a low temperature would favor the products

Issues with the process: Low temperature causes a low

reaction rate High pressure is expensive to

provide A compromise is chosen to maximize

product

COMPROMISE Typical conditions chosen for the Haber

process are pressures in the range of 200-1000 atm and temperatures about 700K

COLLECTING AMMONIA

The reaction is not allowed to reach equilibrium because the reaction rate slows as equilibrium is approached

Usually only about 20% of the reactants are converted to ammonia

The ammonia is condensed and the reactants are recycled to form more ammonia

USES FOR AMMONIA

The manufacture of fertilizers The manufacture of nitrogen-

containing polymers such as nylon Can be oxidized to produce nitric acid Nitric acid is used to produce

explosives such as TNT and dynamite Nitric acid is also used in making

dyes

CONTACT PROCESS

The Contact process is the production of sulfuric acid by the oxidation of sulfur

First, pure sulfur is burned in air S(s) + O2(g) SO2(g)

Then sulfur dioxide is mixed with air and passed over a vanadium (V) oxide catalyst

2SO2(g) + O2(g) 2SO3(g) ∆H = -196 kJ mol-1

CONTINUED

High pressure would favor the formation of sulfur trioxide, but low pressure actually does a good job

Low temperature favors the product, but not too low (slows the reaction)

Compromise: 700-800K, the use of a catalyst (finely divided V2O5) and about 2 atm

CONTINUED

The result is over 90% conversion to sulfur trioxide

The sulfur trioxide must be reacted with water to produce sulfuric acid

SO3(g) + H2O(l) → H2SO4(l)

USES OF SULFURIC ACID

Manufacture of: Fertilizers Polymers Detergents Paints and pigments

Used in the petrochemical industry and in the processing of metals

ANOTHER USE OF SULFURIC ACID

Used as an electrolyte in automobile batteries (battery acid)