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Introduction:

In AS we covered the characteristics of an equilibrium:

Characteristics of the equilibrium state

1) Equilibrium can only be established in a closed system. Matter cannot be exchanged with the surroundings (this will affect the position of the equilibrium), but energy can be exchanged.

2) Equilibrium can be approached from either direction. The products can be used as the reactants to set up the equilibrium – reversible reactions.

3) Equilibrium is a dynamic state – At equilibrium the rate in both directions must be the same.

4) Dynamic equilibrium is stable under fixed conditions but is sensitive to changes in temperature, pH, pressure.

In AS we discovered that Le Chatelier’s Principle was given as a way of determining how changes in the conditions can affect the extent to which a reaction will go.
A more Quantitative approach is needed to understand more fully the ideas of equilibria

The equilibrium law:

The equilibrium law is an expression of the amounts of products : reactants in an equilibrium.

		K_c	=	[PRODUCTS]^p
				[REACTANTS]^r

Consider the reaction:

Where [A], [B], [C] and [D] are the equilibrium concentrations of reactants and products
The indices a,b,c and d are the stoichiometric numbers in the balanced chemical reaction

K_c	=	[C]^c [D]^d
		[A]^a [B]^b

This expression gives us a mathematical way of looking at the proportions of products and reactants in an equilibrium.
At equilibrium these relative proportions will not change.
This means a constant is produced, this is given by - K_c

Approaching equilibrium

Consider the reversible reaction:

N₂O_4(g)

2NO_2(g)

N₂O₄ is a colourless gas, NO₂ is a brown gas.

	Initially:	N₂O₄ molecules decompose into 2 NO₂ molecules, the rate of the forward reaction is fast. Because there are very few molecules of NO₂, the reverse reaction can only happen slowly.
	As the reaction proceeds:	There are now fewer N₂O₄ molecules available to decompose, the rate of the forward reaction decreases. There are now more NO₂ molecules present so the rate of the reverse reaction increases.
	At equilibrium:	Eventually the forward reaction takes place at exactly the same rate as the reverse reaction. Dynamic equilibrium is established and is denoted by D The concentrations of reactants and products now remain constant.

Graph shows how the concentrations of reactants and products change:

Writing expressions for K_c:

At equilibrium the concentrations of NO₂ and N₂O₄ are constant:

N₂O_4(g)

2NO_2(g)

K_c	=	[NO₂]²
		[N₂O₄]

A few other examples:

2SO_2(g)

O_2(g)

2SO_3(g)

K_c	=	[SO₃]²
		[SO₂]² [O₂]

H_2(g)

I_2(g)

2HI_(g)

K_c	=	[HI]²
		[H₂] [I₂]

Units of K_c

These have to be worked out for each K_c.
Just like the rate constant, put the units into each expression:

K_c

[NO₂]²

K_c

[SO₃]²

K_c

[HI]²

[N₂O₄]¹

[SO₂]² [O₂]

[H₂] [I₂]

K_c

mol dm^-3 x mol dm^-3

K_c

(mol dm^-3)²

K_c

mol dm^-3 mol dm^-3

mol dm^-3

(mol dm^-3)²

mol dm^-3

K_c

mol dm^-3 x ~~mol dm~~^-3

K_c

~~(mol dm~~^-3)²

K_c

~~mol dm^-3 mol dm^-3~~

~~mol dm~~^-3

~~(mol dm~~^-3)²

mol dm^-3

~~mol dm~~^-3

K_c

mol dm^-3

K_c

No units

mol dm^-3

K_c

dm³ mol^-1

Qu 1 P127

Calculations using K_c

The equilibrium expression obviously allows you to calculate K_cif you know the concentrations at equilibrium.
You can also calculate K_c and equilibrium concentrations knowing the starting concentrations and one of the equilibrium concentrations.

Determining K_c from equilibrium concentrations:

1) Hydrogen, Iodine, Hydrogen iodide equilibrium:

Reaction:	H_2(g)	+	I_2(g)	D	2HI_(g)
Equilibrium concentrations:	0.140		0.040		0.320

Write the equilibrium expression, put in the values, calculate K_c and work out the units:

K_c	=	[HI]²	K_c	=	[HI]²
		[H₂] [I₂]			[H₂] [I₂]

K_c	=	(0.320)²	K_c	=	mol dm^-3 mol dm^-3
		0.140 x 0.040			mol dm^-3	mol dm^-3

K_c	=	18.3	K_c	=	~~mol dm^-3 mol dm^-3~~
					~~mol dm~~^-3	~~mol dm~~^-3

			K_c	=	No units

2) N₂O₄ / NO₂ equilibrium: Given the number of moles at equilibrium in a volume of 2dm³.

Reaction:	N₂O_4(g)	D	2NO_2(g)
Mole quantities at equilibrium:	0.400		3.20
Equilibrium concentrations:	0.400 / 2		3.20 / 2
Equilibrium concentrations:	0.20		1.60

As the quantities are given in moles, you have to calculate the equilibrium concentrations
Write the equilibrium expression, put in the values, calculate K_c and work out the units:

K_c	=	[NO₂]²		K_c	=	[NO₂]²
		[N₂O₄]				[N₂O₄]

K_c	=	(1.60)²		K_c	=	mol dm^-3 mol dm^-3
		0.20				mol dm^-3

K_c	=	12.8	mol dm^-3	K_c	=	~~mol dm^-3~~ mol dm^-3
						~~mol dm~~^-3

				K_c	=	mol dm^-3

Calculating the quantities and concentrations present at equilibrium:

3) Hydrogen, Iodine, Hydrogen iodide equilibrium:

This time you are told the mole quantities at the start and one equilibrium quantity.
From this you can work out all of the mole quantities at equilibrium.
The final step is to convert to equilibrium concentrations, in this example the reaction is carried out in a 1 dm³ vessel.

Reaction:	H_2(g)	+	I_2(g)	D	2HI_(g)
At start:	0.60		0.40		0.0
At equilibrium:	0.28		0.08		0.64
Reacted:	0.32		0.32		0.64
Equilibrium concentrations:	0.28		0.08		0.64

Write the equilibrium expression, put in the values, calculate K_c and work out the units:

K_c	=	[HI]²	K_c	=	[HI]²
		[H₂] [I₂]			[H₂] [I₂]

K_c	=	(0.64)²	K_c	=	mol dm^-3 mol dm^-3
		0.28 x 0.08			mol dm^-3	mol dm^-3

K_c	=	18.3	K_c	=	~~mol dm^-3 mol dm^-3~~
					~~mol dm~~^-3	~~mol dm~~^-3

			K_c	=	No units

Qu 1 - 2 P129

The equilibrium position and K_c

What is the significance of a K_c value?

K_c is a mathematical representation of the ratio of products : reactants.

If the amount of products is equal to reactants then K_c = 1

Products favoured: K_c > 1

Reactants favoured: K_c < 1

How do changes in temperature affect K_c?

In AS we used Le Chatelier's Principle to predict the how temperature affects the position of equilibrium.
We can use this to predict what happens to K_c.
Apply the direction movement to the formula to work out whether K_c increases or decreases:

1) Endothermic reactions:

Applying LCP means that an increase in temperature will shift the equilibrium to the Products.
This means there will be more Products and less Reactants:

The top number increases and the bottom number decreases.
This makes K_c a larger value.

K_c increases

and vice versa

2) Exothermic reactions:

Applying LCP means that an increase in temperature will shift the equilibrium to the Reactants.
This means there will be more Reactants and less Products:

The top number decreases and the bottom number increases.
This makes K_c a lsmaller value.

K_c decreases

and vice versa

Qu 1 P131

The equilibrium constant, K_c, and the rate constant, k

How does a change in concentration and pressure affect K_c?

In AS we used Le Chatelier's Principle to predict the how concentration and pressure affects the position of equilibrium.
The reasons for these changes in the position of the equilibrium lie with K_c.

K_c is unaltered by changes in concentration and pressure

1) Changes in concentration:

A change in Concentration has no effect on the equilibrium constant.

Remember if a reactant / product is added the equilibrium shifts to the opposite direction to keep the 'proportions' the same - K_c is unchanged:

Reaction:

N₂O_4(g)

2NO_2(g)

[NO₂] = 1.60 Mol dm^-3					[NO₂] = 1.60 Mol dm^-3
[N₂O₄] = 0.20 Mol dm^-3				Double [ ] :	[N₂O₄] = 0.40 Mol dm^-3

	K_c	=	[NO₂]²		K_c	=	[NO₂]²
			[N₂O₄]				[N₂O₄]

	K_c	=	(1.60)²		K_c	=	(3.20)²
			0.20				0.40

	K_c	=	12.8	Mol dm^-3	K_c	=	6.4	Mol dm^-3

If the concentrations are doubled, the system is no longer at equilibrium, K_c = 12.8
To bring K_c back from 6.4 to 12.8:
The system must increase [NO₂] and decrease [N₂O₄]
Remember if a reactant / product is added the equilibrium shifts to the opposite direction to keep the 'proportions' the same - K_c is unchanged:

2) Changes in Pressure:

A change in Pressure has no effect on the equilibrium constant.

If pressure is doubled - the volume is halved - meaning that the concentrations will have doubled

Reaction:

N₂O_4(g)

2NO_2(g)

[NO₂] = 1.60 Mol dm^-3				Double [ ] :	[NO₂] = 3.20 Mol dm^-3
[N₂O₄] = 0.20 Mol dm^-3					[N₂O₄] = 0.40 Mol dm^-3

	K_c	=	[NO₂]²		K_c	=	[NO₂]²
			[N₂O₄]				[N₂O₄]

	K_c	=	(1.60)²		K_c	=	(3.20)²
			0.20				0.40

	K_c	=	12.8	Mol dm^-3	K_c	=	25.6	Mol dm^-3

If the pressure is doubled, the system is no longer at equilibrium, K_c = 12.8
To bring K_c back from 25.6 to 12.8:
The system must increase [NO₂] and decrease [N₂O₄]
Remember if a reactant / product is added the equilibrium shifts to the opposite direction to keep the 'proportions' the same - K_c is unchanged:

How does the presence of a catalyst affect K_c?

A catalyst has no effect on the equilibrium constant.

A catalyst speeds up both the forward and reverse reaction.
Equilibrium is achieved more quickly.

The equilibrium constant K_c, and the rate constant, k

These 2 constants tell us the most important things in the chemical industry:

a)	Equilibrium	How far
b)	Rates	How fast

a) The equilibrium constant, K_c

K_c indicates the position of the equilibrium:

	Large K_c	Products predominate
	Small K_c	Reactants predominate

Remember LCP:

	Endothermic	K_c increases with an increase in temperature (increases products)
	Exothermic	K_c decreases with an increase in temperature (decreases products)

Kc can be written from the balanced chemical equation

b) The rate constant, k

k is a measure of the rate of a reaction:

	Large k	Fast rate
	Small k	Slow rate

Remember:

	k increases with an increase in temperature - Rate increases with an increase in temperature
	k decreases with a decrease in temperature - Rate decreases with a decrease in temperature

k can only be determined experimentally from the rate equation

The importance of compromise:

The 2 desirable outcomes are a) increasing rate and b) increasing the amount:

a) Increasing the rate:

Increasing temperature: Increases the rate of production of product - desirable

Increasing temperature: Decreases the amount of product made - undesirable

b) Increasing the amount:

Decreasing the temperature: Increases the amount of product made - desirable

Decreasing temperature: Decreases the rate of production of product - undesirable

The compromise is:

Moderate temperature:	k increases moderately - Rate increases by a moderate amount
Moderate temperature:	K_c decreases by a moderate amount - allowing a moderate yield

Qu 1 P133 / Qu 4 - 5 P159 / Qu 3 - 4 P161