Reversible Reactions A + B <=>C + D In a reversible reaction as soon as some of

the products are formed they react together, in the reverse reaction, to form the reactant particles.

Example As soon as A + B react forming C + D,

some C+ D react together to produce A + B.

Equilibrium In a reversible reaction the forward and

backward reactions occur at the same time. Therefore the reaction mixture will contain

some reactant and product particles. When the rate of the forward reaction is

equal to the rate of the reverse reaction – we say they are at EQUILLIBRIUM.

Dynamic Equilibrium is when the conditions are balanced and the reaction appears to have stopped.

Factors We can alter the position of equilibrium

by changing: The concentration of reactants or

products. Changing the temperature. Changing the pressure ( in gas

mixtures only)

Le Chatelier’s Principle If a system is at dynamic equilibrium

and is subjected to a change- the system will offset itself to the imposed change.

This is only true when a reversible reaction has reached equilibrium.

Catalysts Catalysts will lower the activation

energy of the forward and reverse reaction by the same rate.

A catalyst increase the rate of the reaction but has no effect on equilibrium position.

Concentration A+B <=> C + D If we add more A or B we speed up the forward

reaction and so more C and D are produced. Equilibrium shifts to RHS

If reduce the amount of C and D – then more A and B will react producing more C and D. Equilibrium shifts to RHS

If we add add more C or D then the reverse reaction will happen – more A and B will be produced. The same will happen if remove some A or B. In both cases equilibrium shifts to LHS.

Temperature In a reversible reaction – one will be

exothermic and the other will be endothermic.

A rise in temperature favours the reaction which absorbs heat – the endothermic reaction.

A drop in temperature favours the reaction that releases heat – the exothermic reaction.

Example N2O4 (g)<=> NO2 (g) ΔH = + (clear) (brown) NO2 is formed when most metal nitrates

decompose or when you add Cu to HNO3. NO2 is a dark brown gas. The forward reaction is endothermic. If we increase the T, it favours the endothermic

reaction and so equilibrium will shift to the RHS. We will see a dark brown gas.

If we decrease the T, it favours the exothermic reaction – the reverse reaction – and so N2O4 will be produced. A colourless gas!

Pressure Changing the pressure will only affect a

gaseous mixture. An increase in P will cause equilibrium

position to shift to the side with the least amount of gaseous molecules.

2 SO2 (g) + 1O2 (g) <=> 2 SO3 (g) 3 moles of gas <=> 2 mole of gas If we increase P – the equilibrium will move to

the RHS since there are fewer gas molecules.

N2O4 (g) <=> 2 NO2 (g) (clear) (brown) I mole of gas <=> 2 moles of gas If we increase the P – equilibrium will move to

the LHS since there are fewer gas molecules. We will see the brown colour vanish.

If we decrease the P – equilibrium will shift to RHS – more gas molecules – we will see the brown NO2.

Catalysts and Equilibrium A catalyst lowers EA and so speeds up

reaction rate. In a reversible reaction it lowers the EA for

the forward and reverse reaction by the same amount.

Therefore they speed up the rate of both reactions by the same amount.

They have no effect on equilibrium position -but a system will reach equilibrium faster.

Equilibrium in Industry The Haber Process Manufacture of NH3

N2(g) +3H2(g)<=>2NH3(g) ΔH=-92kJ The forward reaction is exothermic. Therefore a low T will

move equilibrium to the RHS. ( If T is too low reaction will be slow)

Increasing P will favour equilibrium to shift to the RHS since fewer gas molecules on that side. ( 4moles – 2 moles)

Conditions actually used = 200 atmospheres (P), T = 380 – 400 o C. In continuous processor.

NH3 is condensed – un reacted N2 and H2 recycled.

Acids and Bases The pH scale is a measure of the

concentration of Hydrogen ions. The pH stands for the negative

logarithm: pH = - log10 [H+(aq)] ([ ] = concentration) The pH scale is continuous – (below 1

and above 14)

Water An equilibrium exists with water H2O (l) <=> H+(aq) + OH– (aq) The concentration of both H+ and OH-

are 10 –7 moles l-1. [H+] = [OH-]= 10 –7 mol/l [H+] [OH-] = 10 –7 x 10 –7

• = 10 – 14 mol2 l -2

Calculating concentration [H+] = 10 –14 / [OH-] [OH-] = 10 –14 / [H+] Example Calculate the concentration of OH- ions is a

solution contains 0.01 moles of H+

[OH-] = 10 –14 / [H+] = 10 –14/ 10 –2 ( 0.01 = 10 –2) = 10 –12 mol/l.

More examples Calculate the pH of a solution that

contains 0.1 moles of OH- ions. [H+] = 10 –14 / 10 –1

= 10 –13 mol/l

pH = - log10 [H+]

= - log 10 –13

= 13

pH [H+ ] [OH-]

1 1 x 10 –1 1 x 10 -13

2 1 x 10 –2 1 x 10 -12

3 1 x 10 –3 1 x 10 -11

4 1 x 10 –4 1 x 10 -10

5 1 x 10 –5 1 x 10 -9

6 1 x 10 –6 1 x 10 -8

7 1 x 10 –7 1 x 10 -7

8 1 x 10 –8 1 x 10 -6

9 1 x 10 –9 1 x 10 -5

10 1 x 10 –10 1 x 10 -4

11 1 x 10 –11 1 x 10 -3

12 1 x 10 –12 1 x 10 -2

13 1 x 10 –13 1 x 10 -1

Strong/Weak Acids A strong acid is one where all the

molecules have dissociated (changed into ions)

Example HCl(g) + (aq) —> H+ (aq) + Cl- (aq) (molecules) ( ions) Other strong acids – Sulphuric, Nitric,

phosphoric.

Weak Acids These are acids that have only

partially dissociated ( ionised) in water. Example – carboxylic acids, carbonic

acid, sulphurous acid. The majority of the particles lie at the

molecule side of the equilibrium. CH3COOH (aq) <=> CH3COO- (aq)

(molecules) + H+ (aq) ( ions)

Strong and weak acids differ in: Conductivity, pH and reaction rate. If comparing we must use equimolar solutions I.e.

both same mol/1.

0.1 m HCl 0.1 mol CH3COOH

[H+] 0.1 0.0013

pH 1 2.88

Conductivity High Low

Rate with Mg Fast Slow

Rate with CaCO3 Fast Slow

Strong/Weak Bases Strong base – completely dissociated. Example NaOH(s) + (aq) <=> Na+(aq)+OH-(aq) Other examples – alkali metals. Weak bases are partially dissociated. Example NH3(aq) + H2O <=> NH4

+ (aq)+ OH-(aq)

0.1 mol NaOH (aq) 0.1 mol NH4OH (aq)

[OH-] 0.1 .0013

pH 13 11.12

Conductivity High Low

Affect on equilibrium If we add Sodium ethanoate to

Ethanoic Acid – CH3COOH(aq) <=>CH3COO-(aq) + H+(aq) NaCH3COO(s)+(aq) <=>Na+(aq)+CH3COO-

(aq) We have increased the concentration of the

ethanoate ions (in the system) – equilibrium will shift to the LHS to offset this. Therefore there will be less H+ ions and so pH will rise.

What happens to equilibrium position if we add NH4Cl to NH4OH?

NH4OH(aq) <=> NH4+ (aq) + OH- (aq)

NH4Cl (s) => NH4+ (aq) + Cl- (aq)

The number of NH4+(aq) ions is increasing

on the RHS of the system, equilibrium will shift to the LHS to offset this. The will be fewer OH- (aq) ions and so the pH will decrease.

Salts General Rule

Acid Alkali Salt pH

Strong Strong Neutral

Strong Weak Acidic

Weak Strong Alkaline

Weak Weak Neutral

Explanation! NH4Cl This is the salt of a weak alkali

( NH4OH) and a strong acid ( HCl). When we add it to water: NH4Cl(s) + (aq) <=> NH4

+(aq) + Cl-(aq) H2O (l) <=> H+ (aq) + OH-(aq) The NH4

+ ions and the OH- ions in the system react NH4

+(aq) + OH-(aq) <=> NH3 (aq) + H2O(l) The concentration of OH- ions in the water equilibrium

goes down – the equilibrium shifts to the RHS to offset this – producing more H+ ions and so pH goes down.( acidic!)

NaCH3COO This is the salt of a strong alkali

( NaOH) and a weak acid (CH3COOH). When we add it to water: NaCH3COO(s) + (aq) <=> CH3COO-(aq) + H+ (aq)

H2O (l) <=> H+ (aq) + OH-(aq)

The CH3COO(aq) reacts with the H+ (aq) ion.

CH3COO-(aq) + H+(aq) <=> CH3COOH(aq) The water equilibrium then moves to RHS to offset this – there

are now more OH-(aq) ions and so the pH will increase ( alkaline!)

Soaps Soaps are formed when we hydrolyse fats and oils

using an alkali. They are the salts of weak acids and strong bases

– ph of soaps will be slightly alkaline. CH2 – OCO R CH2 –OH R – COO-Na+

I I CH - OCO R* <=> CH – OH + R* - COO – Na+

I I

CH2 -OCO R** CH2 – OH R** - COO – Na+

Fat/Oil Glycerol Sodium saltsSoaps