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 IIf a system is at dynamic equilibrium and is subjected to a change- the system will offset itself to the imposed change. T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  N 2 O 4 (g) NO 2 (g)ΔH = +  (clear)(brown)  NO 2 is formed when most metal nitrates decompose or when you add Cu to HNO 3.  NO 2 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 N 2 O 4 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 2 SO 2 (g) + 1O2 1O2 2 SO 3 (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 N2O4 (g) 2 NO 2 (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 NO 2.

Catalysts and Equilibrium  A catalyst lowers E A and so speeds up reaction rate.  In a reversible reaction it lowers the E A 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 NH 3  N 2 (g) +3H 2 (g) 2NH 3 (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.  NH 3 is condensed – un reacted N 2 and H 2 recycled.

Acids and Bases  The pH scale is a measure of the concentration of Hydrogen ions.  The pH stands for the negative logarithm:  pH = - log 10 [H + (aq)] ([ ] = concentration)  The pH scale is continuous – (below 1 and above 14)

Water  An equilibrium exists with water  H 2 O (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 mol 2 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 = - log 10 [H + ] = - log 10 –13 = 13

pH[H + ][OH -] 11 x 10 –1 1 x x 10 –2 1 x x 10 –3 1 x x 10 –4 1 x x 10 –5 1 x x 10 –6 1 x x 10 –7 1 x x 10 –8 1 x x 10 –9 1 x x 10 –10 1 x x 10 –11 1 x x 10 –12 1 x 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+ 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.  CH 3 COOH (aq) CH 3 COO - (aq) (molecules)+ H+ 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/ m HCl0.1 mol CH 3 COOH [H+] pH12.88 ConductivityHighLow Rate with MgFastSlow Rate with CaCO 3 FastSlow

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  NH 3 (aq) + H 2 O NH 4 + (aq)+ OH - (aq)

0.1 mol NaOH (aq)0.1 mol NH 4 OH (aq) [OH-] pH ConductivityHighLow

Affect on equilibrium  If we add Sodium ethanoate to Ethanoic Acid –  CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq)  NaCH 3 COO(s)+(aq) Na + (aq)+CH 3 COO - (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 NH 4 Cl to NH 4 OH?  NH 4 OH(aq) NH 4 + (aq) + OH - (aq)  NH4Cl (s) => NH 4 + (aq) + Cl - (aq)  The number of NH 4 + (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 AcidAlkaliSalt pH Strong Neutral StrongWeakAcidic WeakStrongAlkaline Weak Neutral

Explanation!  NH 4 Cl  This is the salt of a weak alkali ( NH 4 OH) and a strong acid ( HCl).  When we add it to water:  NH 4 Cl(s) + (aq) NH 4 + (aq) + Cl - (aq)  H 2 O (l) H + (aq) + OH - (aq)  The NH 4 + ions and the OH - ions in the system react  NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O(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!)

 NaCH 3 COO  This is the salt of a strong alkali ( NaOH) and a weak acid (CH 3 COOH).  When we add it to water:  NaCH 3 COO(s) + (aq) CH 3 COO - (aq) + H + (aq)  H 2 O (l) H+ H+ (aq) + OH - (aq)  The CH 3 COO(aq) reacts with the H + (aq) ion.  CH 3 COO - (aq) + H + (aq) CH 3 COOH(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. CH 2 – OCO R CH 2 –OH R – COO - Na + I I CH - OCO R* CH – OH + R* - COO – Na + I I CH 2 -OCO R** CH 2 – OH R** - COO – Na + Fat/Oil Glycerol Sodium salts Soaps