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Reversible Reactions

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.

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Reversible Reactions

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  1. 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.

  2. 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.

  3. 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)

  4. 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.

  5. 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.

  6. 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.

  7. 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.

  8. 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.

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

  10. 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.

  11. 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.

  12. 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.

  13. 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.

  14. 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)

  15. 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

  16. 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.

  17. 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

  18. 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.

  19. 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)

  20. Strong and weak acids differ in: • Conductivity, pH and reaction rate. • If comparing we must use equimolar solutions I.e. both same mol/1.

  21. 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)

  22. 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.

  23. 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.

  24. Salts • General Rule

  25. 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!)

  26. NaCH3COO • This is the salt of astrong 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!)

  27. 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 salts Soaps

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