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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.
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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.
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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)
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Le Chatelier’s Principle IIf a system is at dynamic equilibrium and is subjected to a change- the system will offset itself to the imposed change. TThis is only true when a reversible reaction has reached equilibrium.
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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.
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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.
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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.
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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.
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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!
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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.
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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.
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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.
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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.
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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)
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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
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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.
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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
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pH[H + ][OH -] 11 x 10 –1 1 x 10 -13 21 x 10 –2 1 x 10 -12 31 x 10 –3 1 x 10 -11 41 x 10 –4 1 x 10 -10 51 x 10 –5 1 x 10 -9 61 x 10 –6 1 x 10 -8 71 x 10 –7 1 x 10 -7 81 x 10 –8 1 x 10 -6 91 x 10 –9 1 x 10 -5 101 x 10 –10 1 x 10 -4 111 x 10 –11 1 x 10 -3 121 x 10 –12 1 x 10 -2 131 x 10 –13 1 x 10 -1
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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.
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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)
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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 HCl0.1 mol CH 3 COOH [H+]0.10.0013 pH12.88 ConductivityHighLow Rate with MgFastSlow Rate with CaCO 3 FastSlow
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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)
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0.1 mol NaOH (aq)0.1 mol NH 4 OH (aq) [OH-]0.1.0013 pH1311.12 ConductivityHighLow
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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.
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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.
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Salts General Rule AcidAlkaliSalt pH Strong Neutral StrongWeakAcidic WeakStrongAlkaline Weak Neutral
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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!)
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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!)
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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
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