Reversible Reactions

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# Reversible Reactions - PowerPoint PPT Presentation

Reversible Reactions. A + B &lt;=&gt; 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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Presentation Transcript
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

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.
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)
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.
• 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
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 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!)
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