Transcript Document

Reversible Reactions
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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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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)
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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.
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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
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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
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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
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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.
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If we decrease the T, it favours the
exothermic reaction – the reverse
reaction – and so N2O4 will be
produced. A colourless gas!
Pressure
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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.
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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
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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
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The Haber Process
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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.
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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 = - log10 [H+(aq)]
([ ] = concentration)
 The pH scale is continuous – (below 1
and above 14)
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Water
An equilibrium exists with water
 H2O (l) <=> H+(aq) + OH– (aq)
 The concentration of both H+ and OHare 10 –7 moles l-1.
 [H+] = [OH-]= 10 –7 mol/l
 [H+] [OH-] = 10 –7 x 10 –7
• = 10 – 14 mol2 l -2
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Calculating
concentration
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[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
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pH
[H+ ]
[OH-]
1
2
3
4
5
6
7
8
9
10
11
12
1 x 10 –1
1 x 10 –2
1 x 10 –3
1 x 10 –4
1 x 10 –5
1 x 10 –6
1 x 10 –7
1 x 10 –8
1 x 10 –9
1 x 10 –10
1 x 10 –11
1 x 10 –12
1 x 10 -13
1 x 10 -12
1 x 10 -11
1 x 10 -10
1 x 10 -9
1 x 10 -8
1 x 10 -7
1 x 10 -6
1 x 10 -5
1 x 10 -4
1 x 10 -3
1 x 10 -2
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.
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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.
 CH3COOH (aq) <=> CH3COO- (aq)
(molecules)
+ 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 HCl
0.1 mol CH3COOH
[H+]
0.1
0.0013
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pH
1
2.88
Conductivity
High
Low
Rate with Mg
Fast
Slow
Strong/Weak Bases
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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)
[OH-]
0.1 mol NaOH (aq) 0.1 mol NH4OH
(aq)
0.1
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pH
13
Conductivity High
11.12
Low
Affect on equilibrium
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If we add Sodium ethanoate to
Ethanoic Acid –
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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.
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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
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General Rule
Acid
Strong
Alkali
Strong
Salt pH
Neutral
Strong
Weak
Acidic
Weak
Strong
Alkaline
Weak
Weak
Neutral
Explanation!
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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!)
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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
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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+
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CH - OCO R* <=> CH – OH + R* - COO – Na+
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CH2 -OCO R**
CH2 – OH R** - COO – Na+
Fat/Oil
Glycerol
Sodium salts
Soaps