Reversible Reactions and Chemical Equilibrium
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Transcript of Reversible Reactions and Chemical Equilibrium
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Reversible Reactions and
Chemical Equilibrium
University of Lincoln presentation
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Outline
• Reversible reactions• Chemical Equilibrium• Le Chatelier’s Principle• Equilibrium constants
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Reversible Reactions
BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)
CH3CO2H + CH3CH2OH ↔ CH3CO2CH2CH3 + H2O
Cr2O72-(aq) + 2OH-(aq) ↔ 2CrO4
2-(aq) + H2O(l)
CH3CO2H(aq) + H2O(l) ↔ CH3CO2-(aq) + H3O+
(aq)
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Chemical Equilibrium
• Reactions not 100% complete– Products and Reactants exist
together• A dynamic equilibrium • Position of equilibrium ???
• Can the position of equilibrium be changed?
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Le Chatelier’s Principle
When an external change is made to a system in equilibrium, the system will respond to oppose the change
External Changes• Concentration • Pressure (gases)• Temperature
Link to external videoLink to external video
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Concentration
1. BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)
2. Cr2O72-(aq) + 2OH-(aq) ↔ 2CrO4
2-(aq) + H2O(l)
How does reaction 1 respond to addition of hydrochloric acid?
How does reaction 2 respond to addition of alkali?How does reaction 2 respond to addition of acid?
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Pressure
N2(g) + 3H2(g) ↔ 2NH3(g)
CO(g) + 2H2(g) ↔ CH3OH(g)
2NO2(g) ↔ 2NO(g) + O2(g)
PCl5(g) ↔ PCl3(g) + Cl2(g)
H2(g) + I2(g) ↔ 2HI(g)
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
How do the above equilibria respond to:An increase in pressureA decrease in pressure
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Temperature
N2(g) + 3H2(g) ↔ 2NH3(g) rH = -92.2 kJ mol-1
H2(g) + I2(g) ↔ 2HI(g) rH = -9.4 kJ mol-1
CO(g) + H2O(g) ↔ CO2(g) + H2(g) rH = -41.2 kJ mol-1
PCl5(g) ↔ PCl3(g) + Cl2(g) rH = 87.9 kJ mol-1
How do the above respond to anIncrease in temperatureDecrease in temperature
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Equilibrium constantsa measure of equilibrium
positionaA + bB ↔cC + dD
ba
dc
c [B][A][D][C]
K
BiCl3(aq) + H2O(l) ↔ BiOCl(s) + 2HCl(aq)
]O][H[BiCl
]][HCl[BiOClK
(l)2(s)3
2(aq)(s)
c
Write the expressions for Kc for the reactions given in previous slides
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Calculating Equilibrium Constants
HNO2(aq) ↔ H+(aq) + NO2-(aq)
Solution
[HNO2(aq)]
mol litre-1
[H+(aq)]
mol litre-1
[NO2-(aq)]
mol litre-1
A 0.090 6.2 x 10-3 6.2 x 10-3
B 0.20 9.3 x 10-3 9.3 x 10-3
C 0.30 11.4 x 10-
3
11.4 x 10-3
The table shows the equilibrium molar concentrations for three solutions of nitrous acid in water at 25 oC
Calculate the equilibrium constant for this reaction at 25oC
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433
2
2c 104.3
0.090106.2106.2
(aq)][HNO(aq)](aq)][NO[H
K
Units of Kc
1
11
c litremol)litre)(mollitre(mol
K
14c litremol104.3K
Solution A
Now try for solutions B and C
1litremol
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Acids and Bases
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Outline
• Definitions• Weak Acids• Dissociation Constants• Weak Bases• Drugs• pH• Buffers
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Acids and Bases• Several definitions available - most common is
Bronsted and Lowry• Acid is a proton donor
– HCl is able to transfer H+
• Base is a proton acceptor
– NH3 is able to accept H+ and become NH4+
• Aqueous solutions
• Proton species is H3O+ (hydroxonium ion)
– HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)
– HCl(aq) H+(aq) + Cl-(aq)
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Strong Acids• Strong acids are fully dissociated
HCl (aq) + H2O(l) H3O+ (aq) + Cl- (aq)
• all dissolved HCl molecules are ionised
• 1 mol dm-3 HCl(aq) there are:
– Approx 1 mol dm-3 H3O+ (aq)
– Approx 1 mol dm-3 Cl- (aq)
DO NOT confuse ‘strong’ and ‘concentrated’1 x 10-4 mol dm-3 HCl (aq) is a dilute solution
of a strong acid
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Other strong acids
• HNO3 (nitric)
• H2SO4 (sulfuric)
• HClO4 (perchloric)
Write equations showing the dissociation of the above acids
Which are monoprotic?Are any diprotic?Chemical equilibrium – K very large
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Weak Acids
• Acids that dissociate in a reversible reaction (e.g. CH3COOH; ethanoic (acetic) acid)
CH3COOH (aq) + H2O(l) ↔ H3O+ (aq) + CH3COO-
(aq)
• Solution of CH3COOH (aq) contains:
– CH3COOH (aq)
– H3O+ (aq)
– CH3COO- (aq)
• CH3COOH is partially dissociated
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How weak is a weak acid?
0.1 mol dm-3 HCl is dissociated 91.4% [H3O+] = 0.091 mol dm-3 pH=1.04
0.1 mol dm-3 CH3COOH is dissociated 1.34%
[H3O+] = 0.0013 mol dm-3 pH=2.87
• Extent given by K
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Weak Acids
HA(aq) + H2O(l) ↔ H3O+(aq) + A-
(aq)
HA Bronsted acidH2O Bronsted base
H3O+ Bronsted acid
A- Bronsted base
O][HA][H]][AO[H
K2
3
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Acid dissociation constant (Ka)
• The higher the Ka value:
– greater degree of ionisation– stronger the acid– Data tables
O][HA][H]][AO[H
K2
3
[HA]]][AO[H
O]K[H 32
[HA]]][AO[H
KO]K[H 3a2
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Ka Values
• HCO2H 1.8 x 10-4 mol dm-3
• CH3CO2H 1.7 x 10-5 mol dm-3
• Are these weak or strong acids?• Which is the stronger acid?
aa logKpK
HCO2H 3.75CH3CO2H 4.77
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pKa values (data tables)
Acid pKa Conjugate base
H3PO4 2.12 H2PO4-
HNO2 3.34 NO2-
H2CO3 6.37 HCO3-
HCN 9.31 CN-
HCO3- 10.25 CO3
2-
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pKa Values
• Controlling the ionisation of weak acids• pH = pKa then [HA] = [A-]• pH > pKa then [A-] > [HA]• pH < pKa then [HA] > [A-]CH3COOH (aq) + H2O(l) ↔ H3O+
(aq) + CH3COO- (aq)
• CH3COOH: CH3COO- at pH = 4.77 ?
• CH3COOH: CH3COO- at pH = 3 ?
• CH3COOH: CH3COO- at pH = 7 ?
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Henderson-Hasselbach
ionised][un[ionised]
logpKpH a
For weak acids
Use the equation with the example in the previous slide. Do you come to the same conclusion regarding the ratio of un-ionised to ionised acid molecules?
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Weak Bases
B(aq) + H2O(l) BH+(aq) + OH-
(aq)
CH3NH2(aq) +H2O(l) ↔ CH3NH3+
(aq) + OH-(aq)
pKa = 10.66 (of conjugate acid) [B]=[BH+]
pH = 10.66pH =8 what happens to CH3NH3
+(aq): CH3NH2(aq)
pH =13 ?
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Henderson-Hasselbach
[ionised]ionised][un
logpKpH a
For weak bases
Use the equation with the example in the previous slide. Do you come to the same conclusion regarding the ratio of un-ionised to ionised acid molecules?
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Acidic drugs
2-[4-(2-methylpropyl)phenyl]propanoic acid
How does this molecule ionise?
pKa=4.5
pH =3 (stomach pH)?
pH=6 (intestine)?
ibuprofen
CH3
CH3OH
O
CH3
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Basic drugs
amphetamine (C6H5CH2CH(NH2)CH3)Write an equation for the reaction of
amphetamine with water.The pKa of the conjugate acid is 9.8. What
will happen to the ratio of ionised to unionised amphetamine at:pH 7pH 12
Why might this be important?
CH3
NH2
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Water
• Can dissociate:H2O(l) ↔ H+
(aq) + OH-(aq)
2H2O(l) ↔ H3O+(aq) + OH-
(aq)
H2O is amphoteric
22
3
O][H]][OHO[H
K
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Water
Kw = [H3O+][OH-]= 1 x 10-14 mol2 dm-6
• Kw the ionic product of water
• In pure water what is [H3O+] and [OH-] ?
• Kw is a very small constant– water is only very partially ionised
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pH
• pH is defined as:pH = -log10[H3O+]
• pH is a measure of the H3O+ concentration in solution and can vary from 1 to 14
• pH=7 – neutral [H3O+] = [OH-]
= 1 x 10-7 mol dm-3 at 25 oC
• pH<7 – acidic [H3O+] >[OH-]
• pH>7 - alkaline/basic [H3O+] <[OH-]
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pH-examples
• 0.1M HNO3
• 0.1M CH3COOH
• What is the pH?
• pH is dependent on the ionisation of the acid
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pH-examples
• What about alkaline solutions?• E.g. 0.1M NaOH solution• Will also depend on degree of
ionisation• use equation: [H+] x [OH-] = 10-14
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Buffers
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Buffers
• A buffer solution resists pH changes on addition of small amounts of acid or base (alkali) to a system.
• Very important– e.g. blood has a pH of 7.4. If it varies by
± 0.4, death can occur
• Buffer solutions rely upon the effects of a weak acid or base and the salt of that acid or base
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Buffers
• Ethanoic acid (a weak acid) and sodium ethanoate (salt) CH3COOH CH3COO- + H+ (1)
CH3COONa CH3COO- + Na+ (2)
• (1)-partially ionised• (2)-fully ionised
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Buffers
[acid][salt]
logpKpH a
Henderson-Hasselbach equation
Acidic buffers
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Making a buffer solution
• Choose a weak acid with a pKa close to the required pH of the buffer.
• Choose an appropriate salt of the weak acid
• Determine [salt]/[acid] ratio needed to give correct pH
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An acidic buffer: Ethanoic acid and sodium ethanoate
[acid][salt]
logpKpH a 04.77pH
What is the [salt] if the acid is 0.1 mol dm-3 to give buffer solutions of
pH = 5
pH = 4
What would be the pH of an ethanoate buffer with equal acid and sodium ethanoate concentrations?
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An alkaline buffer:ammonia solution and ammonium chloride
[salt][base]
logpKpH a
Note the base/salt ratioWhat is the pH of a buffer with base:salt ratio = 1?
9.24log19.24pH
Calculate the base:salt ratios for pH 8.5 and pH 10.5
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Acknowledgements
• JISC• HEA• Centre for Educational Research and Developmen
t• School of natural and applied sciences• School of Journalism• SirenFM• http://tango.freedesktop.org