Topic 4.1 Energetics (II) a “understand the definition of enthalpy of atomisation, ∆ H o at, enthalpy of hydration, ∆ H o hyd and lattice enthalpy, ∆ H o lat” Enthalpy of atomisation, ∆ H o at • Enthalpy change for the formation of one mole of gaseous atoms from its element in its standard state ○ Example: ½ Cl2 (g) → Cl (g) ∆ H o at = +121.1 kJ mol-1 ○ Note: ∆ H o at values are all endothermic Enthalpy of hydration, H o hyd • Enthalpy change for the reaction of 1 mole of gaseous ions with water ○ Example: Mg2+ (g) + (aq) → Mg2+ (aq) ∆ H o hyd = -1920 kJ mol-1 ○ Note: ∆ H o hyd values are all exothermic Lattice enthalpy, H o lat (or UL.E.) • Enthalpy change when one mole of an ionic solid is formed from its constituent gaseous ions ○ Example: Na+ (g) + Cl- (g) → Na+Cl-(s) ∆ H o lat = -776 kJ mol1
○ Note: ∆ H o lat values are all exothermic From Units 1 and 2 and needed in Unit 4: Standard enthalpy of formation, ∆ H o f • Enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (25oC and 1 atmosphere pressure) ○ Example: Na (s) + ½ Cl2 (g) → NaCl (s) ∆ H o f = -411 kJ mol-1 ○ Note: ∆ H o f values are all exothermic First ionisation enthalpy, ∆ H o I.E. (1) • Enthalpy change when one mole of electrons is removed from one mole of gaseous atoms of an element ○ Example:Na (g) Na+ (g) +eH o I.E. (1) = +500 kJ mol-1 ○ Note: H o I.E. (1) values are all endothermic Second ionisation enthalpy, H o I.E. (2) • Enthalpy change when one mole of electrons is removed from one mole of singly positive gaseous ions of an element ○ Example: Mg+ (g) Mg2+ (g) +e- ∆ H o I.E. (2) = +1451 kJ mol-1 ○ Note: ∆ H o I.E. (2) values are always more endothermic than those for ∆ Ho I.E. (1) (Why?)
First electron affinity, H o E.A. (1) • First electron affinity is the enthalpy change for the addition of 1 mole of electrons to one mole of gaseous atoms of an element to form one mole of singly charged anions ○ Example: e- + Cl (g) Cl- (g) ∆ H o E.A. (1) = -349 kJ mol-1 ○ Note: ∆ H o E.A. (1) values are all exothermic Second electron affinity, ∆ H o E.A.(2) • Second electron affinity is the enthalpy change for the addition of 1 mole of electrons to one mole of singly charged anions of an element to form one mole of doubly charged anions ○ Example: e- + O- (g) → O2- (g) ∆ H o E.A. (2) = +798 kJ mol-1 ○ Note: ∆ H o E.A. (2) values are all endothermic (Why?) b “construct a Born-Haber cycle and carry out associated calculations” The Born-Haber cycle is really an application of Hess’ Law • H o f = [∆ H o at (metal) + ∆ H o I.E. (metal (g)) + ∆ H o at (non-metal) + ∆ H o E.A. (non-metal(g))] + ∆ H o latt •
For example, sodium chloride: Na+ (g) + e- + Cl (g)
Energy (kJ mole-1)
[∆ H o at [Cl (g)] +121
-364
[H o E.A. [Cl (g)]
Na+(g) + e- + ½ Cl2 (g) [∆ H o I.E. [Na (g)]
+500
Na (g) + Cl (g) +
-
Na (g) + ½ Cl2 (g) [H o at [Na (g)]
0
+108.4
Enthalpy level diagram for the formation of sodium chloride (units kJ mole-1)
[∆ H o latt [Na+Cl- (s)]
Na(s) + ½ Cl2 (g) [∆ H o f [Na+Cl- (s)]
-411
Na+Cl- (s) Using Hess’s Law: 33691600
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H o f [Na+Cl+ (s)] = [∆ H o at [Na (g)] + ∆ H o I.E. [Na (g)] + ∆ H o f [Cl (g)] +∆ H o E.A. [Cl (g)]]+ H o latt [Na+Cl– (s)]
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Substituting the data (in order to calculate H o latt [Na+Cl– (s)]: (-411) = (+108.4) + (+500) + (+121) + (-364) + H o latt [Na+Cl– (s)] H o latt [Na+Cl– (s)] = -776.4 kJ mole-1 In your own time - use Born-Haber cycles to help you to calculate the following: •
∆ H o L.E. [MgBr2 (s)] given ∆ H o at [Mg (g)] = +147.7 kJ mole-1 ∆ H o at [Br (g)] = +111.9 kJ mole-1 o H E.A [Br (g)] = -324.6 kJ mole-1 ∆ H o f [MgBr2 (s)] = -524.3 kJ mole-1 H o I.E. (1) [Mg (g)] = +738 kJ mole-1 ∆ H o I.E. (2) [Mg+ (g)] = +1451 kJ mole-1
•
H o E.A. (2) [O- (g)]
•
∆ H o f [KH (s)]
given ∆ H o at [Ca (g)] H o at [O (g)] -1 mole H o E.A. (1) [O (g)] = ∆ H o f [CaO (s)] = H o I.E. (1) [Ca (g)] = H o I.E. (2) [Ca+ (g)] = H o latt [CaO (s)] = given H o at
mole
[K (g)]
= +178.2 kJ mole-1 = +249.2 kJ -141.1 kJ mole-1 -635 kJ mole-1 +590 kJ mole-1 +1145 kJ mole-1 -3405 kJ mole-1 = +89.2 kJ
-1
H o at [H (g)] = +218 kJ -1 mole ∆ H o E.A [H (g)] = -72.8 kJ mole-1 ∆ H o I.E.[K(g)] = +419 kJ mole-1 o H latt [KH (s)] = -683.6 kJ mole-1
d “understand the factors that influence the value of lattice enthalpy” •
Lattice enthalpy depends on: • The sum of the ionic radii ○ the lattice enthalpy will be smaller if this sum is large • The charge of the ions ○ the lattice enthalpy will be larger if the cation and /or the anion have a large charge on them
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c “understand that values of lattice enthalpies calculated from the theoretical model may differ from those calculated from a purely ionic model” Theoretical (calculated) lattice enthalpy values are based on a purely ionic model • completely spherical ions • complete transfer of electrons Experimental lattice enthalpy values determined using Born-Haber cycle calculations – a practical route to the value For • • •
NaCl theoretical lattice enthalpy value = -770 kJ mole-1 experimental lattice enthalpy value = -776 kJ mole-1 good agreement between theoretical and experimental values ○ ionic model for NaCl is a good one ○ i.e. a purely ionic model - completely spherical ions and complete electron transfer
For • • •
CdI2 theoretical lattice enthalpy value = -2346 kJ mole-1 experimental lattice enthalpy value = -2050 kJ mole-1 no real agreement between theoretical and experimental values ○ ionic model for CdI2 is a not a good one ○ i.e. there is quite a degree of covalent character in CdI2 and it is not purely ionic ○ the ions are not spherical and there is incomplete transfer of electrons (refer to Unit 1.3 b (ii) – “polarising power of cations and polarisability of anions”)
e “understand the part played by lattice enthalpy and enthalpy of hydration in rationalising the variation in solubilities of the hydroxides and sulphates of Group 2.” Enthalpy of hydration, H o hyd •
always exothermic ○ in positive ions the attraction is between the ion and the negative end of the water dipole (δ - on O) ○ in negative ions the attraction is between the ion and the positive end of the water dipole (δ + on H) • the larger the charge on the ion the larger ∆ H o hyd (i.e. it will be more exothermic) • the larger the size of the ion the smaller ∆ H o hyd (i.e. it will be less exothermic) So as we go down a group the value of ∆ H o hyd will decrease (i.e. it will be less exothermic) 33691600
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The solubility of a substance is governed by the enthalpy change of solution, ∆ H osoln. The more exothermic the ∆ H o soln value the more likely it will be for the substance to dissolve. ∆ H o soln MX (s) + M+ (aq) + Cl(aq) (aq) ∆ H o hyd [M+(g)] + ∆ H o hyd [Cl- g)]
-∆ H o latt
Note: we are reversing the direction of the lattice enthalpy vector ∴ the negative sign Using Hess’s law: g)])
M+ (g) + X- (g) + (aq)
∆ H o soln = (-∆ H o latt) + (H o hyd [M+(g)] + ∆ H o hyd [Cl-
Sulphates of Group 2: • Sulphate ion large - so sum of ionic radii does not change very much as cation size increases • Contribution from ∆ H o latt similar for Group 2 sulphates • H o hyd of the cations gets smaller as the ion size increases (contribution from the sulphate ion the same in all cases) • Solubility of the sulphate decreases as cation size increases because lattice enthalpy is not exceeded so much by the hydration enthalpy (lattice enthalpy is not compensated for by hydration enthalpy) Hydroxides of Group 2: • Hydroxide ion is small – so sum of ionic radii of the cation and anion is affected significantly by the cation size • Lattice enthalpies of the hydroxides decrease as the cation gets larger as the cation–anion distance is not affected by a large anion o • ∆ H latt decreases more rapidly than ∆ H o hyd
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Topic 4.2 Topic 4.2: The Periodic Table II (Period 3 and Group 4) a
Variation of properties across a period
“the variation in properties across Period 3 (sodium to argon) as illustrated by: i reactions of the elements with oxygen, chlorine and water “ ([Ne] = 1s22s22p6) Appearance
Na
Mg
Al
Si
[Ne]3s1
[Ne]3s2
[Ne]3s23p1
Soft shiny grey metal
Shiny grey metal
Metallic
Structure Metallic and bonding Reaction with oxygen Reaction with chlorine
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S (S8) [Ne]3s23p4
Cl2
Ar
[Ne]3s23p2
P (P4) [Ne]3s23p3
[Ne]3s23p5
[Ne]3s23p6
Shiny grey metal
Shiny grey solid
White waxy solid
Yellow powder
Green gas
Colourless gas
Metallic
Giant covalent
Molecular covalent
Molecular covalent
Molecular covalent
Atoms
4Na + O2 ↓ 2Na2O
2Mg + O2 ↓ 2MgO
4Al + 3O2 ↓ 2Al2O3
Si + O2 ↓ SiO2
2Na + Cl2 ↓ 2NaCl
Mg + Cl2 ↓ MgCl2
2Al + 3Cl2 ↓ 2Al2Cl3
Si + 2Cl2 ↓ SiCl4
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P4 + 5O2 ↓ P4O10 2P + 3Cl2 ↓ 2PCl3 (+ Cl2 → PCl5)
7
S + O2 ↓ SO2
2Cl2 + O2 ↓ 2Cl2O
No reaction
2S + Cl2 ↓ S2Cl2
N/A
No reaction
Reaction with water
2Na+H2O Cl2 + H2O Mg + 2H2O Unreactive – ↓ ↓ ↓ 2NaOH + H2 Mg(OH)2 + H2 due to oxide No reaction No reaction No reaction HOCl + HCl layer Very slow Vigorous Dispropn “the variation in properties across Period 3 (sodium to argon) as illustrated by: ii the formulae and acid-base character of the oxides and hydroxides of the metals and oxides of the non-metals” Na Formula
Na2O
Mg
Al
Si
P
S
No reaction
Cl
Ar
MgO
Al2O3
SiO2
P4O10
SO2
SO3
Cl2O
No oxide
Appearan White ce solid Bonding Ionic and structure Na2O + Reaction H 2O with water
White solid
White solid
White solid
White solid Covalent molecular
Colourless gas Covalent molecular
White solid
Colourles s gas Covalent molecular
N/A
Cl2O + H2O
N/A
Type of oxide
Basic
Acidic
N/A
↓ 2NaOH
Basic
Ionic
Intermedia Giant te ionic/ covalent covalent MgO + H2O Insoluble in Insoluble in ↓ water water Mg(OH)2 Amphoteric Acidic
Al2O3 + 6HCl 2AlCl3 + 6HCl Al2O3 + 3H2O + 6NaOH 2Na3Al(OH)6
SO2 + H2O ↓ H2SO3
P4O10 + 6H2O ↓ 4H3PO4
Acidic
Acidic
Covalent molecular SO3 + H2O ↓ H2SO4 Acidic
↓ 2HOCl
∴ Al2O3 acts as a base Al2O3 acts as an acid
SiO2 is insoluble in water and also dilute NaOH (because it is aqueous) but it does dissolve in fused/molten NaOH:
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N/A
SiO2 (s) + 2NaOH (l) Na2SiO3 (s) + H2O (l) Also: Acid-base character of Mg(OH)2 and Al(OH)3 need to be known: Mg(OH)2 + 2HCl → MgCl2 + 2Al(OH)3 + 6HCl → 2AlCl3 + 3H2O H2O ∴ Mg(OH)2 acts as a base Al(OH)3 + OH- Al(OH)4only acid “the variation in properties across Period 3 (sodium to argon) as illustrated by: iii the formulae of the chlorides, and their reactions with water”
Formula Appearan ce Bonding and structure Reaction with water
Na
Mg
Al
Si
P
NaCl
MgCl2
AlCl3
SiCl4
PCl3
White solid
White solid
White solid
Ionic
Ionic
Covalent molecular
Colourless liquid Covalent molecular
NaCl (s) + (aq) ↓ Na+ (aq)+ Cl(aq)
MgCl2 (s) + (aq) ↓ Mg2+(aq) + 2Cl(aq)
AlCl3 + 3H2O ↓ Al(OH)3 + 3HCl
SiCl4 + 2H2O ↓ SiO2 + 4HCl
Cl
Ar
PCl5
S2Cl2
Cl2
Colourles s liquid Covalent molecular
White solid
Liquid
Covalent molecular
Covalent molecular
Green gas Covalent molecular
No chloride N/A
PCl3 + 3H2O
PCl5 + 4H2O
S2Cl2 + 2H2O ↓ H2S +SO2 +2HCl
↓ H3PO3 + 3HCl
Covalently bonded chlorides undergo hydrolysis reactions. Harrow Chemistry
∴ Al(OH)3 acts as an
S
Note: Ionically bonded chlorides undergo hydration reactions.
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∴ Al(OH)3 acts as a base
9
↓ H3PO4 + 5HCl
Cl2 + H2O ↓ HCl + HOCl
N/A N/A
“iv
students should be able to interpret the reactions in (a) (ii) and (a) (iii) in terms of the structure and bonding of the oxides and chlorides”
Covered in the tables above.
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b
Variation of properties down a group
i “recall the reasons for the increase in metallic character with increase in atomic number in Group 4” Going down group 4: •
•
Ca rb on
Non metalli c
Sil ic on
Non metalli c
G er m an iu m
Semi metalli c
Ti n
Metalli c
Le ad
Metalli c
This increasing trend is caused by a decrease in ionisation energy as the cation radius decreases. This makes it energetically more favourable for the elements lower down the group to form ionic compounds
Ease of positive ion formation increases
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•
Ionic character of chlorides increases and there is no significant hydrolysis in PbCl2.
•
Ionic character of oxides increases and the trend is from neutral/acidic to amphoteric/basic ○ CO is a neutral oxide (it so weakly acidic that it is thought of as a neutral oxide) CO2 is an acidic oxide CO2 + NaOH → Na2CO3
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○
SiO2 is an acidic oxide but it needs to react with fused NaOH:
○
PbO is an amphoteric oxide:
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SiO2 (s) + 2NaOH (l) → Na2SiO3 (s) + H2O (l)
As a base: PbO + 2HCl → PbCl2 + H2O As an acid: PbO + 2NaOH + H2O → Na2Pb(OH)4
12
ii “recall that the +2 oxidation state in Group 4 becomes more stable than the +4 oxidation state as the atomic number increases and apply this to the chemistry of tin and lead.” The stability of the +2 oxidation increases down the group; the stability of the +4 oxidation state increases up the group. The smaller elements of group 4 show a wide range of oxidation states in their compounds: C: Sn: Pb
+4 in CO2, +2 in CO, -4 in CH4 +4 and +2 +4 and +2
[+4 more stable than +2 for tin] [+ 2 more stable than +4 for lead]
In carbon the 4 outer electrons are in the second shell and are closely held by the nucleus ∴ compounds of carbon are covalently bonded and carbon is usually +4 in its compounds. In lead the outer electron configuration is 6s26p2. The 6s2 electrons are absorbed into the inner core and are thus held more tightly (in general s electrons are held more tightly); the 6p electrons are thus less firmly held and can be removed more easily.
We use some of the chemistry of tin and lead to illustrate the oxidation state stability trend in group 4. Tin (II) is reducing and becomes tin (IV) which is the preferred oxidation state of tin in its compounds: Sn2+ + Cl2 Sn4+ + 2Cl33691600
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Lead (IV) is oxidising and becomes lead (II) which is the preferred state of lead in its compounds: PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O iii
“recall and explain the structure of carbon tetrachloride” There are eight electrons around the central carbon atom. Thus 4 pairs of electrons all of which are bonding pairs. The 4 electron pairs arrange themselves around the central atom and repel each other arranging themselves as far apart as possible to minimise the repulsion between them. Thus CCl4 is tetrahedral and the bond angle is 109.5o.
iv “recall and explain the behaviour of carbon tetrachloride with water and contrast this behaviour with that of silicon tetrachloride with water”
• •
•
•
CCl4 is resistant to hydrolysis and does not react at all with water: very strong C-Cl bond which has to be broken before reaction can occur there are no vacant orbitals available for the coordination of the lone pair of electrons on the oxygen atom of the water molecule. SiCl4 reacts very rapidly with water at room temperature and undergoes a hydrolysis reaction there are empty energetically accessible 3d orbitals on Si (in SiCl4) [no energetically accessible orbitals available in C (in CCl4)] • the lone pair of electrons on the oxygen atom of the water molecule can coordinate into these vacant energetically accessible 3d orbitals before the Si-Cl bond is broken the Ea value is much lower in SiCl4 compared to that for CCl4 and thus the SiCl4 hydrolyses very readily with water. 33691600
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Topic 4.3: Chemical equilibria II From Unit 2 and needed in Unit 4: Reactions at equilibrium • the rate of the forward reaction is equal to the rate of the reverse reaction. • the concentrations of all the substances at equilibrium are constant (not equal) • the composition at equilibrium can be approached from the reactants or products Two types of equilibrium • homogenous - all of the substances are in the same phase. ○ e.g. HCl (aq) + NaOH (aq) ⇌ NaCl (aq) + H2O(l) •
○
e.g. H2 (g) + I2 (g) ⇌ 2HI (g) heterogeneous - the substances are in different phases. ○ e.g. CaCO3 (s) ⇌ CO2 (g) +CaO (s)
Position of equilibrium • many reactions do not go to completion • reactions reach a state of dynamic equilibrium • the time taken to attain/reach equilibrium is variable • the position of equilibrium is a measure of the extent of a reaction once equilibrium has been established • if a reaction uses more than 50% of the reactants before reaching equilibrium we say that the “position of equilibrium lies to the right” • if a reaction uses more than 50% of the reactants before reaching equilibrium we say that the “position of equilibrium lies to the left” • Note: it must not be assumed that equilibrium has been established when there are 50% of reactants and 50% of products a “define the terms partial pressure and concentration” The partial pressure of a gas in a mixture of gases in a container is the pressure that it alone would exert at a given temperature. In a mixture of nA moles of gas A and nB moles of gas B at a total pressure of P •
the mole fraction (x) of each gas is given by: xA =
•
nA/(nA + nB)
xB =
nB/(nA + nB)
the partial pressure of each gas is given by:
pA = xA X P
pB = xB X P
The concentration of a solution such as HCl (aq) is given the symbol [HCl] and the units are moles dm-3. b “deduce expressions for the equilibrium constants Kc and Kp from given equations and calculate their numerical values with units, given suitable data” e “determine the equilibrium partial pressures in an equilibrium resulting from simple binary gaseous dissociation (eg that of dinitrogen tetroxide) given the value of Kp” The equilibrium constant, Kc Kc is the equilibrium constant in terms of equilibrium concentrations (for purely aqueous or purely liquid systems - single phase). For the reaction: aA (aq) + bB (aq) ⇌ cC (aq) + dD (aq) Kc = [C]c[D]d [A]a[B]b Characteristics of Kc • characterises an equilibrium • measured experimentally independent of the initial composition • temperature dependent • related to the equation for the equilibrium Examples of Kc CH3COOH (l) + CH3CH2OH (l)
⇌
CH3COOCH2CH3 (l)
+ H2O (l)
Assuming a liquid phase equilibrium:Kc = [CH3COOCH2CH3][H2O] [CH3COOH][CH3CH2OH]
N2 (g)
+
3H2 (g)
⇌
2NH3 (g)
Assume a volume V dm3:Kc = [NH3/V]2 [N2/V][H2/V]3
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The gaseous equilibrium constant, Kp Kp is the equilibrium constant in terms of equilibrium partial pressures (for purely gaseous systems - single phase). 2SO2 (g) Kp
=
+
O2 (g)
⇌
2SO3 (g)
(PSO3)2 (PSO2)2 X (PO2)
Calculations involving Kc and Kp 1 1 mole of ethanoic acid and 1 mole of ethanol are mixed together at 100oC and allowed to reach equilibrium. The mixture was rapidly cooled in ice-cold water and then made up to 1 dm3 with distilled water. 25 cm3 portions of this solution required 27.5 cm3 of 0.300 mole dm3 sodium hydroxide for neutralisation. Calculate the value for Kc giving due consideration to its units. Relevant equations:NaOH + CH3COOH → CH3COONa + H2O CH3COOH (l) + CH3CH2OH (l) ⇌ CH3COOCH2CH3 (l) + H2O (l) No. of moles NaOH used in the titration 27.5/1000
= =
0.300 X
8.25 X 10-3 moles
No. of moles ethanoic acid in 25 cm3 of diluted “residue” = No. of moles NaOH used in the titration = 8.25 X 103 moles No. of moles ethanoic acid in 1dm3 of diluted residue = 8.25 X 10-3 x 1000/25 = 0.33 moles This is the number of moles of ethanoic acid present at equilibrium. CH3COOH (l) +CH3CH2OH (l) ⇌ CH3COOCH2CH3 (l) +H2O (l) Start: 1 mole Eqm.: 0.33 mole mole
1 mole 0.33* mole
0 mole
0 mole
0.67 mole
0.67
*0.67 mole acid react with 0.67 mole alcohol ∴ 1 – 0.67 = 0.33 mole alcohol left at equilibrium. 33691600
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Assuming a volume of V dm3 – then the equilibrium concentrations (moles dm-3) are: 0.33/V
0.33/V
0.67/V
0.67/V
CH3COOH (l) + CH3CH2OH (l) ⇌ CH3COOCH2CH3 (l) + H2O (l) Kc = [CH3COOCH2CH3][H2O]
=
(0.67/V)(0.67/V) = 4.12 (no
units [CH3COOH][CH3CH2OH] cancel
(0.33/V)(0.33/V)
- as they out)
2
The equilibrium constant for the reaction: 2A(aq) ⇌ A2(aq) is 6.4 mol-1 dm3 Calculate the equilibrium concentration of A2 if the value for [A] is 0.4 mol dm3. Kc = [A2] [A]2
=
[A2] = (0.4)2
6.4
[A2] = 1.024 mole dm-3 3
Calculate the equilibrium constant for the reaction N2O4 (g) ⇌ 2NO2 (g)
given that 1.00 mole has dissociated by 20% at equilibrium at a pressure of 2 atmospheres. N2O4 (g) ⇌ 2NO2 (g) Start
Eqm.
1.00 moles
0
(1.00 - 0.20)
(2 X 0.20)
moles
0.80 moles
0.40
moles
Total no. of moles at equilibrium = 0.80 + 0.40 = 1.20 moles Partial pressure of N2O4 = mole fraction N2O4 X Ptot = (0.80/1.20) X 2 = 1.33 atm Partial pressure of NO2 = mole fraction NO2 X Ptot = (0.40/1.20) X 2 = 0.67 atm Kp = (PNO2)2 (PN2O4)
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(0.67)2 (1.33)
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=
0.338 atm
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c not
“recall that expressions for Kp and Kc for heterogeneous equilibria do include values for solid and liquid phases”
CaCO3 (s) Kp
⇌ CaO (s)
= (PCaO) X (PCO2) (PCaCO3)
+ CO2 (s) Since solids do not exert partial pressures we assume that (PCaO) and (PCaCO3) equals unity (1) and write out a new constant, Khet, the heterogeneous equilibrium constant
Hence, Khet = (PCO2) d
“recall that the presence of a catalyst does not affect the position of equilibrium”
Catalysts increase the rate of reaction. In an equilibrium: • catalysts increase the rate of the forward reaction, with a subsequent increase in the rate of the forward reaction. • thus the rate at which equilibrium is achieved is faster in the presence of a catalyst. • note: the presence of a catalyst does not affect the equilibrium composition, i.e. has no effect on Kc. f “understand that changes in temperature result in a change in the value of Kc and Kp and that the position of equilibrium will change with change in temperature.” Temperature If the reaction (convention - left to right) is exothermic, the value of Kc (or Kp) will decrease if the equilibrium temperature is increased. This means that the equilibrium position moves to the left and the yield of product falls at higher temperatures. The value of Kc (or Kp) will increase if the temperature is decreased. This means that the equilibrium position moves to the right and the yield of product increases at lower temperatures. “A decrease in temperature favours the exothermic direction of a reaction”. If the reaction (convention - left to right) is endothermic, the value of Kc (or Kp) will increase if the equilibrium temperature is increased. This means that the equilibrium position moves to the right and the yield of product rises at higher temperatures. 33691600
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“An increase in temperature favours the endothermic direction of a reaction”.
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Concentration - using Kc Consider the following equilibrium: aA(aq) + bB(aq) ⇌ cC(aq) + dD(aq) K c= [C]c[D]d [A]a[B]b Kc is fixed at a particular temperature so if [A] and/or [B] are increased then [C] and [D] have to increase in order for the constant value of Kc to be maintained. Pressure An increase in pressure results in a decrease in volume. This means that in an equilibrium the system with the smallest volume will oppose the increase in pressure. E.g. 2SO2(g) + O2 (g) ⇌ 2SO3(g) 2 moles 1 mole 2 moles 2 vols 1 vol 2 vols i.e. 3 vols to 2 vols So, the equilibrium will shift to the right. the formation of SO3 is favoured. Using Kp=
(PSO3)2 (PSO2)2 X (PO2)
Increasing PSO2 and/or PO2 means that PSO3 will increase in order to maintain the constant value of Kp; so, more SO3 formed. Summary Factor Concentratio n Pressure
Effect on Kc Effect on composition None Changes
Temperature Catalyst
Changes None
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None
Changes – for gaseous reactions Changes None
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Topic 4.4: Acid-base equilibria a “recall the Brønsted-Lowry theory and use it to identify acid base behaviour, and identify acid base conjugate pairs and relate them by means of suitable equations” Brønsted-Lowry theory An acid is a proton donor: E.g. HCl (aq) +H2O (l) ⇌ H3O+ (aq) + Cl- (aq) In the forward reaction the HCl is acting as an acid. A base is a proton acceptor: E.g. NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq) In the forward reaction the NH3 is acting as a base. Acid-base conjugate pairs E.g. HCl (aq) + H2O (l) ⇌ H3O+ (aq) + Cl- (aq) Acid 1 Base 2 Acid 2 Base 1 Conjugate base 1 (Cl- ) is related to acid 1 (HCl) because it has been formed by the loss of a proton. Conjugate acid 2 (H3O+) is related to base 2 (H2O) because it has been formed by the gain of a proton. More examples of acid – base pairs. Identify the acid-base conjugate pairs in these equilibria. H2SO4 + H2O ⇌ H3O+ + HSO4HNO3 + H2SO4 ⇌ H2NO3+ + HSO4HClO4 + H2O ⇌ H3O++ ClO4-
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c “understand the terms ‘strong’ and ‘weak’ as applied to acids and bases” Basicity of an acid The basicity of an acid can be thought of as the number of replaceable hydrogens that it has: HCl
monobasic
H2SO4
dibasic
H3PO4
tribasic
CH3COOH
tetrabasic monobasic (why?)
Strong and weak acids Strong acids - fully ionised in aqueous solution. E.g. HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq) The equilibrium lies to the right and the [H3O+] is very high. Strong acids have weak conjugate bases. Weak acids - only partially ionised in aqueous solution. E.g. CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq) The equilibrium lies to the left and the [H3O+] is very low. CH3COO- is the strong conjugate base of the weak acid CH3COOH. Weak acids have strong conjugate bases. Strong and weak alkalis Strong alkalis - fully ionised in aqueous solution. E.g. NaOH (aq) Na+ (aq) + OH- (aq) The equilibrium lies to the right and the [OH-] is very high. Strong alkalis have weak conjugate acids. Weak alkalis - only partially ionised in aqueous solution. E.g. NH4OH (aq) ⇌ NH4+ (aq)+OH- (aq) The equilibrium lies to the left and the [OH-] is very low. NH4+ is the strong conjugate acid of the weak base NH4OH. Weak alkalis have strong conjugate acids.
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b “d efine pH” d “define Ka and Kw and recall their units” e “define pKa and pKw” f “calculate the pH of solutions of strong acids and strong bases” g “calculate pH of solutions of weak acids given Ka and vice versa” Ionic product of water H2O (l) ⇌ H+ (aq) + OH- (aq) Kc = [H+][OH-] [H2O] [H2O] is so large that it remains effectively constant. Kw = [H+][OH-], where Kw incorporates Kc and [H2O] At 298K and 1 atm, Kw = 1 X 10-14 moles2 dm-6 Kw = [H+][OH-] = 1 X 10-14 moles2 dm-6 For a neutral solution [H+]=[OH-]
[pKw = -log10Kw = 14] ([H+]= 1 X 10-7 mole dm-3)
For an acid solution [H+]>[OH-]
([H+]> 1 X 10-7 mole dm-3)
For an alkali solution [H+]<[OH-]
([H+]< 1 X 10-7 mole dm-3)
pH pH= -log10[H3O +]/moles dm-3 More simply,
pH= -log10[H+]
This is the definition of pH!!
Strong acids E.g. calculate the pH of 0.1M HCl. 33691600
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HCl - strong acid fully ionised. So [H+] = 0.1 moles dm-3 pH = - log10(0.1) = 1 E.g. calculate the pH of 0.2M H2SO4. H2SO4 - strong acid ∴ fully ionised* (assume H2SO4 →2H+ + SO42-) So [H+] = 2 X 0.2 moles dm-3 pH = - log10(0.4) = 0.398 Strong alkalis E.g. calculate the pH of 0.01M NaOH. NaOH - strong acid ∴ fully ionised. So [OH-] = 0.01 moles dm-3 Since Kw = [H+][OH-] = 1 X 10-14 moles2 dm-6 ∴ [H+] =(1 X 10-14)/0.01 = 1 X 10-12 moles dm-3 ⇒ pH = 12 Weak acids (1) E.g. calculate the pH of 0.1M CH3COOH given Ka = 1.7 X 10-5 moles dm-3. Eqm. CH3COOH (aq) ⇌ CH3COO-(aq) + H+(aq)
∴ Ka=[CH3COO-][H+] [CH3COOH]
1.7 X 10-5 =[H+]2 (0.1)
Note the assumptions made: ➢ CH3COOH is a weak acid, so only partially ionised ∴ [CH3COOH] is effectively constant/unaltered ➢ [CH3COO-] = [H+]
[H+] = √ (1.7 X 10-5 X 0.1) = 1.3 X 10-3 moles dm-3 ∴ pH = -log10(1.3 X 10-3) = 2.88
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*Dr. Beavon, Chief Examiner says not!! Check his website: www.rod.beavon.clara/net.learning. htm
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Weak acids (2) E.g. calculate the ionisation constant, Ka, of 0.01M CH3CH2COOH given that the pH of the acid is 3.44. What is the pKa value for propanoic acid? Eqm. CH3CH2COOH ⇌ CH3CH2COO- + H+ Ka=[CH3CH2COO-][H+] [CH3CH2COOH] ∴ Ka = [H+]2 [CH3CH2COOH] Ka = (3.6 X 10-4)2 0.01
Note the assumptions made: ➢ CH3CH2COOH is a weak acid, so only partially ionised ∴ [CH3CH2COOH] is effectively constant/unaltered ➢ [CH3CH2COO-] = [H+]
= 1.30 X 10-5 moles dm-3 Since pKa = -log10Ka ∴ pKa = -log10(1.30 X 10-5) = 4.89
(From pH = 3.44) Since pH = -log10[H+] = 3.44 ∴ [H+]= 3.6 X 10–4 moles
Effect of temperature on pH The ionisation of water is slightly endothermic. H2O (l) ⇌ H+ (aq) + OH- (aq) Kw = [H+][OH-] At 25 oC and 1 atm, Kw = 1 X 10-14 moles2 dm-6 At 50 oC and 1 atm, Kw = 1.2 X 10-14 moles2 dm-6 So [H+] and [OH-] increase with increasing temperature. Thus the equilibrium for the ionization of water moves to the right with increasing temperature. This means that the equilibrium must be endothermic in the forward direction. At 25 oC [H+] = (1.0 x 10-14) moles dm-3 At 50 oC [H+] = √(1.2 x 10-14) moles dm-3 So, [H+] increases with increasing temperature. Also pH (=-log10[H+]) decreases as [H+] increases. But the solution does not become more acidic as the temperature increases because there is a simultaneous increase in [OH-] as the temperature increases, thus “cancelling out” the effect of the increase in [H+].
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h “understand the principles involved in acid base titrations” i “recall the sketch curves for the variation in pH during the following titrations; strong acid – strong base, weak acid – strong base and strong acid – weak base” j “use titration curves to determine Ka for a weak acid” k “explain the choice of a suitable indicator for an acid-base titration given pKInd values” Acid-base titrations pH changes during an acid-base titration (0.1 mole dm-3 concentrations) Strong acid vs strong alkali (e.g. HCl and NaOH) 25 cm3 of 0.1M HCl in a conical flask and add small portions of 0.1M NaOH from a burette measuring the pH after each addition. Plot a graph of pH against volume alkali added.
pH of NaOH NaOH in excess
Equivalence point - halfway point of the straightest vertical part of the graph (pH = 7)
pH of 0.1M HCl = 1 30
5
10
15
20
25
Volume of alkali added/cm3
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Weak acid vs strong alkali (e.g. CH3COOH and NaOH) Place 25 cm3 of 0.1M HCl in a conical flask and add small portions of 0.1M NaOH from a burette measuring the pH after each addition. Plot graphs of pH against volume alkali added. Equivalence point - halfway point of the straightest vertical part of the graph (pH = 8.5)
pH of NaOH
Compare this curve with the one for HCl and NaOH
pH of 0.1M CH3COOH ≈ 3
30
5
10
15
20
25
Volume of acid added/cm3
Weak alkali vs strong acid (e.g. NH4OH vs HCl) [Apologies for the “switch” but this was the only set of curves that was available at the time]. Place 25 cm3 of the alkali in a conical flask and add small portions of acid from a burette measuring the pH after each addition. Plot graphs of pH against volume alkali added.
Compare the curve for NaOH and HCl with this one
pH of NH4O H
pH of HCl 30
5
10
15
20
Equivalence points during acid-base titrations Volume of acid added/cm 33691600
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Equivalence point halfway point of the straightest vertical part of the graph (pH = 5.5) 30
You will have noticed that the pH values at equivalence are not always equal to 7. Explanations: HCl vs NaOH: resulting salt NaCl (aq) - the salt of a strong acid and a strong alkali. Ions present in aqueous solution: Na+
Cl- (NaCl(aq)) OHH+ (H2O)
Na+ will not remove OH- as NaOH as NaOH is a strong electrolyte ∴ fully ionised in aqueous solution. Cl- will not remove H+ as Cl- is a weak base ∴ does not remove H+. So [H+] = [OH-] ∴ the salt is neutral (pH = 7 at 25OC) HCl vs NH4OH: resulting salt NH4Cl (aq) - the salt of a strong acid and a weak alkali. Ions present in aqueous solution: NH4+
ClOH-
(NH4Cl(aq)) H+ (H2O)
NH4+ will remove OH- as NH4OH as NH4OH is a weak electrolyte ∴ only partially ionised aqueous solution. Cl- will not remove H+ as Cl- is a weak base does not remove H+. So [H+] > [OH-] ∴ the salt is acidic (pH < 7 at 25OC) CH3COOH vs NaOH: resulting salt CH3COONa (aq) - the salt of a weak acid and a strong alkali. Ions present in aqueous solution: Na+ OH-
CH3COOH+
(CH3COONa (aq)) (H2O)
Na+ will not remove OH- as NaOH as NaOH is a strong electrolyte ∴ fully ionised in aqueous solution. CH3COO- will remove H+ as CH3COO- is the strong conjugate base of a weak acid ∴ removes H+ as unionised CH3COOH. So [H+] < [OH-] ∴ the salt is alkaline (pH > 7 at 25OC). Acid - base indicators 33691600
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An acid-base indicator is (usually) a weak acid whose conjugate base is a different colour from the acid itself. Use HIn as a general formula for an indicator: HIn ⇌ H+ + InColour A
Colour B
During a titration: (1) Acid (e.g. HCl) + indicator in conical flask colour A predominates + as H ions from HCl cause indicator equilibrium to shift to the left. Alkali (e.g.NaOH) added from a burette removes H+ ions from HCl (H+ + OH- H2O) until no more H+ ions from acid left. The next drop of alkali must then remove H+ ions from the indicator equilibrium thus causing the equilibrium to shift to the right ∴ colour B predominates. (2) Alkali (e.g. NaOH) + indicator in conical flask colour B predominates as OH- ions from NaOH cause indicator equilibrium to shift to the right because they remove H+ ions as water (H+ + OH- H2O). Acid (HCl) added from a burette removes OH- ions from NaOH as water until no more OH ions from alkali left. The next drop of acid must then increase the H+ ions in the indicator equilibrium thus causing the equilibrium to shift to the left colour A predominates. Acid - base indicators – quantitative approach HIn ⇌ H+ + InColour A
Colour B
KInd = [H+] X [In-] [HIn] [H+] = KInd X [HIn] [In-] When [Hin] = [In-] [H+] =KInd pH = pKInd So, the solution changes colour when its pH = pKInd. Choice of indicators 3 common indicators: 33691600
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Indicator
pH range 3.2 - 4.5
Colour in acid (Hin) Red
Colour in alkali (In-) Yellow
Methyl orange Phenolphthalein
8.2 - 10.0
Colourless
Pink
Bromothymol blue
6.0 - 7.0
Yellow
Blue
The indicator should change colour anywhere on the straight vertical portion of the curve. So for the three different types of titration: Titration
pH range on curve†
Indicator(s)
Strong acid Strong alkali
4 - 10
All three
Strong acid Weak alkali Weak acid - Strong alkali Weak acid – Weak alkali
4-8
Methyl orange or bromothymol blue 6 - 10 Phenolphthalein or bomothymol blue For weak acids and weak alkalis - equivalence is not very sharply defined difficult to use an indicator (anyway, indicators are weak acids or bases!)
On the straightest vertical portion of the curve.
†
We can obtain the pKa for the weak acid from the weak acid – strong alkali curve: Find the volume of alkali needed for equivalence (e.g. 25 cm3) and divide by 2 (∴ 12.5cm3) and then extrapolate to the pH value (e.g. 4.7)
pKa of CH3COOH = 4.7; since pKa = -logKa then Ka = 1.8 X 10-5 moles dm-
12.5 cm3 alkali added
3
l
Compare this curve with the one for HCl and NaOH
“explain the action of a10buffer15 solution calculate its pH from 5 20 and 25 30
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suitable
data”
Buffer solutions - definition/constituents A buffer solution is one which tends to resist changes in pH when small amounts of acid or alkali are added to it. Constituents of a buffer Weak acid/(usually sodium salt of) its conjugate base e.g. CH3COOH/CH3COO-Na+ Weak base/(salt of) its conjugate base
e.g. NH4OH/NH4+Cl-
Buffer solutions - how do they work? Buffer reactions in (aqueous solution): CH3COO-Na+ → CH3COO- + Na+ (I) CH3COOH ⇌ CH3COO- + H+ (II) (I) - CH3COO-Na+ salt of a gp. I metal fully ionised in aqueous solution. (II) - Equilibrium lies very much to the left because (a) CH3COOH is a weak acid and (b) its ionisation is further suppressed by the CH3COO- ions from (I) which increases the [CH3COO-] in (II) and causes the equilibrium to shift to the left If a small amount of acid ( e.g. H+ from HCl) is added to the buffer: Ethanoate ions from (I) remove the added H+ ions as unionised CH3COOH: CH3COO- + H+ ⇌ CH3COOH Thus maintaining the almost “constant” value of [H+]. If a small amount of alkali (e.g. OH- from NaOH) is added to the buffer: Hydrogen ions from (II) remove the added OH- ions as unionised water: OH- + H+ ⇌ H2O As the H+ ions are removed by the added OH- the equilibrium (II) shifts to the right to replace the H+ removed. Thus maintaining the almost “constant” value of [H+]. The region on the CH3COOH vs NaOH pH curve where buffering takes place: You need to look at the “straightest” horizontal portion of this curve to find this.
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Buffer solutions – calculations Calculate the pH of a buffer solution containing 100 cm3 of 0.1 mole dm-3 CH3COOH and 100 cm3 of 0.2 mole dm-3 of CH3COONa. Ka = 1.8 X 10-5 mole dm-3. Buffer reactions: CH3COO-Na+ CH3COO- + Na+ (I) CH3COOH ⇌ CH3COO- + H+ (II) Ka=[CH3COO-][H+] (for the equilibrium in (II)) [CH3COOH] In a buffer such as the one above (CH3COOH/CH3COO-Na+), CH3COO-Na+ CH3COO- + Na+ (I) CH3COOH ⇌ CH3COO- + H+ (II) [CH3COO-] comes entirely from reaction (I) as the salt is completely ionised and the contribution from (II) is negligible because CH3COOH is a weak acid and does not ionise readily and thus [CH3COO-] = [SALT]. As a further consequence of the latter, [CH3COOH] is the concentration of the acid i.e. [CH3COOH] = [ACID]. Ka= [SALT][H+] [ACID] [H+] = Ka X [ACID] [SALT] [H+] = 1.8 X 10-5 x (0.1/2) (0.2/2)
[Why divide by 2?]
[H+] = 9.0 X 10-6 moles dm-3 pH = -log10(9.0 X 10-6 ) = 5.05 N.B. in a buffer containing equimolar amounts of the weak acid and its salt, [H+] = Ka X [ACID] [SALT] [H+] = Ka ∴ pH = pKa m
“demonstrate an understanding of how the value of enthalpy of neutralisation is related to the strength of acids and bases”.
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Enthalpy considerations H+(aq) + OH- (aq) ⇌ H2O (l) ∆ HN o =-57.1 kJ mole-1. [For example, HCl + NaOH] This is the value obtained for the neutralisation of a strong acid by a strong alkali. In both cases the electrolytes are fully ionised in aqueous solution. Thus, no energy is needed to ionise the acid or alkali. CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2O (l) kJ mole-1.
HN o = -53.2
Here we have the reaction between a weak acid and a strong alkali. So some energy is needed to ionise the acid (endothermic process) i.e. energy is absorbed in order to ionise the weak acid and this is why there is a discrepancy between the two values -57.1 kJ mole-1 vs. -53.2 kJ mole-1. Now compare these values with the following: NH4OH (aq) + HCN (aq) → NH4CN (aq) + H2O (l) 1 .
∆ HN o = -8.4 kJ mole-
Comments?
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Topic 4.5: Organic chemistry II (acids, esters, carbonyl compounds, acid chlorides, nitrogen compounds and further halogenocompounds) From Unit 2 and needed in Unit 4: Questions set on this topic may require the application of knowledge covered in unit 2; specifically the rules for nomenclature and the ideas of isomerism, bond polarity, bond enthalpy and reaction types, including reagents and conditions. a
Further ideas in organic chemistry
i “apply the concept of functional groups to the range of organic compounds found in units 2 and 4” ii ii “recognise and predict the existence of structural isomers within the types of organic compound found in Units 2 and 4” iii
“recognise stereoisomerism as geometrical (cis-trans) or optical”
iv “explain the existence of geometrical (cis-trans) isomerism resulting from restricted rotation about a carbon-carbon double bond” v “understand the existence of optical isomerism resulting from a chiral centre in molecules with a single asymmetric carbon atom, and understand optical isomers as object and non-superimposable mirror images” vi “recall optical activity as the ability of a single optical isomer to rotate the plane of polarisation of plane polarised monochromatic light and understand the nature of a racemic mixture” Isomerism Really developed through the examples in the Organic Chemistry programme. Practise as many examples as you can. However, it may be useful to be aware some of the important aspects of isomerism. Isomerism is where two or more compounds have the same molecular formula but have different spatial arrangements of their atoms and differ in some of their physical properties and, in some cases, chemical properties.
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Two types of isomerism: Structural: Same molecular formula, different structural formula. Chain:
e.g.
CH3CH2CH2CH3 Butane
CH3CH(CH3)CH3 2-methylpropane
Position:
e.g.
CH3CH2CH2OH Propan-1-ol
CH3CH(OH)CH3 Propan-2-ol
e.g.
CH3COCH3 Propanone
Functional group:
CH3CH2CHO Propanal
Stereoisomerism: Same molecular formula, different spatial arrangement of atoms. Geometric: Rotation around C-C restricted because of presence of >C=C< bond and Π -orbital e.g. cis but-2-ene and trans but-2-ene.
Optical: Two isomers having the same molecular and structural formulae but they are non-superimposable object and mirror images of each other. They only differ from each other in the way in which they rotate the plane of plane polarised light. You can tell whether a compound will exhibit optical activity if it possesses an asymmetric carbon atom (4 different groups around a single carbon atom). Such a molecule is said to be chiral - it has an isomer which is a mirror image of itself, the two being nonsuperimposable. E.g. butan-2-ol, CH3CH(OH)CH2CH3 has two optical isomers. Asymmetric carbon atom
The two isomers are: An equimolar mixture of the two optical isomers has no effect on the plane of planepolarised light as the rotational effect of one isomer is cancelled out by the other. Such a mixture is called a 33691600
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racemic mixture.
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b
Further reactions of organic compounds
“Students should be able to recall, in terms of reagents and general reaction conditions, the following reactions and classify reactions in this topic as oxidation, reduction, condensation, nucleophilic substitution or nucleophilic addition” i “halogeno-compounds with magnesium to form Grignard reagents and the reactions of the latter with water, carbon dioxide and carbonyl compounds” Mg (s)/dry ether/I2 (s) crystal/ heat under reflux
H2O Add water or HCl (aq) to the Grignard reagent
HCHO Add HCHO followed by HCl (aq)
Aldehydes other than methanal E.g. add CH3CHO followed by HCl (aq)
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CH3CH2Br + Mg
Mg turnings/ I2 (s)/dry ether → Heat under reflux
CH3CH2MgBr + H2O
CH3CH2MgBr + HCHO CH3CH2CH2OMgBr + H2O
CH3CH2MgBr + CH3CHO CH3CH2CHOMgBrCH3 + H2O
HCl (aq)
HCl (aq)
HCl (aq)
CH3CH2MgBr
CH3CH3 + Mg(OH)Br
CH3CH2CH2OMgBr CH3CH2CH2OH + Mg(OH)Br
CH3CH2CHOMgBrCH3
CH3CH2CHOHCH3 + Mg(OH)Br
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The preparation of a Grignard reagent which is kept in ethereal solution and used straight away This is why anhydrous reagents must be used and the reaction/ product mixture must be kept water free Reaction with methanal (followed by HCl (aq)) gives a primary alcohol Reaction with aldehydes other than methanal (followed by HCl (aq)) gives a secondary alcohol
40
Ketones E.g. add CH3COCH3 followed by HCl (aq)
CO2 E.g. bubble CO2 (g) into the ethereal solution followed by HCl (aq)
CH3CH2MgBr + CH3COCH3
→
CH3CH2CHOMgBrCH3 + H2O
CH3CH2MgBr + CO2
CH3CH2COMgBr(CH3)
CH3CH2COH(CH3)2 + Mg(OH)Br →
CH3CH2CHOMgBrCH3 + H2O
2
HCl (aq) →
HCl (aq) →
CH3CH2COOMgBr
CH3CH2COOH + Mg(OH)Br
Reaction with ketones (followed by HCl (aq)) gives a tertiary alcohol Reaction with CO2 (g) (followed by HCl (aq)) gives a carboxylic acid. “Dry ice” (CO2 (s)) could be used as an alternative form of CO2.
iv “carbonyl compounds with hydrogen cyanide, 2,4dinitrophenylhydrazine, alkaline ammoniacal silver nitrate solution, Fehling’s solution, iodine in the presence of alkali (or potassium iodide and sodium chlorate(I)), sodium tetrahydridoborate(III) (sodium borohydride) and lithium tetrahydridoaluminate(III) (lithium aluminium hydride)” HCN KCN/H2SO4 carefully buffered at pH 4/heat under reflux
2,4-DNPH 2,4dinitropheny l-hydrazine at room temperature
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CH3CHO + HCN
Carbonyl + 2,4-DNPH
KCN/H2SO4 pH 4 → Heat under reflux
→ r.t
CH3CH(CN)OH
2,4 – dinitrophenylhydrazone derivative
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Mechanism: Nucleophilic addition
The test for a carbonyl compound as in aldehydes or ketones
41
I2/NaOH “the iodoform test”
“CH3CHO
I2/NaOH →
CHI3 (s) + HCOO- “ Iodoform and the salt of a carboxylic acid with one less carbon atom than the parent compound.
“CH3COCH3
KI/NaOCl →
CHI3 (s) + CH3COO- “ Iodoform and the salt of a carboxylic acid with one less carbon atom than the parent compound.
I2 (aq) followed by NaOH (aq)/r.t. or gentle warming
KI/NaOCl “the iodoform test” KI (aq) and NaOCl (aq)/r.t. or gentle warming
NaBH4 NaBH4/H2O/ Heat under reflux
LiAlH4 LiAlH4/dry ether/ Heat under reflux
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CH3CHO + 2[H]
CH3COCH3 + 2 [H]
NaBH4/H2O → Heat under reflux LiAlH4/dry ether → Heat under reflux
CH3CH2OH
CH3CHOHCH3
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The test for the presence of a methylketone functional group (CH3CO-) e.g. in ethanal, propanone, butanone, pentan-2-one; it is also the test for the presence of a methyl secondary alcohol functional group (CH3CHOH-) e.g. in ethanol, propan-2-ol, butan-2-ol.
Reducing agent specific to carbonyl compounds Reducing agent can be used with carbonyls, acids, nitriles, amides, etc.
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K2Cr2O7/ H2SO4 K2Cr2O7 (aq)/H2SO4 (aq)/heat under reflux
Ammoniacal silver nitrate [aldehyde s only]
CH3CHO + [O]
K2Cr2O7 (aq) /H2SO4 (aq) → Heat under reflux
CH3CH2CHO + [O]
AgNO3/ NH3 (aq) → Heat
CH3CHO + [O]
“Cu2+” → Heat
CH3COOH
CH3CH2COOH
AgNO3 (aq)/NH4OH (aq)/heat
Fehling’s solution [aldehyde s only]
CH3COOH
Fehling’s solutions – essentially Cu2+
Ketones cannot be oxidised; this reaction applies to aldehydes as well as primary and secondary alcohols – the mixture turns from orange to green Ammonical silver nitrate contains Ag+ which are reduced to Ag (s) – a silver mirror
The solution turns from blue to a brick red ppt. (Cu2O)
ii “carboxylic acids with alcohols, lithium tetrahydridoaluminate(III) (lithium aluminium hydride), phosphorus pentachloride, sodium carbonate and sodium hydrogencarbonate” Na2CO3 Solid or aqueous Na2CO3/r.t.
NaHCO3 Solid or aqueous NaHCO3/r.t.
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2CH3COOH + Na2CO3
Na2CO3 (s) → r.t.
CH3COOH + NaHCO3
NaHCO3(s ) →
2CH3COONa + CO2 + H2O
CH3COONa + CO2 + H2O
r.t.
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Shows the acidic nature of carboxylic acids Shows the acidic nature of carboxylic acids
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PCl5 (s) Dry PCl5 (s)/ r.t.
CH3COOH + PCl5
PCl5 (s) →
CH3COOH + CH3CH2OH
Conc. H2SO4 → Heat
CH3COCl + HCl (g) + POCl3
The test for the presence of the C-OH group (as in alcohols and carboxylic acids)
CH3COOCH2CH3 + H2O
Esterification reaction
r.t.
[ANHYDROUS CONDITIONS]
ROH Alcohol, conc. H2SO4, heat then pour into aq. NaHCO3
LiAlH4 (s) LiAlH4 in dry ether and heat under reflux
NaOH Aqueous NaOH
CH3COOH + 4[H]
CH3COOH + NaOH
LiAlH4 (s)/ dry ether → Heat under reflux
→
CH3CH2OH + H2O
CH3COONa + H2O
iii “esters with acids and alkalis in aqueous solution” HCl Aqueous HCl/heat under reflux
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CH3COOCH2CH3 + H2O
HCl/H2O Heat under reflux →
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CH3COOH + CH3CH2OH
This is an equilibrium reaction because the reactants are an ester and water and the products are an acid and an alcohol
44
NaOH Aqueous NaOH/heat under reflux
CH3COOCH2CH3 + NaOH
NaOH/H2O → Heat under reflux
CH3COONa + CH3CH2OH + H2O
This is not an equilibrium reaction as the reactants are an ester and “water” but the products are an acid salt and an alcohol
v “ethanoyl chloride with water, alcohols, ammonia and primary amines” H2O H2O/r.t.
ROH Alcohol/r.t.
NH3 NH3/r.t.
CH3NH2 CH3NH2/r.t.
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CH3COCl + H2O
→ r.t.
CH3COOH + HCl
Acid chlorides hydrolyse in moist air
CH3COCl + CH3CH2OH
→ r.t.
CH3COOCH2CH3 + HCl
A much more vigorous esterificatio n reaction because of the reactivity and volatility of the acid chloride
CH3COCl + NH3
→ r.t.
CH3CONH2 + HCl
CH3COCl + CH3NH2
→ r.t.
CH3CONHCH2CH3 + HCl
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vii “nitriles undergoing hydrolysis and undergoing reduction” HCl
CH3CN
HCl (aq) /heat under reflux
NaOH
CH3CN
NaOH (aq) /heat under reflux
HCl/H2O → Heat under reflux
NaOH/H2O → Heat under reflux
CH3COOH
CH3COO-
Acidification of the mixture will give the free carboxylic acid: CH3COO- + H+ HCl (aq) CH3COOH
LiAlH4 LiAlH4/dry ether/heat under reflux
CH3CN + 4[H]
LiAlH4/ dry ether → Heat under reflux
CH3CH2NH2
viii “amides with phosphorus(V) oxide and bromine in aqueous alkali” P4O10
CH3CONH2
Distil over P4O10
Br2/ NaOH Br2 at r.t. then NaOH and warming
CH3CONH2 + 2NaOBr (from Br2 + NaOH)
P4O10 → distil
Br2 at r.t. then NaOH and warming
CH3CN + H2O
A dehydratio n reaction
CH3NH2 + CO2 + 2NaBr + 2H2O
This is a degradatio n reaction
vi “primary amines with aqueous hydrogen ions, acid chlorides” 33691600
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HCl followed by NaOH
CH3CH2NH2 + HCl CH3CH2NH3+Cl+ NaOH
RNH2 RNH2/r.t.
CH3COCl + CH3CH2NH2
CH3CH2NH3+Cl-
-→ r.t. -→ r.t.
CH3CH2NH2 + NaCl + H2O CH3CH2 (CONH)CH2CH3 + H2O
- r.t.
-CONH- is the amide link
ix “amino acids with acids and bases, their zwitterion structures.”
H2NCH2COOH
With acids: H+ + H2NCH2COOH
+
⇌
→
+
H3NCH2COO-
Zwitterion equilibrium – amino acids behave more like ionic salts in many ways
H3NCH2COOH
→ H2NCH2COO- + H2O H2NCH2COOH + OHWith bases: “formation of polyesters and polyamides” – condensation polymerisation
Polyesters: We need a dicarboxylic acid and a diol in order to produce a polyester: …..COOH HOCH2CH2OH HOOC(CH2)4COOH HO……
- H2O
….(COOCCH2CH2OCO(CH2)4COO)….. The reason for the diol and dicarboxylic acid is that we need active sites at either end of the chain in order to effect further polymerisation. The reactions are quicker if a dicarboxylic acid chloride is used in place of the dicarboxylic acid as the dicarboxylic acid chloride is more volatile and more reactive than the dicarboxylic acid: …..COCl HOCH2CH2OH ClOC(CH2)4COCl HO…… ↓
- HCl
….(COOCCH2CH2OCO(CH2)4COO)….. Polyamides: We need a dicarboxylic acid chloride and a diamine in order to produce a polyamide such 6:6-nylon: 33691600
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…..COCl H2N(CH2)6NH2 ClOC(CH2)4COCl H2N…… ↓
- HCl
….(CONH(CH2)6NHCO(CH2)4CONH)….. The reason for the diamine and dicarboxylic acid chloride is that we need active sites at either end of the chain in order to effect further polymerisation.
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