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Edexcel GCE Student Conference 2006

EdexcelA2 Chemistry



Edexcel GCE Student Conference 2006 – A2 Chemistry

Enthalpy

Know your definitions• All are enthalpy (or energy) change per mole.

• Atomisation is for 1 mol of atoms from the element in itsstandard state.

½ Br 2(l) → Br(g) (endothermic)

• Lattice energy is for 1 mol of an ionic solid being formed

from its gaseous ions.Mg2+(g) + 2Br -(g) → MgBr2(s) (exothermic)




Solubility

1 = -LE 2 = ∆Hhyd (Mg2+)

3 = 2 x ∆Hhydr (Cl-) 4 = ∆H

soln = (by Hess’s Law) 1 + 2 + 3

∆H soln = -LE + ∆H hyd (Mg2+) + 2 x ∆H hydr (Cl-)

Given any three, calculate the fourthThe more exothermic ∆ H

soln , the more soluble

Mg2+(g) + 2Cl -(g)

MgCl2(s) Mg2+(aq) + 2Cl -(aq)

1 2 3

4




Changes of solubility in a Group• Lattice energy gets less exo (decreases) as the ionic

radius increases.

• Hydration energy also gets less exo (decreases) as

ionic radius increases.

• If anion big (sulphate), decrease in HE > decrease inLE, so solubility decreases.

• If anion small (hydroxide), decrease in LE > decreasein HE, so solubility increases.




Periodic Table• Know the equations for the reactions in the

syllabus.• Stability is a comparison.• Tin is more stable in the +4 state than in the +2 state,

but the other way round for lead.• Tin(II) is a reducing agent; tin(IV) is not an oxidising

agent.• Lead(IV) is an oxidising agent; lead(II) is not a

reducing agent.




Equilibrium• For both Kc and Kp first calculate equilibrium moles

using stoichiometry in equation, then

Calculate value and give unitsCalculate value and give units

Multiply by total pressure andsubstitute into Kp expression

Substitute into Kc expression

Divide by total equilibriummolesDivide by volume

Kp

Kc




• PCl5 is 25% dissociated at a pressure of 2.0 atm and at acertain temperature. Calculate the value of Kp.

Units are

p(PCl 5)

==p(PCl 3) x p(Cl 2)Kp =

Partial pressure

Mole fraction

Equilibrium moles

+ Cl2(g)⇌ PCl3(g)PCl5(g)




pHA weak acid is slightly ionised

HA ⇌ H+ + A- & Ka = [H+] x [A-][HA]

Assumptions made

• pH of a weak acid:[H+] = [A -] & [HA] eq = [HA] initially

• Buffer solutions[A-] = [salt] & [HA] = [weak acid]




Organic• Know your reactions, reagents and conditions.

• Carbonyl compounds+ HCN (at pH 8) adds H to O and CN to C+ 2,4-dnp to give an orange ppt with a >C=N- group+ NaBH4(aq) or LiAlH 4 (in dry ether) to give an alcohol

• Aldehydes reduce Fehlings to a red ppt and ammoniacal silvernitrate to a silver mirror

• Methyl ketones and ethanal do the iodoform reaction




Redox• Electrode potential equations always written asreduction.

Calculation of E 0 and feasibility• Rewrite one equation so that the reactant is on the

left – change the sign of its E 0 value.

• Add to the other electrode potential equation.

• Add E0

values: if positive the reaction is feasible.

• Overall equation: make sure electrons cancel.




Will dichromate(VI) ions oxidise chloride ions in acidsolution?

Cr2O

7

2- + 14H + + 6e - → 2Cr3+ + 7H2O E0 = +1.33V

Cl2 + 2e - → 2Cl- E0 = +1.36V

Which are the reactants? – they must end up on the left




Transition elements• Definition: they have one or more ions with an half-

filled d -orbital.

Coloured complex ions

• Ligand splits d -orbitals into two different energylevels.• Some of the white light is absorbed and its energy

promotes an electron to the higher level.• Light going through becomes coloured because some

colours removed from the white light.




Deprotonation and ligand exchange• Addition of base to hydrated ion - deprotonation

[Cu(H 2O)6]2+

+ 2OH- →

[Cu(H 2O)4(OH)2] + 2H 2OOr

[Cu(H 2O)6]2+ + 2NH3 → [Cu(H 2O)4(OH)2] + 2NH 4

+

Blue precipitate formed

• Addition of ammonia to precipitate – ligand exchange

[Cu(H 2O)4(OH)2] + 4NH 3 →

[Cu(NH3)4(H2O)2]2+

+ 2OH-

+ 2H2ODeep blue solution formed




Kinetics – initial rates• Consider the reaction: A + 3B → 2C + D

18 x 10 -40.300.203

2.0 x 10 -40.100.202

1.0 x 10 -40.100.101

Rate/mol dm -3 s-1

[B]/ mol dm -3

[A]/ mol dm -3

Experiment

• Deduce the order of reaction with respect to A and BWhen [A] doubles, the rate also doubles, so the order with respect to A

is one.When [B] goes up by a factor of 3, the rate increases by nine times, sothe reaction is third order with respect to B, as 3 x 3 = 9.

Full marks????




Half-lives• Use the graph below to calculate two successive half-lives

and hence deduce the order of the reaction.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 10 20 30 40 50

time

A




Arrhenius equationUse the graph to calculate the slopeand then the activation energy.

[R = 8.3 J K -1 mol -1]T1

RE

Alnkln a⎥⎦

⎤

⎢⎣

⎡−=

lnk

-9.5

-9

-8.5

-8

-7.5

-7

0.003 0.0031 0.0032 0.0033 0.0034

1/T

l n k




Please take a short break

and be back, seated, by12 noon




Synoptic1. A synoptic question can be based on any part of the

specification, AS as well as A2.

2. It often draws different parts of the specificationtogether in the same question, for example:

Kinetic data of the hydrolysis of a halogenoalkaneused to find order, then mechanism asked which is

consistent with the order you have just calculated.




Synoptic in papers 5 and 6BPaper 6245

• Any of AS and A2 organic maybe asked, but the emphasiswill be on Unit 5 organic.

• Questions may link differenttopics within Unit 5, forexample kinetics andmechanisms.

Paper 6246B – The synoptic paper

• All AS and all A2.

• Section A based on a quantitativeexperiment.

• Section B a choice of questiondesigned to test your thinking skills.

• Questions will link different topics,for example organic and weak acidpH.




Mechanisms – addition reactions

Add the curly arrows

Nucleophilic addition (to carbonyl compounds)

(CH3)2C=O

:CN-

(CH3)2C

CN

O-

H CN

+ CN-

(CH3)2C

O-

CN (CH3)2C

O-H

CN




Electrophilic addition (to alkenes)


+ Br-H Br

H2C = CH2

H3C CH2

+

H3C CH2+

:Br-

H3C CH2Br




Electrophilic substitution (to benzene)

FeBr3

+ Br2

→ Br+ + FeBr4

-

Br+

+

H

Br

Br

+ H++

H

Br




Nucleophilic substitution (to halogenoalkanes)


HO:-

SN2 HHC

CH3

Cl HO ----C-----Cl

HH

CH3

CH3CH2OH + Cl -




Nucleophilic substitution (to halogenoalkanes)


SN1

C

CH3

CH3

H

Cl C+

HCH3

CH3

+ Cl-

C

CH3

CH3

H

OH

C+

HCH3

CH3

:OH -

Step 2




Synoptic organic problem

A compound C 4H8O2 was tested as follows:1. It gave no precipitate with Brady’s reagent.

2. It turned bromine water from brown to colourless.

3. When excess PCl 5 was added to it 2 mol of HCl was givenoff per mol of the organic compound.

4. The compound has three peaks in its nmr spectrum.

5. Suggest a formula for the compound. Justify youranswer.




Synoptic multi topic question

a) Write the structural formulae of the organic products of

reacting 3-oxobutanoic acid, CH 3COCH2COOH, with:

i. Lithium aluminium hydride in dry ether (2)

ii. Hydrogen cyanide at pH = 8 (1)

iii. Aqueous iodine and sodium hydroxide (2)

Then ……




b) Write the formula of the conjugate base of 3-oxobutanoic acid.

Calculate the pH of a solution made by adding 25 cm 3 of 1.0 mol dm -3 3-oxobutanoic acid to 25 cm 3 of 0.50 moldm -3 sodium hydroxide. K

a for the acid = 2.6 x 10 -4 moldm -3 .




Shapes and boiling points

Shapes• Count number of bond and lone pairs.

• These electron pairs repel to a position of maximumseparation. This gives the shape.• lp/bp repulsion (if applicable) > bp/bp. This would reduce

the bond angle.

Boiling points• Identify the intermolecular forces.• Compare their strengths.• Relate strength to energy required to separate particles.




a) Explain the shape of the sulphur(II) chloride, SCl2,

molecule and its bond angle.

No of bond pairs =

No of lone pairs =

Shape is

Bond angle is

Then ….




Therefore …Whichstronger?

Imf in HClImf in HF

Therefore …Which

stronger?

Imf in SF 2Imf in SCl 2

b) Explain why SF2

has a lower boiling temperature than SCl2but HF has a higher boiling temperature than HCl.




I hope that this has been of

interest and use to youGOOD LUCK IN YOUR EXAMS

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