A2 Chemistry Areas of Difficulty Misconceptions

8/8/2019 A2 Chemistry Areas of Difficulty Misconceptions

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EDEXCEL CHEMISTRY

A2: areas of difficulty and

misconceptions



Energetics• Definitions of lattice energy and enthalpy of atomisation.

• Factors which effect lattice energy (forget size match).

• Born Haber calculations .

• Solubility including Hess’s Law diagram

Group 2 hydroxide.Group 2 sulphates.



Solubility of calcium bromideThis is a balance between lattice energy and the hydration energies

of the ions.

Ca2+(g) + 2Br-(g)

-∆Hlatt ∆Hhyd 2 x ∆Hhyd

∆Hsolution

CaBr2(s) Ca2+(aq) + 2Br-(aq)

∆Hsoln = -∆Hlatt + ∆Hhyd(Ca2+) + 2 x ∆Hhyd(Br-)



Periodic Table• Know the equations.

• Explain reactions in terms of metallic character and

bonding.

• Understand the meaning of “amphoteric”.

• Hydrolysis of CCl4 and SiCl4 in terms of d orbitals (carbonhas no 2d orbitals) and steric factors.



Equilibrium• Expressions for Kc and for Kp.

•

• Units of Kc and Kp.

• What to ignore.

• Calculations – do it in a table.

You must use equilibrium moles.

For Kc convert moles to concentrations.

For Kp convert moles to mole fraction to partialpressure.



Question10 mol of SO2 and 10 mol of O2 were allowed to reach

equilibrium under certain conditions in a vessel of

pressure 2.0 atm. 80% of the SO2 reacted. Calculate thevalue of Kp under these conditions.

2SO2(g) + O2(g) ⇌ 2SO3(g)



Answer

2SO2 + O2 ⇌ 2SO3

Initial moles 10 10 0

Equilibrium moles 10 - 8 = 2 10 - 4 = 6 8

Total moles 2 + 6 + 8 = 16

Mole fraction at

equilibrium

2/16 = 0.125 6/16 = 0.375 8/16 = 0.5

Partial pressure = mole fraction x total pressure (2 atm)

0.25atm 0.75atm 1.0atm

K p = p(SO3)

2

= 1.0

2

= 21 atm

-1

p(SO2)2 p(O2) 0.252 x 0.75



Effect of conditions on the value of K

• K only equals fraction when at equilibrium.

• Increase in temperatureFor exothermic reactions K decreases.

For endothermic reactions K increases.

So fraction has to alter and so position moves.

• Increase in pressure

No effect on K, but fraction changed, so position shifts until

fraction again equals K.

• Catalyst

No effect on K nor on fraction.



Acid - base equilibrium

• pH = - lg[H+]

• pOH = - lg[OH-

]

• pKa = - lg Ka

• [H+] = 1 x 10-14 / [OH-] or pH = 14 – pOH

• Ignore [H2O] in Ka expression.

• Base all calculations on this expression.

• If solution of acid [H+] = [A-]

• If buffer [HA] = [weak acid] and [A-] = [salt]



Titration curves

• Read the question carefully & estimate the following

• The pH at the start (acid or base; strong or weak).

• Rule of 2.

• The pH at the equivalence point.• The pH range of the vertical part of the graph.

• The volume at the equivalence point.

• The final pH (strong or weak).• Choice of indicator.



Organic chemistry

• Grignard reagents

• Made by reaction of halogenoalkanes withmagnesium in dry ether.

• React with carbonyl compounds to form alcohols.

• React with solid carbon dioxide to form carboxylic

acid.

• Hydrolysis with dilute acid necessary.



Carbonyl compounds

• Aldehydes and ketones.

• Prepared by oxidation of alcohols.

• Test with 2,4-dinitrophenylhydrazine.

• Both react with HCN and with LiAlH4.

• Aldehydes with Fehlings and Tollens.

• Iodoform reaction.



Organic nitrogen compounds

• Amines

• Preparation from halogenoalkanes and from nitriles.• Reaction with H+.

• Reaction with acid chlorides.

• Be able to write repeat unit for polyamides (andpolyesters).

• Amino acids

• Formula.

• Reactions with acids and with bases.

• Zwitterion and physical properties.



Reaction mechanisms

• Meaning of curly arrowsStart on atom or bond.

End between atoms forming a bond or on an atom

forming a negative ion.

• Smiley benzene ring.



Redox equilibrium

• Overall equations always have both reactants on the

right.

• Half equations normally written as reductions (electrons

on left).

• Calculation of Ecell

• Feasibility of reaction; Ecell > 0



Transition metals

• Origin of colour : light absorbed as electron promoted.

• Deprotonation reactions of aqua ions (including amphoteric

behaviour)

[Cu(H2O)6]2+

+ 2OH- →

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

• Ligand exchange reactions

[Cu(H2O)4(OH)2] + 4NH3→ [Cu(NH3)4]2+ + 2OH- + 4H2O



Kinetics

• Give reasons when calculating order from initial rate

data.

• Don’t forget the rate constant k in rate equations.

• Total order = sum of the powers to which the

concentrations are raised in the rate equation.



Organic synthesis

• If the number of carbon atoms in the chain increases,use KCN or HCN or a Grignard.

• Halogenoalkanes react with KCN not HCN.

• Alcohols react with neither HCN nor KCN.

• If the number of carbon atoms decreases, try Hofmann

degradation (of amides), or oxidising an aromatic sidechain to COOH, or the iodoform reaction.

• If you cannot see how to start, work backwards.



Spectra

• NMR gives peaks due to hydrogen atoms being in

different environments. The chemical shift depends

on the environment and the peak height on thenumber of H atoms in that environment.

• IR gives peaks due to the stretching of chemicalbonds. Look out for C=O (at around 1700 cm-1) and OH

(at around 3200 cm-1).

• Mass spectra. Look out for the peak due to the

molecular ion. This gives you the Mr. Also peak at

(Mr – 15) caused by loss of CH3.



Resources

• Your teacher.

• Your notes.

• The specification (syllabus).

• Make the Grade (Nelson Advanced Science ISBN 0 7487

7281 2).

• Do Brilliantly A2 (Harper Collins ISBN 0 00 7124 21 X).



and finally

Good luck in your exams

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A2 Chemistry Areas of Difficulty Misconceptions