A-Level Chemistry Helpful Tips/Exam technique UPDATED

• Way to remember in calculations if its (products - reactants) or (reactants - products) for enthalpy change of formation ,standard entropy ,combustion and bond enthalpy

  FPAR(products-reactants) ΔfH

  SPAR(products-reactants) ΔS

  CRAP(reactants-products) ΔcH

  BRAP(reactants-products) Δ bond enthalpy
• Conversion between cm^3 dm^3 m^3


  CDM remember this (Ngolo Kante)

  
  

  cm^3 → dm^3 → m^3

  
  

  going forwards multiply by 1000

  going backwards divide by 1000

  
• Converting between mg and g and understanding how much a tonne is

  mg → g

  going forwards divide by 1000 from mg to get into g

  
  

  1 tonne= 1000kg = 1,000,000g
• Explaining shapes/bond angle- EXAM TECHNIQUE

  -Electron pairs repel eachother as far away as possible

  -This molecule(s) has/have x bonding pairs and y lone pairs

  -Therefore the shape is...

  -Lone pairs repel more than bonding pairs

  -Therefore the bond angle is ….

  -If there’s a double bond involved, refer to areas of electron density instead of electron pairs. If there’s a double bond involved, talk about x single bonds, y double bonds and z lone pairs.
• Explaining why a molecule is (non) polar- EXAM TECHNIQUE

  if the bonds are polar the molecule is asymmetrical

  – so the dipoles don’t cancel out

  - if the bonds are non-polar the molecule is symmetrical

  – so the dipoles cancel out
• How do London forces arise- EXAM TECHNIQUE

  -Electrons move

  -They can get into an uneven distribution

  -This creates an instantaneous dipole in an atom/molecule

  -This induces dipoles in neighbouring atoms/molecules.
• Explaining the trend in first ionisation energies- EXAM TECHNIQUE

  • Moving from ….. to …..
  • The nuclear charge increases/decreases(however?)
  • The distance between the nucleus and the outermost electron(s) increase/decreases
  • Shielding remains the same/increases/decreases
  • Nuclear attraction for the outermost electron increases/decreases
  • The energy needed to remove an electron increases/decreases.
• Explaining dip from Beryllium to Boron (or Mg to Al) - EXAM TECHNIQUE


  outer electron is in a p subshell, which is higher energy than the s subshell of Be/Mg therefore less energy is required to remove the electron.

  
• Explaining the dip from Nitrogen to Oxygen(or P to S)- EXAM TECHNIQUE

  outer electron in N/P is in a p orbital on its own (p3 ) whereas O/S has a p4 configuration with two electrons in one orbital. Spin pair repulsion reduces the energy needed to remove an electron.

  
• Explaining the electrical conductivity of aqueous ionic compounds

  Aqueous salts conduct because they have mobile ions.

  Ensure to say mobile(free mobile charged carriers)
• Why is it unlikely that a reaction could take place in one step- 2 REASONS

  -stoichiometry in the rate equation does not match the stoichiometry in the overall equation

  -collisions are unlikely with more than 2 or more ions/species/particles
• Kc change with concentration/temperature- EXAM TECHNIQUE


  1. Kc does(n’t) change – it only changes with temp

  2. Increase/decrease of [X] increases/decreases the numerator/denominator of the Kc expression

  3. (Refer to exo/endothermic reaction if relevant)

  4. So more X/Y is made to decrease/increase the numerator/denominator to restore/increase /decrease the equilibrium Kc.
• Kc change with pressure- EXAM TECHNIQUE

  1. Kc doesn’t change – it only changes with temp

  2. The concentrations of (all reactants and products) increase/decrease

  3. However, the numerator/denominator increases/decreases more/less

  4.So more X/Y is made to decrease/increase the numerator/denominator to restore the equilibrium Kc.
• Factors to consider when explaining reactivity down/up groups 2 and 7

  -Distance between nucleus and electron(s) to be lost/gained

  -Increased/decreased shielding

  -Increased/decreased nuclear attraction

  -Easier/harder to remove/gain an electron
• Explaining indicator colour changes HA <=> H+ + A--EXAM TECHNIQUE


  Acid: [H+] increases

  H+ + A- -> HA

  equilibrium shifts to the left therefore becomes (red)

  Alkali: [H+] decreases/ [OH-] increases

  H+ + OH- -> H2O

  HA -> H+ + A-

  equilibrium shifts to right therefore becomes (blue)
• How does a buffer solution work- EXAM TECHNIQUE

  Acid: [H+] increases

  H+ reacts with A- to form HA

  equilibrium shifts to the left removing most of the H+ ions

  Alkali: [OH-] increases

  OH- reacts with H+ to form H2O

  [H+] decreases and HA dissociates to form more H+

  equilibrium shifts to right to restore the concentration of H+ ions
• What effect will concentration changes have on a half-cell's electrode potential value- EXAM TECHNIQUE

  1) [X] increases/decreases

  2) equilibrium shifts to the right/left

  3) therefore reducing/oxidising [X]

  4) and absorbing/releasing more electrons

  5) and making E more positive/negative

  6) if talking about cell potential, talk about the difference whether its smaller or larger
• What makes a fuel cell different from an ordinary cell? (COMMON QUESTION)

  A fuel cell uses energy from the reaction of a fuel with oxygen to generate a voltage
• Haemoglobin ligand substitution- EXAM TECHNIQUE

  -In the lungs, the iron (II) ion of haemoglobin forms a coordinate bond with O2

  -O2 is then released in the tissues (which is then replaced immediately by H2O/CO2)

  -CO binds to same site as O2, replacing the O2

  -CO binds more strongly than O2, so the ligand substitution is irreversible: the haemoglobin is useless

  -Tissues become starved of oxygen
• Reducing percentage uncertainty- 2 REASONS

  -use greater quantities e.g. a larger mass or larger volume of water

  -use equipment with greater precision
• When you calculate Delta S (entropy) what must you remember to do


  Divide by 1000 to get it from J to kJ

  
• Purifying a solid- EXAM TECHNIQUE

  -Filter reaction mixture under reduced pressure

  -(recrystallisation)Dissolve impure product in minimum volume of hot solvent

  -Leave to cool and allow crystals to form

  -Filter crystals under reduced pressure again

  -Rinse them with ice -cold distilled water

  -Leave them to dry on a dish

  -Decant

  -Record the melting point if you want to identify the product
• Purifying a liquid- EXAM TECHNIQUE

  -Allow to cool

  -Decant into separating funnel and run off aqueous layer

  -Add Na2CO3 until bubbling stops to remove excess acid and run off aqueous layer again

  -Run into beaker

  -Add anhydrous MgSO4 to dry

  -Redistil and collect at boiling point
• Benzene's relative resistance to bromination compared to alkenes- EXAM TECHNIQUE

  1. The pi- electrons are delocalised in benzene, but localised in alkenes
  2. The pi- electrons are therefore more spread out benzene (lower pi-electron density)
  3. Therefore, bromine is polarised less by benzene …4.… and attracted less. (less susceptible to an electrophilic attack)
• Why phenol does bromination more readily than benzene- EXAM TECHNIQUE

  1. The lone pair from the -OH group in phenol is delocalised into the benzene ring. (spelling)
  2. So the electron density increases…
  3. .…and the electrophile is more strongly attracted.(more susceptible to an electrophilic attack)
• Mass spec fragments: you must always...

  input the + charge next to the fragment

  e.g. CH3CH2+, CH3CO+ etc.
• Explaining stereoisomerism (or lack of it) in alkenes- EXAM TECHNIQUE

  -There is restricted rotation around the C=C group

  -Each C atom in that group is/n’t bonded to two different atoms/groups
• Markownikoff's rule: explain why one product is more abundant than the other- EXAM TECHNIQUE

  the major product is formed from a secondary/tertiary carbocation intermediate.

  The minor product is formed from a primary/secondary carbocation intermediate, which is less stable.