Chem 5 things to learn

some flashcards of things you have to rote learn for chem 5

HideShow resource information
  • Created by: Ella Shaw
  • Created on: 17-06-12 09:02

Finding the enthalpy of solution

Lattice dissociation enthalpy (always endothermic- watch out they may trick you by giving enthalpy of lattice formation which is exothermic)

Hydrating the positive ions (exothermic)

Hydrating the negative ions (exothermic

1 of 8

Periodicity of Period 3 (part 1)

Atomic Radius- Decreases. An extra election being added to the same principal shell. Nuclear charge increases. No extra shielding. Attraction between positive nucleus and negative electrons greater.

First Ionisation energy- Increase from Na-Ar. No extra shielding. More attraction between positive nucleus and negative out electron. More energy required to remove. Al is lower than Mg. Al the outer electron goes from 3p to 3s. Mg removing from 3s. 3p has higher energy than 3s in Mg. so easier to remove from 3p as there is less attraction. S is lower than P. S has 2 paired electrongs of opposite spin in 3p sub-level. electrons repel each other so it takes less energy to remove one of them than to remove the unpaired from P as it is more easily lost. 

Electronegativity- Increases. Nuclear charge increasing. no extra shielding. more attraction between positive nucleus and bonding pair of electrons

Electrical conductivity- Na-Al high conductivity. they have a large number of free, delocalised outer electrons which are free to move and carry the charge as they have metallic bonding. Si-Cl low conductivity. Non-metals. no free electrons to carry the charge. 

2 of 8

Periodicity of Period 3 (part 2)

Electrical conductivity- Na-Al increase. Na atoms only one outermost electrong. Mg two, Al three. Biggest in Al as it has most delocalised electrons to be able to move and carry the charge

Melting points and boiling points- Na-Al increase. Particles have to break free of the forces holing them together. electrostatic attraction for Na-Al between delocalised electrons and positive ions. Melting point increases as attraction increases. Na+ V Al3+ greater charge on cation and Na 1 V Al 3 delocalised electrons. Si very high.  Giant covalent structure. held together by many strong covalent bonds. need to break which needs a lot of energy. P4, S8, Cl2. simple molecules. melting point determined by VdW forces. S8 is biggest and has stroger VdW forces. Ar lowest. Noble gas. Inert. Exists as single atoms. Very few VdW forces between atoms. only small amount of energy needed to break them 

3 of 8

Redox equilibria- Altering the reaction conditions

Changing the cell current- if current is allowed, emf drops until i reaches zero when chemical equilibrium has been established. e.g Cu2+ + Zn --> Cu + Zn2+. Cu2+ RHS, Zn LHS. if current is allowed, Zn (reducing agent) loses electrons to form Zn2+. eventually conc. so high that it sends the reaction backwards to equilibrium. At the same time Cu2+ --> Cu. conc of Cu eventually becomes so high that it sends the reaction backwards to equilibrium 

Changing the temperature- if the temp is increased above 298K equilibria will shift towards the endothermic reaction. most cell reactions with positive e.m.f values are exothermic so an increase will favour the backwards reaction and therefore decrease the e.m.f

Changing concentration: Effect on e.m.f- Cu2+ + Zn --> Cu + Zn2+ . Inc Zn, inc e.m.f. inc Cu2+, inc e.m.f. Inc Cu, dec e.m.f. Inc Zn2+, dec e.m.f.                                                                   : Effect on electrode potential- Inc. Cu2+ more positive electrode potential. better at gaining electrons. Inc. Zn more negative electrode potential. Better at losing electrons 

4 of 8

Commercial cells (part 1)

Primary- irreversible and not rechargeable by an electric current

Secondary- reversible and is designed to be recharged by an electric current after they have been run down

Fuel Cell- generated electricity from the continuous oxidation of gaseous or liquid fuels

Non-rechargeable cells- the daniel cell (zinc/copper). a wet cell with emf 1.10V, impractical for portable devices due to liquid. The Leclanche cell (zinc/carbon). a dry cell emf of up to 1.5V, anode is a zone can containing ammonium chlorides and zinc chloride paste, cathode is a mixture of powdered manganese oxide and graphite surrounding the graphite rod, at the anode zinc is oxidised and the zinc ions react with ammonia which is produced when hydrogen ions are released from ammonium ions in the electrolyte, at the cathode manganese is reduced and involves the hydrogen ions that are released from the ammonium ions. Alkaline dry cells. still have emf of up to 1.5V but last longer as zinc not attacked by acidic electrolyte, KOH electrolyte not ammonium/zinc chloride, at anode zinc is oxidised and zinc ions react with hydroxide ions, at cathode manganese is reduced

5 of 8

Commercial cells (part 2)

Non-rechargeable cells- lithium cell. used in digital cameras and watches, light due to Li being light, at cathode Mn is reduced, at anode Li is oxidised, large emf go 3.5-4.0V as Li has a very negative reduction potential

Rechargeable cells- provide current to an external circuit whilst discharging (galvanic action) and use current whilst charging (electrolytic action). lead-acid batteries. Used in starter motors of cars, consist of 6 2V cells to give 12V, each cell is made up of a positive lead plate coated with PbO2 and a negative plate of lead, plates are dipped in a solution of sulphuric acid, on recharging the reverse reactions take place. Nickel Cadmium. most common portable battery used in tv remotes, emf 1.2V, anode made from cadmium, cathode from nickel hydroxide, potassium hydroxide is the electrolyte. Lithium ion. light so used in phones and cameras, high emf of 3.7V, cell reactions complex including lithium cobalt oxide, anode LiC6, cathode LiCoO2.

Fuel cell- uses hydrogen and oxygen fuels to produce electricity, reactants can be supplied continuously, alternative to burning fossil fuels, supply of hydrogen is usually from methane, it is possible for it to be carbon neutral if acidified water is electrolysed using renewable energy. 

6 of 8

Transition metals

Planck's equation- E=hv where E is the difference in energy in J, h is planck's constant, v is the frequency of the electromagnetic radiation in Hz

The colour of the transition metal complex depends upon the degree of splitting of the 3d orbital and hence E, this is affected by: the transition element, The oxidation state of the transition metal, the ligand, the coordination number

Colorimetry method- A calibration graph is produced by measuring the absorbance of solutions of the ion at known concentrations. the absorbance of the sample is then measured and the concentration can then be read off the calibration graph. if the concentration of the solution is too low to absorb enough light, the intensity of colour can be increased by adding a ligand e.g EDTA 

Finding the formula of a transition metal complex using colorimetry- 2 solutions are mixed in different proportions one containing the metal ion one containing the ligand, when they are mixed in the same ratio as they are in the complex there is the maximum concentration of complex in the solution and the absorbance will be highest i.e the absorbance doesn't increase any more. 

7 of 8

Transition metals- catalysis

The Haber Process- manufacture of ammonia: N2 + 3H2 --> 2NH3. the catalyst is iron in pea sized lumps to increase the surface area, it lasts about 5 years before it is poisoned by sulphur impurities in the methane gas stream 

The Manufacture of Methanol: CO + 2H2 --> CH3OH. catalyst is Cr2O3 in a mixture with zinc and/or copper oxides, methanol is important in the production of plastics e.g. Bakelite

The Contact Process- Manufacture of Sulphuric Acid: 2SO2 + O2 --> 2SO3. the catalyst is vanadium (V) oxide V2O5, the reaction occurs in 2 steps. step 1: SO2 + V2O5 --> V2O4 + SO3. step 2: V2O4 + 1/2O2 --> V2O5. each of the two steps has lower activation energy than the uncatalysed single step so reaction goes faster

The oxidation of iodide ions by peroxodisulfate ions catalysed by Fe2+: the reaction is so slow because both ions are negatively charged and so repel each other. there are 2 steps which can occur in either order 

8 of 8


No comments have yet been made

Similar Chemistry resources:

See all Chemistry resources »See all resources »