Chemistry Page

Linear Shape

• BeH2
• 180°
• 2 bonding pairs

Trigonal Planar

• BF3
• 120°
• 3 bonding pairs

Tetrahedral

• CH4
• 109.5°
• 4 bonding pairs

V-Shaped

• H2O
• 104.5º
• 2 bonding pairs and 2 lone pairs

Pyramidal

• NH3
• 107º
• 3 bonding pairs and 1 lone pair

Electron Pair Repulsion Theory the electron pairs in the outer shell of the central atom repela each other and end up as far apart as is geometrically possible

Since lone pairs are closer to the nucleus of the central atom, they are closer together, so their mutual repulsion is greater than that between bond pairs >>>> distorted shape

Order of Strength of Repulsions

1. lone pair:lone pair
2. lone pair:bond pair
3. bond pair: bond pair

Symmetry and Polarity

Symmetrical Atoms

• Linear
• Tetrahedral
• Trigonal Planar
• Non- polar (even if bonds are polar)

E.g. BF3

Boron (central atom) is slightly positive >>> centre of positive charge is here

Each flourine is slightly negative >>> centre of positive charge is the central point between these atoms

Centre of positive coincides with centre of negative >>> non-polar

Non-Symmetrical Atoms

• V-Shaped
• Pyramidal
• Polar

E.g. NH3

Nitrogen (central atom) is slightly negative >>> centre of negative charge will be at apex of pyramid

Hydrogen atoms areslightly positive >>> centre of positive charge at base of pyramid

Centre of negative does not coincide with centre of positive >>> polar

Chemistry Page

Catalysts work by providing an alternative reaction route with a lower activation energy.

Intermediate Formation Theory of Catalysis

The reactant molecules and the catalyst form and unstable intermediate complex that breaks up to form products and regenerate the catalyst

How does the reaction of aqueous cobalt (II) chloride between H2O2 and potassium sodium tartrate give evidence for the intermediate formation theory of catalysis?

1. Initial solution is pink
2. During reaction, there is a colour change from pink to green [intermediate complex formed]
3. Frothing and bubbling reaction [fast rate of reaction - products formed]
4. Reaction returns to pink [catalyst regenerated]

Surface Adsorption Theory of Catalysis

The reactant molecules adsorb onto solid catalyst where the greater local concentration leads to a quick reaction – bonds formed must be strong enough to adsorb and increase concentration, but weak enough to decompose quickly and form products

How does the oxidation of methanol using a hot platinum catalyst provide evidence for the surface adsorption theory of catalysis?

1. Series of mild explosions and glowing platinum [fast exothermic rate of reaction]
2. H atoms are removed more quickly by reactant adsorbing to surface of catalyst which weakens and breaks bonds.

Catalytic Poisons

When the active sites of the catalyst are blocked by substances bonding to it more strongly than the reactants, the catalyst is poisoned

e.g. Lead, arsenic, sulfur

Chemistry Page

A catalyst is a substance that alters the rate of a chemical reaction but is not consumed in the reaction

Heterogeneous catalysis involves reactants in different physical states i.e. liquid reacting with solid e.g.  MnO2 on H2O2

Homogeneous catalysis involves reactants in same physical states i.e. both in aqueous solution e.g. potassium iodide on H2O2

Enzymes are homogeneous biological catalysts e.g. amylase on starch

Autocatalysis occurs when the product of a reaction increases the reaction rate i.e. reaction makes its own catalysts e.g. reduction of manganate (VII) ions with Fe 2+

Activation Energy

Activation energy is the minimum energy required by particles colliding to cause a reaction

Exothermic reactions give out heat [Energy of: products < reactants ]

Endothermic reactions take in heat [Energy of: products > reactants ]

Average kinetic energy of particles is directly proportional to the temperature – greater the energy, greater the speed, greater the reaction rate. This means:

1. the number of collisions per second increases
2. each collision is more energetic and a higher proportion of collisions has the necessary activation energy

The second factor is more significant

Pollution and Catalytic Converters

Engines produce harmful CO, NO, NO2 and hydrocarbons.

Catalytic converters (e.g. palladium and platinum) speed up reactions to reduce harmful emissions

e.g. 2CO + O2 >>> 2CO2

 2CO + 2NO >>> 2CO2 + N2

This is an example of heterogenous catalysis.

Chemistry Page

Rate of Reaction is the change in concentration in unit time of any one reactant or product

Average Rate – Change in Concentration

                     Time taken

Instantaneous Rate of Reaction is the rate at a particular point in time during the reaction

To calculate the instantaneous rate of reaction

1. Draw a tangent to the curve
2. The tangent is the hypotenuse to a right angled triangle
3. θ = angle by tangent to the horizontal axis
4. Slope = tan θ = 18.5/1.2 = 15.4 cm³/min

On graphs:

• The reaction time is inversely proportional to concentration (x axis = time) i.e. the shorter the time, the greater the concentration
• When using the reciprocal (1/time) the reaction time is directly proportional  to concentration i.e. the greater the time, the greater the concentration

Factors Affecting Reaction Rate

1. Temperature

• Increase in temperature – increase in reaction rate
• In some cases, an increase of 10 K can up to double reaction rate

2. Concentration

• Increase in concentration – increase in reaction rate

3. Particle Size

• Smaller particles – increase in reaction rate
• Finely divided solids – Greater surface area for reaction to occur over

Dust Explosions

The factors for a dust explosion to occur are:

1. Finely divided combustible particles
2. An enclosed space
3. Enough oxygen to sustain combustion
4. A spark to ignite combustion

4. Nature of Reactants

• Covalent bonds – slow reaction
• Bonds must be broken before reaction can occur – Rate will depend on bonds (Single bonds easier to break than double bonds)
• Ionic bonds – Quick in aqueous solution
• Ions pulled apart by water – instant reaction

5. Presence of Catalysts

• Can speed up or slow down reaction
• Catalyts work by lowering the energy required for a reaction to occur

Chemistry page

Oxidation Number the charge an atom appears to have when electrons are distributed according to certain rules

Rules for assigning charges

• Free elements: 0 e.g. O2 – o
• Sum of oxidation numbers: 0 e.g. H2O – o
• Oxidation numbers equals charge on ion e.g. Fe3+ – +3
• Sum of oxidation numbers in a complex ion equal to charge e.g. NO3- = +5 +3(-2) = -1
• Hydrogen: +1 except hydrides: -1
• Oxygen: -2 except H2O2 – O: -1   and   F2O – O:+2
• Group 1: +1, Group 2: +2…etc.
• Halogens: -1 (when bonded to less electronegative) e.g. NaCl
• Transition metal may have many oxidation numbers

What is the oxidation number of C in C6H12O6?

C6H12O6

6x+12(+1)+6(-1)=0

6x+12-12=0

x=0

Oxidation is an increase in oxidation number

Reduction is a decrease in oxidation number

What is oxidised and what is reduced in ZnS + 2O2 >>> ZnSO4

ZnS + 2O2 >>> ZnSO4

+2 -2  2(0)>>>+2 +6 4(-2)

S: -2>>-6 oxidised

O: 0>>-2 reduced

Balance the equation MnO4- + Fe2+ + H+ >>> Mn2+ + Fe3+ +H2O

MnO4- + Fe2+ + H+ >>> Mn2+ + Fe3+ +H2O

+7 4(-2)  +2      +1       +2     +3     2(+1)-2

Mn: +7 >> +2 : +5 e-

Fe: +2 >> +3 : -1e-

Mn + 5e- >> Mn

5Fe – 5e- >> Fe

Mn + 5Fe    Ratio 1:5

MnO4- + 5Fe2+ + H+ >>> Mn2+ + 5Fe3+ + H2O

MnO4- + 5Fe2+ + 8H+ >>> Mn2+ + 5Fe3+ + 4H2O

Chemistry page

This is the theory section of acid-base titrations.
Experiments make up the majority of this section.

1 molar = 1 mole per litre e.g. 1 mole Na2CO3 = 106g/L : 0.1 molar = 1.6g/L

Concentration is the amount of solute in a specified amount of solution e.g. moles per litre or grams per litre

Ways of Expressing Concentration

1. Percentage weight per volume (w/v) e.g. 3% NaCl solution = 3g NaCl in 100cm³ solution
2. Percentage volume per volume (v/v) e.g. 3% alcohol solution = 3cm³ alcohol in 100cm³ solution
3. Percentage weight per weight (w/w) e.g. 3% sugar solution = 3g sugar in 100g solution
4. Parts per million (p.p.m.) e.g. 2 p.p.m. solution has 2mg substance per litre

Standard solution a solution whose concentration is accurately known

Primary standard a water-soluble substance that is stable and available in pure form

Formula for Titration Problems

V1 x M1 = V2 x M2
    n1          n2                           Where V=volume, M=molarity, n=moles of solution present

If 20cm³ of 0.3 molar NaOH are neutralised by 25cm³ of H2SO4 solution, find the concentration of H2SO4 in (i) moles/L (ii)g/L according to the equation 2NaOH + H2SO4 >>> Na2SO4 + 2H2O

V1=25cm³  
M1=M1  n1=1    
V2=20cm³ 
M2=0.3  n2=2

25 x M1 = 20 x 0.3
   1            2

M1 = 0.12 molar

(i) 0.12 moles/litre    
(ii) 0.12 x 98 g = 11.76 g/litre

Chemistry page

Equipment:

• 0.02M potassium iodate solution
• 0.5M potassium iodide solution
• 1M sulfuric acid
• Sodium thiosulphate solution
• Starch indicator solution
• Deionised water
• Pipette and filler
• Burette
• Funnel
• COnical flask
• Ahite tile
• Graduated cylinder
• Retort stand and clamp
• Beakers

Experiment:

1. Wash pipette, burette and conical flask with deionised water
2. Rinse pipette with potassium iodate and burette with iodine thiosulphate
3. Pipette 25 cm³ potassium iodate into conical flask, with 20cm³ sulphuric acid and 10cm³ potassium iodide
4. Fill burette with sodium thiosulphate. Titrate against iodine solution.
5. When colour of solution fades to pale yellow, add starch indicator. Blue-black colour appears.
6. Continue titration until colour changes from blue-black to colourless.
7. Carry out two more accurate titrations
8. Calculate concentration of thiosulphate solution.

Chemistry Page

Equipment:

• Comparator
• DPD No.1 Tablets
• Sample of water

Experiment:

1. Rinse compartments of comparator with samples of water
2. Fill compartments with sample to calibrated mark
3. Add DPD No.1 tablets to each compartment and dissolve completely with a stirring rod
4. Fit lid and shake until dissolved completely
5. Compare colour of solutions to pre calibrated colours on comparator. Note concentration.

Possible Questions

1. What makes up free chlorine?

Chloric acid (HOCl) and the chlorate ion (ClO-)

Chemistry Page

Equipment:

• Water trough
• 250 cm³ reagent bottle
• Dropper
• Deionised water
• Manganese (II) sulfate solution
• Alkaline potassium iodide solution
• Concentrated sulfuric acid
• Sodium thiosulfate solution
• Starch indicator
• Pipette and filler
• Burette
• Conical flask
• Retort stand and clamp
• River water sample

Experiment:

1. Rinse the reagent bottle with deionised water, with vigorous shaking to avoid air bubbles
2. Completely fill bottle underwater with sample, making sure to avoid trapped air bubbles
3. Using a dropper placed well below surface of water, add 1 cm³ each of alkaline potassium iodide and manganese (II) sulfate solution
4. Stopper bottle. Some solution will overflow at this point
5. Invert bottle and allow precipitate to settle out
6. Run 1 cm³ sulfuric acid down inside of bottle, stopper and invert to dissolve precipitate. If precipitate does not dissolve, repeat this step.
7. Rinse burette, pipette and conical flask with deionised water and rinse burette with thiosulfate and flask with iodine solution
8. Fill burette to the 0 mark with thiosulfate and pipette 25cm³ iodine solution into the flask
9. Carry out one rough and two accurate titrations, adding starch indicator when solution is pale yellow. Endpoint when colour changes from blue-black to colourless.
10. Calculate concentration of iodine solution

Calculating Total Dissolved Oxygen

Lets take the concentration of the thiosulfate as 0.005M and the titre as 12.5cm³

The ratio of oxygen:thiosulfate is 1:4

0.005 x 12.5 = 25 x M

    4                1

M = 0.000625 moles/litre

0.000625 x 16 = 0.01g/litre

0.01g x 1000 = 10 p.p.m.

Possible Questions

1. Why is the reagent MnSO4 used?

Dissolved oxygen will not react completely in its absence. The use of MnSO4 results in the formation of Mn(OH)₂, which reacts completely with dissolved oxygen.

2. Why is concentrated H2SO4 used?

To enable the Mn(IV) species to release the free iodine needed for the redox reaction.

Chemistry Page

Equipment:

• Water sample
• Universal Indicator

Experiment:

1. Add a few drops indicator to the sample
2. Compare colour of sample to colour on chart to determine pH