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
- Draw a tangent to the curve
- The tangent is the hypotenuse to a right angled triangle
- θ = angle by tangent to the horizontal axis
- 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:
- Finely divided combustible particles
- An enclosed space
- Enough oxygen to sustain combustion
- 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
- Percentage weight per volume (w/v) e.g. 3% NaCl solution = 3g NaCl in 100cm³ solution
- Percentage volume per volume (v/v) e.g. 3% alcohol solution = 3cm³ alcohol in 100cm³ solution
- Percentage weight per weight (w/w) e.g. 3% sugar solution = 3g sugar in 100g solution
- 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:
- Comparator
- DPD No.1 Tablets
- Sample of water
Experiment:
- Rinse compartments of comparator with samples of water
- Fill compartments with sample to calibrated mark
- Add DPD No.1 tablets to each compartment and dissolve completely with a stirring rod
- Fit lid and shake until dissolved completely
- 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:
- Rinse the reagent bottle with deionised water, with vigorous shaking to avoid air bubbles
- Completely fill bottle underwater with sample, making sure to avoid trapped air bubbles
- Using a dropper placed well below surface of water, add 1 cm³ each of alkaline potassium iodide and manganese (II) sulfate solution
- Stopper bottle. Some solution will overflow at this point
- Invert bottle and allow precipitate to settle out
- Run 1 cm³ sulfuric acid down inside of bottle, stopper and invert to dissolve precipitate. If precipitate does not dissolve, repeat this step.
- Rinse burette, pipette and conical flask with deionised water and rinse burette with thiosulfate and flask with iodine solution
- Fill burette to the 0 mark with thiosulfate and pipette 25cm³ iodine solution into the flask
- 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.
- 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:
- Sample of filtered water
- Beaker
- Oven
Experiment:
- Find mass of clean dry beaker
- Place 100 cm³ water in beaker and evaporate in oven
- Find mass of cool beaker + dissolved solids
- By subtraction, calculate mass of dissolved solids
Calculating Total Dissolved Solids in p.p.m.
Lets take mass of dissolved solids as 1.5g/100 cm³
Total Dissolved Solids: 1.5g/100 cm³ = 15g/L
Total Dissolved Solids (p.p.m.) = 15 x 1000 = 15000 p.p.m
Possible Questions
1. Why is it necessary to use filtered water?
If unfiltered water is used, we do not get an accurate measure of the total dissolved solids. We get a measure of total dissolved solids + total suspended solids.
Chemistry Page
Equipment:
- Sample of water with suspended solids
- Filter paper
- Filter funnel
- Beaker
Experiment:
- Find mass of dry filter paper
- Filter 250 cm³ water through the filter paper in funnel
- Allow filter paper to dry overnight
- Find mass of filter paper + suspended solids
- By subtraction, calculate mass of suspended solids
Calculating Total Suspended Solids in p.p.m.
Lets take mass of suspended solids as 0.05g /250cm³
Total Suspended Solids: 0.05g/250cm³ = 0.2g/L
Total Suspended Solids (p.p.m.) = 0.2g x 1000 = 200 p.p.m
Chemistry Page
Equipment:
- Pipette and filler
- Burette
- Conical Flask
- Edta solution
- pH buffer 10
- Deionised water
- Hard water sample
- Eriochrome Black T Indicator
- Retort stand and clamp
Experiment:
- Rinse pipette, burette and flask with deionised water. Rinse pipette with hard water and burette with edta solution.
- Fill burette to 0 mark with edta, making sure below tap is filled and bottom of meniscus lies on the 0 mark.
- Pipette 50 cm³ hard water into conical flask with 2-3 cm³ pH buffer 10.
- Add minimal amount of solid indicator to the flask until a wine red colour is achieved. Do not use excess indicator.
- Carry out one rough and two accurate titrations. End point at colour change from wine red to blue.
- Calculate total hardness.
Calculating Total Hardness
Lets take concentration of edta as 0.01 M and the titre as 22.15 cm³
50 x M = 22.15 x 0.01
1 1
M = 0.00443 moles/litre Mg+ and Ca+
Total Hardness = 0.00443 x 100g/l CaCO3
= 0.443g/l CaCO3
= 0.443 x 1000 p.p.m
= 443 p.p.m. CaCO3
Possible Questions
1. Why is pH buffer 10 used?
The experiment requires a pH > 9 to occur
2. Why is it important to avoid using excess indicator?
Excess indicator will make it very difficult to detect an accurate endpoint
3. Why is total hardness expressed as CaCO3?
This is for convenience. Total hardness is actually a measure of Ca+ ions and Mg+ ions but the edta solution does not allow for differentiation between these ions.
