Sunday, October 3, 2010
Friday, October 1, 2010
Thursday, September 16, 2010
Monday, September 13, 2010
REDOX - electrolysis as aqueous of molten?
Only copper ions, silver ions and gold ions can undergo electrolysis as an aqueous solution.
Every other metal ion needs to be in its molten state.
Why?
Because in an aqueous solution there is also H+ and OH- ions present H2O --> H+ + OH-
and so the H+ ions compete for the cathode with the metals ions.
Because all metals except Cu, Ag, Au are more reactive than H, the H+ ions will be reduced to H2 before the others.
(remember the more reactive the metal is the more stable the metal ions is, and so the harder it is to reduce.)
Every other metal ion needs to be in its molten state.
Why?
Because in an aqueous solution there is also H+ and OH- ions present H2O --> H+ + OH-
and so the H+ ions compete for the cathode with the metals ions.
Because all metals except Cu, Ag, Au are more reactive than H, the H+ ions will be reduced to H2 before the others.
(remember the more reactive the metal is the more stable the metal ions is, and so the harder it is to reduce.)
REDOX - electrolysis
This is used to make a chemical take back electrons when it rather wouldn't.
example:
Cu + Cl2 --> 2CuCl2 This is spontaneous. It is a REDOX.
Cu --> Cu2+ + 2e- (LEO-oxidation)
Cl2 + 2e --> 2Cl- (GER- reduction)
BUT
CuCl2 --> Cu + Cl2 is not spontaneous but can be achieved via electrolysis.
example:
Cu + Cl2 --> 2CuCl2 This is spontaneous. It is a REDOX.
Cu --> Cu2+ + 2e- (LEO-oxidation)
Cl2 + 2e --> 2Cl- (GER- reduction)
BUT
CuCl2 --> Cu + Cl2 is not spontaneous but can be achieved via electrolysis.
REDOX -terms
When answersing questions on REDOX, you must always think about what is happening to the electrons, what chemical is gaining and what chemical is losing electrons.
Oxidised = has lost electrons (LEO)
Reduced = has gained electrons (GER)
Oxidising agent/ oxidant = the chemical that is itself reduced, so it has gained electrons.
Reducing agent/ reductant = the chemical that is itself oxidised, so it has lost electrons.
Oxidised = has lost electrons (LEO)
Reduced = has gained electrons (GER)
Oxidising agent/ oxidant = the chemical that is itself reduced, so it has gained electrons.
Reducing agent/ reductant = the chemical that is itself oxidised, so it has lost electrons.
REDOX -Halogens as reducing agents
Colours of the halogens:
All halogen ions (Cl-, Br-, I-) are colourless.
Fluorine (F2) - Pale yellow
Chlorine (Cl2) - Green
Bromine (Br2) - Orange
Iodine (I2) - Brown
F2 is the strongest reducing agent because it is the most reactive and 'wants' to be F- ions more than the other halogens.
F2 is able to displace any other halide ion from solution.
example:
F2 + 2NaBr --> 2NaF + Br2
(F atoms) (Br- ions) (F- ions) (Br atoms)
Pale yellow Orange
All halogen ions (Cl-, Br-, I-) are colourless.
Fluorine (F2) - Pale yellow
Chlorine (Cl2) - Green
Bromine (Br2) - Orange
Iodine (I2) - Brown
F2 is the strongest reducing agent because it is the most reactive and 'wants' to be F- ions more than the other halogens.
F2 is able to displace any other halide ion from solution.
example:
F2 + 2NaBr --> 2NaF + Br2
(F atoms) (Br- ions) (F- ions) (Br atoms)
Pale yellow Orange
REDOX - Oxidation numbers
NOTE:
There are some things you MUST put into your answers to questions about Oxidation numbers.
1. If the oxidation number goes up - the chemical has been oxidised and so has lost electrons.
2. If the oxidation number goes down - then the chemical has been reduced and so has gained elections.
Assigning Oxidation numbers
There are some things you MUST put into your answers to questions about Oxidation numbers.
1. If the oxidation number goes up - the chemical has been oxidised and so has lost electrons.
2. If the oxidation number goes down - then the chemical has been reduced and so has gained elections.
Assigning Oxidation numbers
Monday, September 6, 2010
Sunday, August 29, 2010
Monday, June 28, 2010
The titration method
The titration sequence of events:
1. Rinse a conical flask with water.
2. Rinse the pipette with the HCl solution and then pipette, a 25 mL sample of the hydrochloric acid and place it in a conical flask, (an aliquot). Repeat so that you have 3 conical flasks set up and ready to go.
A pipette is a piece of apparatus which accurately delivers a given volume -before filling it is always rinsed with a small amount of the solution.
3. A few drops of an appropriate acid-base indicator, in this case phenolphthalein, is added to the flask. The solution will remain colourless.
4. A burette is then rinsed with a sample of sodium hydroxide (the solution it is to be filled with) and then filled to just below the 0.00 mL mark
A burette is another piece of glassware that accurately measures out a volume of liquid.
5. The initial volume of aqueous NaOH in the burette is carefully read (to 2 decimal places - giving an accuracy of ± 0.02 mL).
6. Carefully add the NaOH to the aqueous HCl, finishing the titration as soon as the first permanent pink colour is observed.
The colour change is referred to as the end-point, and assuming the correct indicator has been used, is the point when the acid and base have reacted in the molar ratio given by the balanced neutralisation equation.
7. By taking the difference between the initial and final volumes in the burette, the total volume of NaOH(aq) added can be carefully determined.
8. The titration is repeated at least 3 times, or until you have obtained 3 concordant results i.e. burette volumes (titres) that, ideally, agree to within ± 0.2 mL.
1. Rinse a conical flask with water.
2. Rinse the pipette with the HCl solution and then pipette, a 25 mL sample of the hydrochloric acid and place it in a conical flask, (an aliquot). Repeat so that you have 3 conical flasks set up and ready to go.
A pipette is a piece of apparatus which accurately delivers a given volume -before filling it is always rinsed with a small amount of the solution.
3. A few drops of an appropriate acid-base indicator, in this case phenolphthalein, is added to the flask. The solution will remain colourless.
4. A burette is then rinsed with a sample of sodium hydroxide (the solution it is to be filled with) and then filled to just below the 0.00 mL mark
A burette is another piece of glassware that accurately measures out a volume of liquid.
5. The initial volume of aqueous NaOH in the burette is carefully read (to 2 decimal places - giving an accuracy of ± 0.02 mL).
6. Carefully add the NaOH to the aqueous HCl, finishing the titration as soon as the first permanent pink colour is observed.
The colour change is referred to as the end-point, and assuming the correct indicator has been used, is the point when the acid and base have reacted in the molar ratio given by the balanced neutralisation equation.
7. By taking the difference between the initial and final volumes in the burette, the total volume of NaOH(aq) added can be carefully determined.
8. The titration is repeated at least 3 times, or until you have obtained 3 concordant results i.e. burette volumes (titres) that, ideally, agree to within ± 0.2 mL.
Calculations
Rules and regulations:
You must record your calculations to as many sif figs as are on the calculator.
These sig figs must then be round down to 3 sig fig for your final answer.
You must record your calculations to as many sif figs as are on the calculator.
These sig figs must then be round down to 3 sig fig for your final answer.
Titrations
What is a titration?
In an acid-base titration the neutralisation reaction between an acid and base is used to determine the concentration of one of the reactants, if the concentration of the other is accurately known.
Using the reaction between a solution of hydrochloric acid (unknown concentration) and sodium hydroxide (standard solution - concentration known) is an example.
For achievement with excellence:
• at least three titre values must fall within a range of 0.2 mL; the average titre value must be within 0.2 mL of the expected outcome
• a titration calculation where the stoichiometry is not one-to-one must be carried out correctly using only concordant titre values. The final answer must have correct units and an appropriate number of significant figures.
In an acid-base titration the neutralisation reaction between an acid and base is used to determine the concentration of one of the reactants, if the concentration of the other is accurately known.
Using the reaction between a solution of hydrochloric acid (unknown concentration) and sodium hydroxide (standard solution - concentration known) is an example.
For achievement with excellence:
• at least three titre values must fall within a range of 0.2 mL; the average titre value must be within 0.2 mL of the expected outcome
• a titration calculation where the stoichiometry is not one-to-one must be carried out correctly using only concordant titre values. The final answer must have correct units and an appropriate number of significant figures.
Tuesday, June 15, 2010
Measuring Reaction Enthalpy
Calculate the enthalpy change for a reaction when 10g of NaOH is added to 250 mL of water and the temperature changes from 25 oC to 29.8 oC.
Remember you are calculating the enthalpy for the surroundings, so the reactions enthalpy will be the opposite sign.
Remember you are calculating the enthalpy for the surroundings, so the reactions enthalpy will be the opposite sign.
Equilibrium
Kc Equilibrium constant:
eg for: N2 + 3H2 --> 2NH3
Kc = [product]/[reactant]
Kc = [NH3]2/[N2][H2]3
This is a ratio of product concentration to reactant concentration at equilibrium.
If Kc>1 then the equuilibrium position favours the product side
If Kc<1 then the equilibrium position favours the reactant side.
* Solids do NOT get included into the Kc expression
Le Chateliers Principle:
States that:
When a system is at equilibrium, that system will shift to minimise any changes made to it.
[When answering questions on equilibrium, you need to name the chemical(s) affected by the change]
Things that can change the position of equilibrium:
1. Temperature:
When heat is added, the equilibrium will shift in the endothermic direction.
When cooled the equilibrium will shift in the exothermic direction.
Kc will change as it is temperature dependant.
2. Concentration:
When adding or removing chemicals from an equilibrium you are changing the concentration. The equilibrium will shift in the direction to either replace any chemicals removed or use up any chemical added.
Kc remains the same.
3. Pressure:
When the pressure is increased (by decreasing the volume of the container) the equilibrium will shift to the side with the least number of moles of gas.
When the pressure is decreased (by increasing the volume of the container) the equilibrium will shift to the side with the most moles of gas.
Kc remains the same.
4. Catalyst:
A catalyst will not affect equilibrium position. It will just increase the rate at which equilibrium is reacted and it will increase the rate of the forward and reverse reactions equally.
A catalyst is not used up in the reaction.
eg for: N2 + 3H2 --> 2NH3
Kc = [product]/[reactant]
Kc = [NH3]2/[N2][H2]3
This is a ratio of product concentration to reactant concentration at equilibrium.
If Kc>1 then the equuilibrium position favours the product side
If Kc<1 then the equilibrium position favours the reactant side.
* Solids do NOT get included into the Kc expression
Le Chateliers Principle:
States that:
When a system is at equilibrium, that system will shift to minimise any changes made to it.
[When answering questions on equilibrium, you need to name the chemical(s) affected by the change]
Things that can change the position of equilibrium:
1. Temperature:
When heat is added, the equilibrium will shift in the endothermic direction.
When cooled the equilibrium will shift in the exothermic direction.
Kc will change as it is temperature dependant.
2. Concentration:
When adding or removing chemicals from an equilibrium you are changing the concentration. The equilibrium will shift in the direction to either replace any chemicals removed or use up any chemical added.
Kc remains the same.
3. Pressure:
When the pressure is increased (by decreasing the volume of the container) the equilibrium will shift to the side with the least number of moles of gas.
When the pressure is decreased (by increasing the volume of the container) the equilibrium will shift to the side with the most moles of gas.
Kc remains the same.
4. Catalyst:
A catalyst will not affect equilibrium position. It will just increase the rate at which equilibrium is reacted and it will increase the rate of the forward and reverse reactions equally.
A catalyst is not used up in the reaction.
Wednesday, May 26, 2010
Factors affecting rate of reaction
The collision theory
The collision theory states that for a chemical reaction to occur the chemicals have to collide, they have to collide with enough energy to react and they have to collide with the correct orientation.
Factors that affect the rate of reaction.
Temperature:
With an increase of temperature the particles have more kinetic energy, this leads more collisions per second.
This means that the rate of reaction increases.
Temperature also changes to amount of particles that have the required activation energy to react. When heated, a greater number of the particles have the required activation energy and so there is a faster reaction rate.
Concentration:
A more concentrated reactant will have more particles in a given volume. This leads to more collisions per second and therefore a faster rate of reaction.
Surface area:
Powdered solid has a greater surface area as there is more surface exposed to collisions. This means that there are more collisions per second so a faster rate of reaction.
Catalyst
A catalyst lowers the activation energy needed for a reaction to take place. This means that a greater number of particles have the required energy needed to react and hence a faster rate of reaction.
A catalyst does not get used up during the reaction.
The collision theory states that for a chemical reaction to occur the chemicals have to collide, they have to collide with enough energy to react and they have to collide with the correct orientation.
Factors that affect the rate of reaction.
Temperature:
With an increase of temperature the particles have more kinetic energy, this leads more collisions per second.
This means that the rate of reaction increases.
Temperature also changes to amount of particles that have the required activation energy to react. When heated, a greater number of the particles have the required activation energy and so there is a faster reaction rate.
Concentration:
A more concentrated reactant will have more particles in a given volume. This leads to more collisions per second and therefore a faster rate of reaction.
Surface area:
Powdered solid has a greater surface area as there is more surface exposed to collisions. This means that there are more collisions per second so a faster rate of reaction.
Catalyst
A catalyst lowers the activation energy needed for a reaction to take place. This means that a greater number of particles have the required energy needed to react and hence a faster rate of reaction.
A catalyst does not get used up during the reaction.
Saturday, May 8, 2010
Internal sample questions and answers
Achieved
1 Calculate the amount (in moles) of carbon dioxide, CO2, in 23.0 g of the gas?
M(CO2) = 44.0 g mol-1
n(CO2) = = 0.523 mol
Achieved
2 Calculate the mass of 2.65 mol of sodium hydroxide, NaOH.
M(NaOH) = 40.0 g mol-1
m = 2.65 mol x 40.0 g mol–1 = 106 g
Achieved
3 24.5 mL of a solution contains 0.00257 mol of HCl. Calculate the concentration, in mol L–1, of the HCl solution.
c = = 0.105 mol L–1
Achieved
4 Calculate the amount (in mole) of NaOH present in 15.6 mL of 0.152 mol L–1 solution.
n = 0.152 mol L–1 x 15.6 x 10–3 L = 2.37 x 10–3 mol
Merit
5(a) Caffeine is a stimulant which has a mass composition of 49.5% carbon, 5.20% hydrogen, 28.9% nitrogen and 16.5% oxygen.
Calculate the empirical formula of caffeine.
(a) n(C) = = 4.13 mol n(H) = = 5.20 mol
n(N) = = 2.06 mol n(O) = = 1.03 mol
empirical formula C4H5N2O
Amount in moles of each element correct.
Correct process, but incorrect empirical formula used. (achieved)
(b) If the molar mass of caffeine is 194 g mol-1, use your answer to part (a) above to determine the molecular formula of caffeine.
molecular formula C8H10N4O2
Empirical and molecular formulae correct. (merit)
Merit
6. Calculate the percentage of carbon and oxygen in sucrose, C12H22O11.
M(C12H22O11) = 342 g mol–1
% C= x 100 = 42.1 %
% O = x 100 = 51.5 %
Process correct but incorrect molar mass of sucrose used. (Achieved)
% of C and O correct(Merit)
Merit
7. 500 mL of 0.253 mol L-1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L-1 NaHCO3 solution.
What is the concentration of the final solution?
n(NaHCO3)
= (0.253 mol L-1 x 0.500 L) + (0.824 mol L-1 x 0.800 L)
= 0.786 mol
Total volume = 1.30 L
concentration of final solution = = 0.604 mol L-1
Correct total amount of NaHCO3
OR
correct process for calculating concentration using a volume of 1.3 L. (Achieved)
Correct determination of concentration of final solution. (Merit)
Excellence
8. A chemist found that 4.69 g of sulfur combined with fluorine t
o produce 15.81 g of gas. Determine the empirical formula of the compound.
n(S) = = 0.146 mol
m(F) = 15.81 - 4.69 = 11.1 g
n(F) = = 0.584 mol
Hence S : F = 1 : 4 and empirical formula is SF4.
Correct amount of S OR F. (achieved)
Correct process with one minor error. eg Incorrect mass of F (Merit)
Correct determination empirical formula. (excellence)
Excellence
9. What mass of CO2 is produced in the complete combustion of 36.5 g of ethanol according to the following equation?
C2H5OH + 3 O2 2 CO2 + 3 H2O
M(C2H5OH) = 46.0 g mol-1
(i) n(C2H5OH) = 36.5 g / 46.0 g mol-1 = 0.793 mol
(ii) n(CO2) = 2 x n(C2H5OH) = 1.59 mol
(iii) m(CO2) = 1.59 mol x 44.0 g mol-1 = 69.8 g
Correct process used in either step (i) or (iii). (achieved)
Correct process with one minor error. eg An incorrect molar mass. (Merit)
Correct answer for mass of CO2 produced. (excellence)
Excellence
10 .What mass of iron can be produced if 50.0 g of carbon monoxide react with iron(III) oxide according to the following equation?
Fe2O3 + 3 CO 2 Fe + 3 CO2
(i) n(CO) = = 1.79 mol
(ii) n(Fe) = x n(CO) = 1.19 mol
(iii) m(Fe) = 1.19 mol x 55.9 g mol-1 = 66.5 g
Correct process used in either step (i) or (iii).(achieved)
Correct process with one minor error.eg An incorrect molar mass.(Merit)
Correct answer for mass of Fe produced. (Excellence)
Excellence
11. Hydrated magnesium sulfate is heated in a crucible. The following data is collected:
mass of crucible and lid = 26.49 g
mass of hydrated magnesium sulfate = 2.13 g
mass crucible, lid and magnesium sulfate after first heating = 27.55 g
mass of crucible, lid and magnesium sulfate after second heating = 27.53 g
Use these results to determine the formula of hydrated magnesium sulfate.
M(MgSO4) = 120.3 g mol-1 M(H2O) = 18.0 g mol-1
mass of anhydrous MgSO4 = 1.04 g
mass of H2O = 1.09 g
amount of MgSO4 = = 8.65 x 10-3 mol
amount H2O = = 6.06 x 10-2 mol
ratio n(H2O) / n(MgSO4) = 7
formula is MgSO4.7H2O
Correct determination of masses of MgSO4 and H2O.(Achieved)
Correct determination of either amount of MgSO4 or H2O. (Merit)
Correct formula for hydrated salt.(Excellence)
1 Calculate the amount (in moles) of carbon dioxide, CO2, in 23.0 g of the gas?
M(CO2) = 44.0 g mol-1
n(CO2) = = 0.523 mol
Achieved
2 Calculate the mass of 2.65 mol of sodium hydroxide, NaOH.
M(NaOH) = 40.0 g mol-1
m = 2.65 mol x 40.0 g mol–1 = 106 g
Achieved
3 24.5 mL of a solution contains 0.00257 mol of HCl. Calculate the concentration, in mol L–1, of the HCl solution.
c = = 0.105 mol L–1
Achieved
4 Calculate the amount (in mole) of NaOH present in 15.6 mL of 0.152 mol L–1 solution.
n = 0.152 mol L–1 x 15.6 x 10–3 L = 2.37 x 10–3 mol
Merit
5(a) Caffeine is a stimulant which has a mass composition of 49.5% carbon, 5.20% hydrogen, 28.9% nitrogen and 16.5% oxygen.
Calculate the empirical formula of caffeine.
(a) n(C) = = 4.13 mol n(H) = = 5.20 mol
n(N) = = 2.06 mol n(O) = = 1.03 mol
empirical formula C4H5N2O
Amount in moles of each element correct.
Correct process, but incorrect empirical formula used. (achieved)
(b) If the molar mass of caffeine is 194 g mol-1, use your answer to part (a) above to determine the molecular formula of caffeine.
molecular formula C8H10N4O2
Empirical and molecular formulae correct. (merit)
Merit
6. Calculate the percentage of carbon and oxygen in sucrose, C12H22O11.
M(C12H22O11) = 342 g mol–1
% C= x 100 = 42.1 %
% O = x 100 = 51.5 %
Process correct but incorrect molar mass of sucrose used. (Achieved)
% of C and O correct(Merit)
Merit
7. 500 mL of 0.253 mol L-1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L-1 NaHCO3 solution.
What is the concentration of the final solution?
n(NaHCO3)
= (0.253 mol L-1 x 0.500 L) + (0.824 mol L-1 x 0.800 L)
= 0.786 mol
Total volume = 1.30 L
concentration of final solution = = 0.604 mol L-1
Correct total amount of NaHCO3
OR
correct process for calculating concentration using a volume of 1.3 L. (Achieved)
Correct determination of concentration of final solution. (Merit)
Excellence
8. A chemist found that 4.69 g of sulfur combined with fluorine t
o produce 15.81 g of gas. Determine the empirical formula of the compound.
n(S) = = 0.146 mol
m(F) = 15.81 - 4.69 = 11.1 g
n(F) = = 0.584 mol
Hence S : F = 1 : 4 and empirical formula is SF4.
Correct amount of S OR F. (achieved)
Correct process with one minor error. eg Incorrect mass of F (Merit)
Correct determination empirical formula. (excellence)
Excellence
9. What mass of CO2 is produced in the complete combustion of 36.5 g of ethanol according to the following equation?
C2H5OH + 3 O2 2 CO2 + 3 H2O
M(C2H5OH) = 46.0 g mol-1
(i) n(C2H5OH) = 36.5 g / 46.0 g mol-1 = 0.793 mol
(ii) n(CO2) = 2 x n(C2H5OH) = 1.59 mol
(iii) m(CO2) = 1.59 mol x 44.0 g mol-1 = 69.8 g
Correct process used in either step (i) or (iii). (achieved)
Correct process with one minor error. eg An incorrect molar mass. (Merit)
Correct answer for mass of CO2 produced. (excellence)
Excellence
10 .What mass of iron can be produced if 50.0 g of carbon monoxide react with iron(III) oxide according to the following equation?
Fe2O3 + 3 CO 2 Fe + 3 CO2
(i) n(CO) = = 1.79 mol
(ii) n(Fe) = x n(CO) = 1.19 mol
(iii) m(Fe) = 1.19 mol x 55.9 g mol-1 = 66.5 g
Correct process used in either step (i) or (iii).(achieved)
Correct process with one minor error.eg An incorrect molar mass.(Merit)
Correct answer for mass of Fe produced. (Excellence)
Excellence
11. Hydrated magnesium sulfate is heated in a crucible. The following data is collected:
mass of crucible and lid = 26.49 g
mass of hydrated magnesium sulfate = 2.13 g
mass crucible, lid and magnesium sulfate after first heating = 27.55 g
mass of crucible, lid and magnesium sulfate after second heating = 27.53 g
Use these results to determine the formula of hydrated magnesium sulfate.
M(MgSO4) = 120.3 g mol-1 M(H2O) = 18.0 g mol-1
mass of anhydrous MgSO4 = 1.04 g
mass of H2O = 1.09 g
amount of MgSO4 = = 8.65 x 10-3 mol
amount H2O = = 6.06 x 10-2 mol
ratio n(H2O) / n(MgSO4) = 7
formula is MgSO4.7H2O
Correct determination of masses of MgSO4 and H2O.(Achieved)
Correct determination of either amount of MgSO4 or H2O. (Merit)
Correct formula for hydrated salt.(Excellence)
Monday, March 29, 2010
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