Catalytic Converter

Catalytic converters are an excellent technology that helps the environment reduces the harmful chemicals, especially the ones coming out of motor vehicles. The catalytic converter is positioned halfway between the exhaust stream and the exhaust pipe. This device may not seem to be much, however, it is in fact a very technologically complex system which also happens to cost quite a lot. The cost is high due to the uses of the precious metals working inside of the catalysts such as platinum and palladium. For the cost issue, most manufacturers use some optimising techniques to bring the cost low. The catalytic converter has two main catalysts. One is the reduction catalyst while the other is the oxidation catalyst.

If we look at how an engine works to make the car move, we can say that the power for the wheels to move come straight from the explosion that occurs in the pistons. The cylinder which the piston is located in takes in oxygen and petrol and ignites them by a spark via the spark plug. For a car that is doing around 6000 rounds per minute; the time for the piston to take in the oxygen and petrol to when it pushes the burnt waste out, takes approximates 5 milliseconds. How can the car gain so much power within such a little time to make it move? Well, because an engine has 4 strokes in each round, it allows it to accelerate some reactions. For example as it takes in the oxygen and petrol in the first stroke, it compresses the gases in the second stroke. The compression increases which means the gases are moving at a higher pressure while also having a smaller volume to move around in. This makes the chances of the gases to collide perfectly higher. Also because the gases enter at such a high speed, increases the chances of a successful collision between the oxygen and the petrol. After being compressed the spark plug sparks, which makes the gases combust and create a mechanical force to push down the piston and make the car move. However, if the piston did not compress the gases, the reaction rate would have been slower and the mechanical power that the explosion does would have been weaker.

A three-way catalytic converter does three things.
1. Oxidation of unburnt hydrocarbons HC to carbon dioxide and water: C_xH_2x+2+ [(3x+1) / 2]O₂
= xCO₂ + (x+1)H₂O
2. Oxidation of carbon monoxide to carbon dioxide: 2CO + O₂ = 2CO₂
3. Reduction of nitrogen oxides to nitrogen and oxygen: 2NO_x  = xO₂ + N₂

These reactions occur most efficiently when the catalytic converter receives exhaust from an engine running with approximately 15 parts air to 1 part petrol. Cars that are fitted with a three-way catalytic converter are equipped with a computerised close loop feedback fuel injection system using one or more oxygen centres. These oxygen centres are where the catalytic converter stores oxygen from the exhaust stream when the air to fuel ratio is lean. Because of this, when the air to fuel ratio is rich, the catalytic converter releases some of the oxygen stored to make the ratio just right. How can the exhaust gases be rich or lean? It is because even though cars have a computerised fuel and air injection system, the procedure takes an extremely little time, which means there are chances of mistakes from the computer. A rich ratio is when there is an excess fuel to the available oxygen, while lean means there is excess oxygen to the available fuel. A place where the catalyst may not be very efficient is when for example the car doing certain manoeuvers such as a hard acceleration. During this manoeuver, nitrous oxide escapes the catalytic converter. Note: nitrous oxide is 300 times more dangerous than carbon dioxide.

A two way catalytic converter only does the first two things that the three way catalytic converter does. This means it does not separate nitrogen oxides. Two way catalytic converters are widely used on diesel engines, it was also used on gasoline engines too until 1981 which was then replaced by the three way converter.

A three way catalytic converter has two different types of catalysts at work. One is a reduction catalyst and the other is an oxidation catalyst. Both of these catalysts have a ceramic structure coated with metals. Usually these metals are platinum, rhodium and palladium. These metals are extremely expensive, that is why newer converters use a gold mix in the catalyst as well instead of the metals states before. Gold is actually increases oxidation, the reaction that reduces pollution, by 40%. The first phase, the reduction catalyst separates the nitrogen oxides to become nitrogen and oxygen. Once this compound is separated, it becomes non-toxic. The oxidation catalyst which is the second phase of the catalytic converter reduces the unburnt hydrocarbons and carbon monoxide by burning them over a platinum and palladium catalyst. These catalysts aid at reducing the toxic gases such as the carbon monoxide by mixing it with oxygen to create carbon dioxide. Carbon dioxide is bad for the environment however; it is not as bad and deadly as carbon monoxide.

The centre of the catalytic converter is made up of a honeycomb structure for the gases to move from the first catalyst to the second. The honeycomb structure helps by exposing as much surface area possible. As we all know, the more surface area exposed, the higher the chances of a successful collision between the gas and the rhodium coating on the honeycomb passageway. This technique allows the catalytic converter to be more compact which ultimately means the less coating there needs to be which finally means the cost to manufacture it decreases. Therefore it’s a win-win situation both for the manufacturer and the customer who purchases the motor vehicle.

References:

1. https://en.wikipedia.org/wiki/Catalytic_converter#Types
2. http://www.howstuffworks.com/catalytic-converter2.htm
3. http://www.youtube.com/watch?v=LyFKVqLsPiA
4. http://www.youtube.com/watch?v=KIVdPaTSot8
5. http://www.youtube.com/watch?v=nbCe68ck6qg
6. http://www.youtube.com/watch?v=umdoG7qdWWs
7. http://www.youtube.com/watch?v=3wjFEUD8P7I
8. http://www.chytecs.com/cata-converters.html
9. http://answers.yahoo.com/question/index?qid=20081120032312AADaMIP
10. http://www.molecularsieve3a.com/honeycomb-ceramic-catalytic-converter.html
11. http://en.wikipedia.org/wiki/Air%E2%80%93fuel_ratio
12. http://www.chemguide.co.uk/physical/basicrates/pressure.html

Redox Reactions

Experiment 19:

Aim:

To prepare potassium permanganate solution

Equipment:

Refer to STAWA book page 148

Procedure:

Refer to STAWA book page 148

Processing of Results and Questions:

The solution was first boiled for 10 minutes to speed up the rate of reaction. This is because at room temperature, the solution has a very slow rate of reaction.

Experiment 20:

Aim:

To prepare standard oxalic acid solution

Equipment:

Refer to STAWA book page 149

Procedure:

Refer to STAWA book page 149

Processing of Results and Questions:

m(Oxalic Acid) = 1.58 g
M(Oxalic Acid) = 126.068 gmol ^-1
n(Oxalic Acid) = 0.0125 mol
V = 0.25 L
c(Oxalic Acid) = n ÷ V
c = 0.0125 ÷ 0.25
c(Oxalic Acid) = 0.05013 mol L  ^-1

Experiment 21:

Aim:

Standardising potassium permanganate solution

Equipment:

Refer to STAWA book page 150

Procedure:

Refer to STAWA book page 151

Results:

	Rough Estimate	Accurate 1	Accurate 2	Accurate 3
Final Reading (mL)	40.5	40.5	40.7	40.6
Initial Reading (mL)	20	20	20	20
Titration Volume (mL)	20.5	20.5	20.7	20.6

Processing of Results and Questions:

1.       c(Oxalic Acid) = 0.05013
V(Oxalic Acid) = 0.02
n(Oxalic Acid) = 0.001 mol
n(Permanganate) = [n(Oxalic Acid) ÷ 5] x 2
n(Permanganate) = 0.000401 mol
V(Permanganate) = 0.0206 L
c(Permanganate) = n(Permanganate) ÷ V(Permanganate)
c(Permanganate) = 0.000401 ÷ 0.0206
c(Permanganate) = 0.0195 mol L ^-1

2a)    Condy’s Crystals, a name given to potassium permanganate has a pinkish, purple colour.

2b)  Manganese oxide has a pale yellowish brown colour

2c)   MnO₄^- + 8H ⁺ + 5e ^-  --> Mn²⁺ + 4H₂O
H₂S --> S + 2H⁺ + 2e ^-
2MnO₄^- + 6H ⁺ + 5H₂S --> 2Mn²⁺ + 8H₂O + 5S

2d)  A very horrendous smell of skunk scent and other smells such as garlic.

3.       Yes oxalic acid can be used as a primary standard to standardise a basic solution such as sodium hydroxide. This is because oxalic acid is pure and can be accurately weighed out on a scale.

Experiment 22:

Aim:

Analysis of iron content

Equipment:

Refer to STAWA book page 152

Procedure:

Refer to STAWA book page 152-153

Results:

	Rough Estimate	Accurate 1	Accurate 2	Accurate 3
Final Reading (mL)	43.0	43.7	43.5	43.6
Initial Reading (mL)	30	30	30	30
Titration Volume (mL)	13.0	13.7	13.5	13.6

Processing of Results and Questions:

1.       MnO₄^- + 8H ⁺ + 5e ^-  --> Mn²⁺ + 4H₂O
Fe²⁺ --> Fe³⁺ + e ^-
MnO₄^- + 8H ⁺ + 5Fe²⁺   --> Mn²⁺ + 4H₂O + 5Fe³⁺

2.       c(Permanganate) = 0.0195 mol L ^-1
V(Permanganate) = 0.0136 L
n(Permanganate) = c(Permanganate) x V(Permanganate)
n(Permanganate) = 0.000265 mol

3.       n(Iron) = n(Permanganate) x 5
n(Iron) = 0.001324 mol in 20 ml

4.       n(Iron) = 0.016548 mol in 250 ml        ==           ([0.001324 ÷ 20] x 250)
M(Iron) = 55.85 gmol ^-1
m(Iron) = n(Iron) x M(Iron)
m(Iron) = 0.924224 g

5.       0.924224 ÷ 1 = 0.924224
0.924224 x 100 = 92.4%

6.       Iron can be oxidised from its surroundings. Oxygen is a main source of this impurity.

7.       Fe + H₂SO₄ --> FeSO + H₂
Fe + 2H⁺ --> Fe²⁺ + H₂

Investigation 12:

Aim:

To determine the concentration of hydrogen peroxide

Equipment:

·         Beakers

·         Conical Flask

·         Funnel

·         Volumetric Flask (500ml)

·         Pipette

·         Pipette Filler

·         Burette and Stand

·         Sulfuric Acid

·         Potassium Permanganate

·         Hydrogen Peroxide

·         Distilled Water

·         Safety Equipment

Procedure:

1.       Put on safety equipment

2.       Take out 2 ml of concentrated hydrogen peroxide

3.       Mix 20 ml of sulphuric acid with 30 ml of potassium permanganate solution

4.       Add 2 ml intervals of the solution to the hydrogen peroxide

5.       Do this until you see the first undissolved purple colour in the hydrogen peroxide solution

6.       Place 5 ml of hydrogen peroxide in volumetric flask and dilute it (should be by a factor of 100), we diluted ours by a factor of 102

7.       Titrate with sulphuric acid and potassium permanganate solution

Results:

	Rough Estimate	Accurate 1	Accurate 2	Accurate 3
Final Reading (mL)	44.0	43.7	43.7	43.8
Initial Reading (mL)	20	20	20	20
Titration Volume (mL)	24	23.7	23.7	23.8

 

 

Processing the Data:

1.       MnO₄^- + 8H ⁺ + 5e ^-  --> Mn²⁺ + 4H₂O
H₂O₂ --> O₂ + 2H⁺ + 2e ^-
2MnO₄^- + 6H ⁺ + 5H₂O₂  --> 2Mn²⁺ + 8H₂O + 5O₂

2.       c(Permanganate) = 0.0195 mol L ^-1
V(Permanganate) = 0.023733 L
n(Permanganate) = c(Permanganate) x V(Permanganate)
n(Permanganate) = 0.000462 mol
n(Hydrogen Peroxide) = [n(Permanganate) ÷ 2] x 5
n(Hydrogen Peroxide) = 0.001155 mol in 20 ml
n(Hydrogen Peroxide) = 0.029456 mol in 510 ml         ([0.001155 ÷ 20] x 510)
c(Hydrogen Peroxide) = n(Hydrogen Peroxide) ÷ V(Hydrogen Peroxide)
V(Hydrogen Peroxide) = 0.51 L
c(Hydrogen Peroxide) = 0.058912 mol L ^-1 (dilute)
c(Hydrogen Peroxide) = 6.01 mol L ^-1               (concentrated)                                ([0.057757 ÷ 5] x 510)

3.       c(Hydrogen Peroxide) x M(Hydrogen Peroxide) = g mol ^-1
M(Hydrogen Peroxide) = 34.016 g mol ^-1
c(Hydrogen Peroxide) = 204.4 gL ^-1

Evaluating the Investigation:

1.       Based on our results and the time it took us to finish the investigation, I would say that we worked fairly efficiently. Therefore, I do not see any places to make improvements. The only minor ‘problem’ we hit was the fact that we diluted the hydrogen peroxide too much. After filling the volumetric flask with distilled water, we found out that we had 510 ml of standard solution instead of 500 ml. This only changed our calculation number.

2.       [c(Hydrogen Peroxide) ÷ 1000] x 100 = % by mass
% by mass = 20.44%
The bottle stated a 30%, however, we got 20%. This can easily be explained through chemistry. As soon as the seal cap is opened, gases start to escape the bottle. Therefore the percentage drops over time since the gases escape. This is the reason why our results are lower than the one stated on the bottle.

3.       We were the first group to finish this experiment.  Compared to the other groups, our results were more accurate since we used the hydrogen peroxide sooner; therefore, fewer gases had escaped. Overall, the only error that may have occurred in our experiment would be from the hydrogen peroxide solution; since it wasn’t seal closed when we used the solution.

Finding The Unknowns

Aim:

To find the unknown chemicals that are in the labelled bottles

Equipment:

·         Test Tubes

·         Safety Glasses

·         Distilled Water

·         Eye Dropper

·         Calcium Hydroxide

·         Copper Sulfide

·         Silver Sulfate

·         Potassium Iodide

·         Silver Chloride

·         Copper Sulfate

·         Silver Nitrate

Results:

Set 1: (answers simulated as no actual results could be found)

bottle	A	b	c	d
Result	BaCl2	CuSO4	H2SO4	NaCl

Set 3: (answers are from actual results)

bottle	M	n	o	p	q	r
Result	H2SO4	Na2CO3	BaCl2	HCl	Pb(NO3)2	KOH

Processing Data:

set 1:

Looking at all the solutions in a test tube, we can immediately recognise one of the solutions as being blue. Now from the data sheet, we can see that copper ions have a blue colour in an aqueous solution, and copper sulfate is soluble in water. Therefore this means that once copper sulfate is added to water, it will completely dissociate to make copper ions and sulfate ions. As said before, copper ions in an aqueous solution makes a blue colour. Therefore, bottle B has to be copper sulfate. Now, by eliminating the copper sulfate, we are left with 3 bottles of unknown solution. If we add calcium hydroxide (limewater) to the solutions, we will see that one of the solutions will form a white precipitate. Looking at the data sheet, we can see that the only solution which would slightly form white precipitates would be calcium sulfate.
H₂SO_{4 (aq)} + Ca(OH)_{2 (aq)} --> CaSO_{4 (s)} + H₂O _(l)
Now, this would be the second solution that is eliminated. Now by adding copper sulfide to the two remaining solutions, we will again see that one of the test tubes forms a white precipitate. From the data sheet, we can see that the only solution that is insoluble in water is BaS.
BaCl_{2 (aq)} + CuS _(aq) --> BaS _(s) + CuCl_{2 (aq)}.
Therefore, we can say that bottle A is BaCl₂. Now from the process of elimination we can see that the only remaining solution is NaCl. However, just to prove that the remaining solution is NaCl, we just need to add silver sulfate to the solution to form a white precipitate.
2NaCl _(aq) + Ag₂SO_{4 (aq)} + Na₂SO_{4 (aq)} + 2AgCl _(s)

Set 3:

By adding potassium iodide to the solutions we can see that one of the precipitates has a yellow colour. This shows us that the precipitate formed is lead iodide.
Pb(NO₃)_{2 (aq)} + 2KI _(aq) --> PbI_{2 (s)} + 2KNO_{3 (aq)}
Now by adding silver chloride to the remaining solutions, again we can see that one of the precipitates formed has a yellow colour. From the data sheet, we can confirm that the precipitate formed is silver carbonate.
Na₂CO_{3 (aq)} + 2AgCl _(aq) --> Ag₂CO_{3 (s)} + 2NaCl _(aq)
By adding copper sulfate to the remaining solutions, would leave two of the solution with a green colour. However, one of those solution that is green will also have precipitates formed as well. This shows us that the solution that formed a precipitate is BaCl₂ and the others are HCl and H₂SO₄.
BaCl_{2 (aq)} + CuSO_{4 (aq)} --> BaSO_{4 (s)} + CuCl_{2 (aq)}
2HCl _(aq) + CuSO_{4 (aq)} --> H₂SO_{4 (aq)} + CuCl_{2 (aq)}
H₂SO_{4 (aq)} + CuSO_{4 (aq)} --> H₂SO_{4 (aq)} + CuSO_{4 (aq)}
Now to find out which of the two solutions is HCl and H₂SO4, we just need to add silver chloride to the two solutions. One of the solutions will form a white precipitate. Therefore, according to our data sheet we can confirm that the solution that has the precipitate is HCl (2 mol) and the other solution has to be H₂SO₄ based on elimination.
CuCl_{2 (aq)} + 2AgCl _(aq) --> CuCl_{2 (aq)} + 2AgCl _(s)
Now the only solution that is left is KOH. To prove that the solution is KOH, we can just add silver nitrate to the solution which would form silver hydroxide precipitates
KOH _(aq) + AgNO_{3 (aq)} --> KNO_{3 (aq)} + AgOH _(s)

Evaluation:

From the results above, I can say that the way we uncovered each of the unknown solutions; was done in the least number of steps given the solutions that we were provided. If we were given objects such as matches, would have simplified the procedures. For example, instead of testing for precipitates, we could just test for gases. This would have been a great use mainly in set 3. However, based on our results and how we got each one by proving exactly what they were, I am confident that our results are accurate. Finding what each precipitate was just purely looking at the data sheet to see what the compound was.

Conclusion:

To conclude, the results here have been modified because we had not written down how each solution was proven to be what. Therefore, we also didn’t write down what we mixed each solution with. This is why I had to work out how to get each of the solution based purely on the data sheet.

Electrolysis: Potassium Iodide (Molten)

Aim:

To find the oxidising and reducing agents of molten potassium iodide

Equipment:

1x U-tube made out of copper
A power supply
2x wires connecting power pack to carbon electrodes
2x carbon inert electrodes
Molten potassium iodide

Safety:

·         Wear protective gear such as thick boots, high durable safety glasses, lab coat and extremely durable gloves

·         Avoid contact with any molten potassium iodide because 680^oC > is a hot temperature

Procedure:

1.       Collect all equipment needed and make sure all safety precautions are done or set.

2.       Setup the apparatus the same way as the previous experiment (Electrolysis: Potassium Iodide (solution))

3.       Add molten potassium iodide to the copper U-tube and turn the power supply on.

4.       Observe and record all results.

Predicted Results:

Anode	Cathode
Collecting the gas and cooling it down allows us to see a purple/black coloured substance at room temperature	Silvery solid starts forming over the carbon electrode until there is no molten potassium iodide is left

Redox:

2I⁻ _(l) → I₂ _(g) + 2e⁻   Oxidising

2K⁺ _(l) + 2e⁻   → 2K_(s) Reducing

2I⁻ _(l) +2K⁺ _(l) → I₂ _(g) +2K_(s) Redox Equation

Discussion:

From the procedures of this experiment we can see right at the start that it is very dangerous because of many parameters. First of all, the temperature of the molten potassium iodide is extremely high. This is because of its ionic bond creating a very strong bond within the ions, and breaking these bonds requires high amount of energy such as heat. Therefore, because of these dangerous temperatures, safety equipment is compulsory.

Our group could not experiment this theory as the extreme temperatures, lack of equipment and lack of space all stopped us because of the high risks.

To conclude, we can say that this experiment is not practically possible to do in a school environment; however, this experiment is done on a massive scale in the mining industry allowing the separation of a compound to its elemental form such as potassium iodide.

Electrolysis: Potassium Iodide (Solution)

Aim:

To find the oxidising and reducing agent in an aqueous solution of potassium iodide

Equipment:

1x U-tube
Phenolphthalein indicator
0.5 mol L⁻ Potassium Iodide Solution
About 15ml of distilled water
A power supply
2x carbon inhert electrodes
2x wires connecting power pack to carbon electrodes

Procedure:

1.       Collect all equipment needed and make create an observation and reaction table

2.       The diagram below shows the basis of the apparatus’ setup.

3.       Once the setup looks like the diagram below add a mix of 15 ml of potassium iodide and 15 ml of the distilled water

4.       Turn the power (6V supply should be enough)

5.       Once you see some reaction take place, add a few drops of the phenolphthalein indicator to check for any type of bases that may be in the solution

6.      

Record all results and observations.

Results:

Anode	Cathode
Solution around anode becomes yellowish/brown in colour.	Solution around cathode bubbles and is clear until Phenolphthalein indicator is added then solution becomes purple.

Redox:

2I⁻ _(aq) → I₂ _(aq) + 2e⁻   Oxidised
2H₂O _(l) + 2e⁻   → H₂ _(g) + 2OH⁻ _(aq) Reduced
2I⁻ _(aq) + 2H₂O (l) → I₂ _(aq) + H₂ _(g) + 2OH⁻ _(aq) Redox Equation

Discussion:

This experiment has shown and taught us a few lessons. First off we can see that water is a stronger oxidising agent than potassium. For this reason water reacts at the cathode instead of potassium, therefore forming hydrogen gas and hydroxide ions. To prove this we added a few drops of phenolphthalein indicator which resulted the solution to turn to purple colour at the cathode, in other words meaning hydroxide ions as the phenolphthalein turns purple when it comes in contact with a basic solution and colourless with an acidic solution.

The anode has a very different story as the water is not a stronger reductant than iodine, therefore, we can say that iodine reacts at the ₂ is formed at the anode however our group could not confirm this to be Iodine gas as the solution turned brownish/yellow, and the WACE WA data sheet stated that I₂ in an aqueous solution turns to a brownish/yellow colour. Therefore, this would mean that the I₂ is actually an ion.

anode instead of the water. We can see that I

Our experiment resulted in near perfect results it was conducted more than once and all steps being performed as perfectly possible.

To conclude, we can see that water is a stronger oxidising agent compared to potassium, however it is a weaker reducing agent compared to iodine