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An Investigation into the Decomposition of Hydrogen Peroxide
Aim: To investigate the rate of decomposition of H2O2 with different
amounts of
catalyst (MnO2).
Hypothesis: When H2O2 and a catalyst are mixed together, the catalyst
would break down H2O2 into water and oxygen. This will result in
bubbles being produced. With the data of these oxygen bubbles, the
rate at which H2O2 decomposed could be found out.
2H2O2 (l) à2H2O + O2
The controlwould be to maintain the same temperature (room
temperature) and to use the same amount of hydrogen peroxide (10ml) in
all the tubes.
The variableis the use of different amount of catalyst in different
test tubes.
Equipment: 10% v/v solution of H2O2
Catalyst - MnO2
5 test tubes
Stopwatch
Beaker with distilled water
Glass tube
5 single-holed rubber stoppers
Procedure:
1) Fill all 5 test tubes with 10 ml of H2O2
2) Measure 0.01g of MnO2 using a filter paper and pour this into the
first test tube. Immediately close the test tube with the single-holed
rubber stopper in which one end of a glass tube must be inserted. The
other end of the glass tube must lead to a beaker with distilled
water.
3) We can see oxygen bubbles being let through the glass tube into the
beaker. This shows the decomposition of H2O2 . Start the stopwatch and
also start counting the no. of bubbles produced.
4) Note down the results for every minute, up to 5 minutes.
5) Repeat steps 2, 3 and 4 by using different amounts of MnO2. i.e.
0.02g, 0.03g, 0.04g and 0.05g.
Data Collection:
For data collection we counted the number of oxygen bubbles that is
being produced for each amount of MnO2. We did this for each minute up
to five minutes because we thought that this would give a wider range
of data, which would later be used in processing and interpreting. We
have also mentioned the two areas of uncertainties.
2-ethyl-1,3-hexanediol. The molecular weight of this compound is 146.2g/mol. It is converted into 2-ethyl-1-hydroxyhexan-3-one. This compounds molecular weight is 144.2g/mol. This gives a theoretical yield of .63 grams. My actual yield was .42 grams. Therefore, my percent yield was 67%. This was one of my highest yields yet. I felt that this was a good yield because part of this experiment is an equilibrium reaction. Hypochlorite must be used in excess to push the reaction to the right. Also, there were better ways to do this experiment where higher yields could have been produced. For example PCC could have been used. However, because of its toxic properties, its use is restricted. The purpose of this experiment was to determine which of the 3 compounds was formed from the starting material. The third compound was the oxidation of both alcohols. This could not have been my product because of the results of my IR. I had a broad large absorption is the range of 3200 to 3500 wavenumbers. This indicates the presence of an alcohol. If my compound had been fully oxidized then there would be no such alcohol present. Also, because of my IR, I know that my compound was one of the other 2 compounds because of the strong sharp absorption at 1705 wavenumbers. This indicates the presence of a carbonyl. Also, my 2,4-DNP test was positive. Therefore I had to prove which of the two compounds my final product was. The first was the oxidation of the primary alcohol, forming an aldehyde and a secondary alcohol. This could not have been my product because the Tollen’s test. My test was negative indicating no such aldehyde. Also, the textbook states that aldehydes show 2 characteristic absorption’s in the range of 2720-2820 wavenumbers. No such absorption’s were present in my sample. Therefore my final product was the oxidation of the secondary alcohol. My final product had a primary alcohol and a secondary ketone
When the flame was blown out and the glowing wooden splint was placed halfway into the test tube containing H2O2 and MnO2 crystals, the splint reignited and caught flame once again. This demonstrates the decomposition of H2O2 into water and hydrogen. MnO2 is a catalyst that increases the rate at which H2O2 decomposes. Adding oxygen to a fire will cause it to burn faster and hotter and the oxygen rich test tube allowed the splint to reignite.
Cu (aq) + 2NO3 (aq) + 2Na+ (aq) + 2OH- (aq) → Cu(OH)2 (s) + 2Na+ (aq) + 2NO3(aq)
This enzyme speeds up the break down of hydrogen peroxide into water and oxygen, as enzymes are biological catalysts. [IMAGE]The reaction: Hydrogen peroxide Water + Oxygen Catalase -------- [IMAGE] 2H2O2 2H2O + O2 Apparatus: Hydrogen Peroxide, Several sticks of celery, Stand, boss and clamp, 100ml conical flask, 25cm3 burette, 1800cm3 beaker, Rubber bung with delivery tube, Distilled water, Large container filled with water, 10cm3 measuring cylinder, 10cm3 syringe, 20cm3 syringe, Blender, Knife, Ceramic tile, Electronic balance (correct to 2 decimal places), Sieve, Stopwatch/timer. The variables: There are many possible variables in this investigation, such as pH, temperature, the concentration of substrate and the concentration of the enzyme.
== == == = This is what I'm going to be changing in the experiment and this will be the temperature and the concentration of the yeast. There are several variables in this experiment, they are: · Amount Used - Too much or too little of the hydrogen peroxide causes the reaction to speed up/slow down producing different amounts of oxygen.
2H2O(aq) à 2H2O(l) + O2 (g) It is able to speed up the decomposition of hydrogen peroxide because the shape of its active site matches the shape of the hydrogen peroxide molecule. This type of reaction were a molecule is broken down into smaller pieces is called an anabolic reaction. I will be studying the effect of concentration of the catalase in this reaction. Hypothesis Hydrogen peroxide will breakdown to oxygen in water in the presence of catalase.
Investigating the Factors Influencing the Rate of Reaction Between Sodium Thiosulphate and Dilute Hydrochloric Acid
9. Get your stopwatch ready and drop the Alka-Seltzer tablet at the same time you started the timer. 10. When it finishes dissolving (you can see through the water and there is no more fizzing.) stop the timer and record the results. 11.
While measuring the effect of the use of a catalyst and temperature on the reaction rate, several factors must be kept constant. During the reaction with a catalyst, the temperature will be kept constant (at room temperature), concentration, pressure and since the reaction involves liquids, the surface area will be kept constant at all times, however they must be mixed the same.
The first experiments investigate the order of reaction with respect to the reactants; hydrogen peroxide, potassium iodide and sulphuric acid by varying the concentrations and plotting them against 1/time. An initial rate technique is used in this experiment so ‘the rate of reaction is inversely proportional to time.’ To find the order of reaction in respect to the reactants, 1/time is plotted against the concentration of Hydrogen Peroxide using the equation:
I shall be measuring how much gas is given off. This will be done by measuring the amount of froth on the surface of the liquid. The oxygen released is collected in the form of these bubbles. The equation for the reaction is: (catalase) [IMAGE] H2O2 2H2O + O2 (hydrogen peroxide) (2 part water) (oxygen) I will change the concentration of H2O2 and O2 (making sure the volume stay the same, when one part of a H2O2 particle is taken, an O2 particle is added. Prediction
Investigate the Effect of pH on Immobilised Yeast Cells on the Breakdown of Hydrogen Peroxide
2. In the large beaker, put water and boil it completely. After that, remove the beaker from heat. 3. Sample tubes (A-D) should be labeled and capped tightly.
tube. Add 6 mL of 0.1M HCl to the first test tube, then 0.1M KMnO4 and
In this experiment three different equations were used and they are the Stoichiometry of Titration Reaction, Converting mL to L, and Calculating the Molarity of NaOH and HCl (Lab Guide pg. 142 and 143).