• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13

To determine the rate law for a chemical reaction among hydrogen peroxide, iodide and acid, specifically by observing how changing each of the concentrations

Extracts from this document...


Experiment 3 - Chemical Kinetics Objectives 1. To determine the rate law for a chemical reaction among hydrogen peroxide, iodide and acid, specifically by observing how changing each of the concentrations of H2O2, and H+ affects the rate of reaction. 2. To observe the effects of temperature and catalyst on the rate of reaction. Introduction Generally, two important questions may be asked about a chemical reaction: (1)How far do the reactants interact to yield products, and (2) how fast is the reaction? "How far?" is a question of chemical equilibrium which is the realm of chemical thermodynamics. "How fast?" is the realm of chemical kinetics, the subject of this experiment. In this experiment we will study the rate of oxidation of iodide ion by hydrogen peroxide which proceeds according to the following reaction: H2O2 (aq) + 2 I-(aq) + 2H+(aq)-->I2(aq) + 2H2O(l) By varying the concentrations of each of the three reactants (H2O2, I- and H+), we will be able to determine the order of the reaction with respect to each reactant and the rate law of the reaction, which is of the form: Rate = k [H2O2]x[I-]y[H+]z By knowing the reaction times (?t) and the concentrations of H2O2 of two separate reaction mixtures (mixtures A & B), the reaction order of H2O2, x, can be calculated. x = log(?t2/ ?t1) / log ( [H2O2]1/[H2O2]2 ) The same method is used to obtain the reaction order with respect to I- (mixtures A & C) and H+ (mixtures A & D). Procedures Part I) Standardization of H2O2 Solution 1. A stand, a burette clamp and a white tile were collected to construct a titration set-up. 2. A burette was rinsed with deionized water and then with 0.05 M Na2S2O3 solution. 3. The stopcock of the burette was closed and the sodium thiosulphate solution was pour into it until the liquid level was near the zero mark. ...read more.


dm3 / [(30+30+45+25+5+5+10) /1000 dm3] = 0.01 mol dm-3 Since Ka = [H+] [CH3COONa] / [CH3COOH] 1.8x10-5 = [H+] (0.01) / (0.1) [H+] = 1.8x10-4mol dm-3 Reaction Mixture Time / s Acetic acid concentration/ M Sodium acetate concentration/ M Calculated H+ Concentration/ M A 172 0.01 0.01 1.8x10-5 D 169 0.1 0.01 1.8x10-4 The order of reaction respect to iodide: Since x = log(?t2/ ?t1) / log([H+]1/[H+]2) x = log(169/172)/log[(1.8x10-5)/( 1.8x10-4)] x = 0.0076 = 0 It is zero order reaction respect to H+. 2(a) From your measurements, what are the reaction orders with respect to each of the three reactants H2O2, I- and H+? Ans: It is zero order respect to H+, first order respect to iodide, I- , and first order respect to hydrogen peroxide, H2O2. (b) What is the rate expression (rate law) for the uncatalyzed reaction? What is the total reaction order? Ans: The rate law is: Rate = k[H2O2][I-] where k is rate constant Therefore it is second order respect to the reaction. (c) Is the rate law consistent with the idea that the mechanism of the reaction is H2O2 (aq) + 2 I-(aq) + 2H+(aq)--> I2 (aq) + 2H2O(l) (all in one step)? Explain why or why not. Ans: If the mechanism involves only one step only, then it means that this step is the rate determining step. The rate law is Rate = k[H2O2][I-]. The rate determining step should include only one hydrogen peroxide and one iodide molecule only, but the above reaction involve one hydrogen peroxide , two iodide and two hydrogen ion molecule. Also, the hydrogen ion should not be present in the rate determining step as it is zero order and the rate law does not include it. Actually, according to collision theory, it is not favour that 3 molecules collides together and proceed a reaction. Collision of two or one molecule is more favourable. ...read more.


(iii). Since the color change from iodine to iodide cannot be used to accurate the end point (the change in color brown to yellow to colourless is difficult to observe), starch is used as the indicator. Starch + iodine --> blue complex (iv). The rate equation for a reaction may help to deduce the mechanism of the reaction. The order of a reaction is an important implication of the rate determining step. From the order, we can deduce the mechanism and the rate determining step. The orders are experimentally found, they are not necessarily equal to the coefficient of the reactant in the equation unless the reaction is a single reaction. (v). A catalyst is a substance which alters the rate of a particular chemical reaction, but itself is chemically unchanged at the end of the reaction. A catalyst can change the rate of a reaction by providing an alternative pathway for the reaction, usually with a pathway of lower activation energy than that of the uncatalyzed reaction. There are some improvements in this experiment. First, hydrogen peroxide is unstable, it decomposes to water and oxygen by time. Therefore do the titration as quick as possible. 2H2O2(aq) --> 2H2O(I) + O2(g) Second, the concentration of iodine increase is due to the iodide can be oxidized by oxygen which promoted by acids. Therefore do the titration as quick as possible. 4I-(aq) + O2(g) + 4H+(aq) --> 2I2(aq) + 2H2O(aq) Third, as for the human error, the problem can be minimized by performing the titration by the same person. So, the reading can be taken by the same person and the color change can be observed by the same person. Conclusion In the experiment, the reaction was found to be zero order respect to (H+), it is first order respect to iodide, (I-) , it is first order respect to hydrogen peroxide, (H2O2). Hence the rate law is Rate = k[H2O2][I-]. The rate of reaction is increase when the temperature is increase and the rate is increase when a positive catalyst is added to the reaction. ?? ?? ?? ?? 1 ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Patterns of Behaviour section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Patterns of Behaviour essays

  1. Rates of Reaction - The Iodine Clock

    I am therefore required to perform three different experiments, changing the concentration of a single variable in each one. This will mean that the variables, independent and controlled, will change in each 'set'. All experiments should be performed at standard temperature and pressure (S.T.P.) - 298K and 1 ATM pressure.

  2. The Iodine Clock Investigation

    Bearing in mind that there is limited lab time, this is also not ideal. Therefore, in the main experiment, a thermostatic water bath will be used that will accurately maintain the temperature of the water, and also be capable of holding a large volume of test tubes at any given time.

  1. The Decomposition of H2O2 using Catalase, in yeast as a catalyst.

    This is why there is still oxygen given off at 60 and 70?C, but the rate of reaction is not highest at these temperatures. So from this experiment I can conclude that the optimum temperature for the enzyme catalase is around 40?C, I can also conclude the rate of reaction,

  2. The Effect of Catalase in the Breakdown of Hydrogen Peroxide

    Therefore; there are less available active sites to prolong the reaction. In the peeled potato reaction, the trials for both concentrations of catalase were slow. Though both had a pattern that showed continuality throughout the experiment, however; there was a small different between the 50% and the 100% when there should actually be a doubled increase.

  1. Activity of Diastase On Starch

    This is called denaturisation. Other forces that also disrupt these bonds will have the same effect: extremes of pH, mixing a detergent, extreme concentrations of salt and so on. I like the analogy of a car built for road use that you take off-road.

  2. The Iodine Clock

    Fair Test: To make sure that I keep the test fair I will keep the following the same throughout the experiments: * Keep the amounts of Potassium iodide and Ammonium persulphate the same each time as these will affect the speed at which the experiment takes place.

  1. An investigation to see the difference in the rate of reaction when catalyse is ...

    I shall always be checking that the stopper on the test tube is always firmly on tight because of the gas pressure, and any other equipment that could be faulty. Keep the same apparatus, because apparatus differs for instance on test tube may be more air tight than the other.

  2. Investigating the rate of a reaction

    At intervals of 20, 40, 60, 80, 100, 120 seconds, which can be seen from the stopwatch, note down the amount of carbon dioxide released from the reading on the gas syringe in cm3. Note the result down in the table of results.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work