Lab Report Kinetics of Chemical Reactions Kinetics of chemical reactions is how fast a reaction occurs and determining how the presence of reactants affects reaction rates. In this experiment the rate of reaction for Fe+3 and I- is determined. Because the rate of chemical reactions relates directly to concentration of reactants, the rate law is used to find the rate constant, and calculated with specified temperatures. Two catalyst reactants are used in the experiment, thiosulfate and starch, to dictate the time of reactions. The order with respect to Fe+3 and I- is also determined by graphing the slope of the log rate initial as a function of the log (Fe+3) or (I-). The activation energy is also graphed with the rate constant as a function of the inverse of the temperature. Procedure The volumes of solutions were obtained and placed into two separate beakers as shown in the tables below. Reagent Fe+3 HNO3 KI S2O3-2 Starch Volume needed 150 mL 150 mL 100 mL 100 mL 50 mL Beaker 1 Beaker 2 Combo .04 M Fe+3 .15 M HNO3 H2O .04 M KI .004 M S2O3-2 Starch H2O 1 10.00 20 20 10.00 10.00 5 25 2 20.00 10 20 10.00 10.00 5 25 3 30.00 0 20 10.00 10.00 5 25 4 10.00 20 20 5.00 10.00 5 30 5 10.00 20 20 15.00 10.00 5 20 Part A: The two beakers were allowed to chill for 15 minutes. Their contents were then mixed and put back on ice. Combination #1 was run at room temperature on a separate trial. The temperature was recorded at 23 0C. When the solution turned blue, the time was recorded. Finally, combination #1 was run at 45 0C and the solution was monitored until it became blue. Part B: Combinations #2 - #5 were all run at room temperature. The temperature varied slightly for each combination... ... middle of paper ... ...I- rate initial log I- initial Log rate initial 1. .004 2.03 x 10-6 -2.40 -5.69 4. .002 5.94 x 10-6 -2.70 -6.23 5. .006 4.01 x 10-6 -2.22 -5.40 A) Order of reaction with respect to [Fe+3] » 1 B) Order of reaction with respect to [I-] » 2 C) Rate law = K [Fe+3]1 [I-]2 2.03 x 10-6 = K [.004]1 [.004]2 K = 31.7 M-2s-1 D) ln K = ln 31.7 = 3.46 Rate initial for hot and cold: HOT: Rate initial = ½ (.0004) / 5.22s Rate initial = 3.83 x 10-5 3.83 x 10--5 = K [.004]1 [.004]2 K = 598 M-2s-1 COLD: Rate initial = ½ (.0004) / 2043.47s Rate initial = 9.79 x 10-8 9.79 x 10-8 = K [.004]1 [.004]2 K = 1.53 M-2s-1 1/T ln K Cold .00364 .425 Room .00338 3.46 Hot .00315 6.39 Ea M = - Ea/R M = -18001 -Ea = -18001 x .008314 -Ea = -149.7 Ea = 149.7 kJ
Two equations were used in this experiment to determine the initial temperature of the hot water. The first equation
The solution was heated for 10 minutes and distilled water was added to it periodically to ensure that the volume of solution in the beaker stayed the same. The solution was then spilt equally into two centrifuge tubes and the solution was centrifuged until there was no more suspended precipitate. The supernatant was then decanted into a centrifuge tube. This was the stock solution. The logic tree from part two was used to analyze and identify the unknown
The Iodine Clock Investigation Introduction This is an investigation into the rate of a reaction and the factors that contribute to how fast a reaction will take place. Through the recording and analysis of raw data, this investigation also allows us to apply generally accepted scientific rules and to test them against results gained from accurate experimental procedures. Aim The aim of this experiment is to investigate the rate at which iodine is formed when the concentration and temperature of the reactants are varied, and to attempt to find the order and activation energy. The Chemistry 'THE IODINE CLOCK' - This is the experiment that will be used to investigate reaction rates, and it is a reaction between acidified hydrogen peroxide and potassium iodide: 2H+(aq)
8. Continue stirring. Record the temperature at which crystals begin to appear in the solution.
Objective: The objective of the experiment is to determine what factors cause a change in speed of a reaction. It is also to decide if the change is correlated with the balanced equation of the reaction and, therefore, predictable. To obtain a reaction, permanganate, MnO_4^(1-), must be reduced by oxalic acid, C_2 O_4 H_2. The balanced equation for the reaction is:
I used Cheerios, distilled water, and a pestle and mortar. I ground the Cheerios until they had a fine, sand-like texture and consistency. I then added distilled water and mixed until I was left with a thin, runny solution, that was beige in color. Once I had the stock solution made, I was able to perform my first experiment, beginning with Benedict’s reagent. For this experiment I used a hot plate, beaker, and three test tubes, one labeled + (positive control), - (negative control), and Cheerios. Two milliliters of each solution was then added to the tubes they were labeled to go into. In this experiment, the positive control was a glucose solution. I then added two milliliters of Benedict’s reagent to each tube. Once a boiling bath had been made using water, the beaker, and the hot plate, each of the three test tubes were places, sitting upwards, into the boiling bath. A timer was set for three minutes, and I recorded the color
Repeat step 4 after another minute continue this for 5 minutes Beaker Start 1 2 3 4 5 Temperature change 1 59°c 57°c 56°c 55°c 52°c 50°c -9°c 2 72°c 71°c 66 °c 63°c 60°c 57°c -15°c 3 86°c 71°c 64°c 58°c 56°c 52°c -34° c 4 72°c 68°c 65°c 60°c 57°c 53°c -19°c Main Investigation ------------------ Aim To find out weather a beaker with a larger surface area cools quicker than one with a smaller surface area. Fair test To make it a fair test we will keep the following the same: Colour of tin - we will use clear beakers Amount of water - we will use 100ml
Once the mixture had been completely dissolved, the solution was transferred to a separatory funnel. The solution was then extracted twice using 5.0 mL of 1 M
Abstract: This week we experimentally determined the rate constant k for the reaction 2HCl (aq) +Na2S2O3 (aq) → S (s) + SO2 (aq) + H2O (l) + 2NaCl (aq). In order to do this the average reaction time was recorded in seconds during two trials. The data from the experiment shows this reaction is in the first order overall: rate=.47s-1 [HCl]0 [Na2S2O3]1. These findings seem to be consistent with the expected results
- Temperature was measured after and exact time i.e. 1 minute, 2 minutes, 3 minutes.
The aim of this experiment was to investigate the affect of the use of a catalyst and temperature on the rate of reaction while keeping all the other factors that affect the reaction rate constant.
The aim of this investigation is to: 1) find the rate equation for the reaction between hydrogen peroxide, potassium iodide and sulphuric acid by using the iodine stop clock method and plotting graphs of 1/time against concentration for each variable. Then to find the activation energy by carrying out the experiment at different temperatures using constant amounts of each reactant and then by plotting a graph of in 1/t against I/T, 3) to deduce as much information about the mechanism as possible from the rate equation.
The purpose of the experiment is to identify and understand reactions under kinetic and thermodynamic control. A reaction under kinetic and thermodynamic control can form two different types of products. A reaction under kinetic control is known to be irreversible and the product is formed quickly. A reaction under thermodynamic control is known to require rigorous conditions. It is also reversible. The final product is more stable than the product made by kinetic control. The chart below shows the two types of reaction coordinates:
has on the rate of reaction. I will do this by recording the time it
There are five factors which affect the rate of a reaction, according to the collision theory of reacting particles: temperature, concentration (of solution), pressure (in gases), surface area (of solid reactants), and catalysts. I have chosen to investigate the effect of concentration on the rate of reaction. This is because it is the most practical way to investigate. Dealing with temperatures is a difficult task, especially when we have to keep constant high temperatures. Secondly, the rate equation and the constant k changes when the temperature of the reaction changes.