User:LkNsngth/chem

Rate Order Determination

In this lab you will determine the rate law for the reaction of iodine ion, hydrogen peroxide and hydrogen ion. The reaction that you will be observing is:

$$H_2O_{2(aq)} + 3I^-_{(aq)}+2H^+_{(aq)}\rightarrow I^-_{3(aq)}+2H_2O_{(aq)} $$ The rate law for this reaction would have the form

$$ R=k[H_2O_2]^x[I^-]^y[H^+]^z$$ where the rate of the reaction would be the moles/L of I-3(aq) that are formed per second. You are to use the method of initial rates to determine the value of x,y,z, and k. By changing the concentration of hydrogen peroxide, iodide and hydrogen ion one at a time, as you measure the reaction rate you should be able to determine the x, y, and z.

You will time the reaction using a built-in chemical alarm clock that causes the reaction solution to suddenly change color after a certain amount of I3- has been formed. This effect is achieved by adding thiosulfate and starch. The thiosulfate reacts with the tri-iodide converting it to iodide according to the following reaction: $$ 2 S_2 O_{3(aq)}^{2-} + I_{3(aq)}^{-} \rightarrow 3I^{-}_{(aq)}+S_4O_{6(aq)}^{2-}$$

So, as long as there is thiosulfate in the solution all the trio-iodide formed will be very quickly concerted to iodine. Once all the thiosulfate is used up then trio-iodide will be formed and react with starch to form a blue color.

So this alarm clock will allow us a way to time the reaction, what we need to worryabout is the number of moles of I3- that were formed. From the reaction above, we know that for ever 2 moles of thiosulfate thatwe add 1 mole of I3- wil be used up. Given the amount of thiosulfate to start with we can therefore determine the amount of tri-iodide produced, dividing that by the time will give the reaction rate.

Part A
1A. Obtain 2 clean 400mL beakers and label them 1 and 2. To each beaker add all the reactants except hydrogen peroxide for the corresponding experiment according to the table bellow. Measure the volumes as accurately as possible. Record exact reactant concentrations on your data sheet.

2A. Measure 10.0 mL of H2O2 into a separate clean small beaker.

Table 1A: initial volumes of reactants

Beaker 1: 100mL H2O; 50mL 0.05M Kl; 5mL 0.05M Na2S2O3; 5mL 0.1% starch; 30.0 PH 4.7 Buffer; Total Volume = 200; 10mL 0.8M H2O2 in separate beaker.

Beaker 2: 125mL H2O; 25mL 0.05M Kl; 5mL 0.05M Na2S2O3; 5mL 0.1% starch; 30.0 PH 4.7 Buffer; Total Volume = 200; 10mL 0.8M H2O2 in separate beaker.

3A. Measure the temperature of the solution in each beaker. Temperatures should be within 1 degree C of each other. If they are not wait until they are.

We measured 22.0 C and 21.9 C

4A. Beaker 1: have a clock ready. Check and record the temp. Add H2O2 and start timing; stir continuously and watch the solution for the first appearance of color. We recorded 46 seconds and it ended at 22.1 C 5A. Beaker 2: First, measure 10 mL of H2O2 into the clean small beaker. Repeat as in step 4A. We recorded 1:36 for time and the beaker started at 22.0 C and ended 22.1 C

Part B
1B obtain a clean 400mL beaker and label it 3. Add the reactants listed in table 1B except for H2O2. Measure the volumes as accurately as possible. Record exact reactant concentrations on your data sheet. Beaker 3: 115mL H2O; 25mL 0.05M Kl; 5mL 0.05M Na2S2O3; 5mL 0.1% starch; 30.0 PH 4.7 Buffer; Total Volume = 200; 20mL 0.8M H2O2 in separate beaker. 2B. Measure the temperature of the solution. 3B. Carefully measure the correct amount of H2O2 into a clean small beaker. Carry out the measurements of 4A. We recorded 33 seconds and the temperature went from 21.4 C to 21.5 C (there is a line separating this from a second set of data that reads time: 58 seconds and the temperature remains at 22.2 C)

Part C
In parts A and B, all reactions were buffered to pH=4.7, so the hydrogen ion concentration was 2.00*10-5M. Since the hydrogen ion is a reactant we need to vary it's concentration while holding the others constant. In experiment 4 we will add 25.0 mL of 0.30M acetic acid (CH3COOH). This increases the H+ by a factor of ten. The new concentration is ''2.00*10-4 with a pH of 3.7. 1C. As in A and B, use a clean 400mL beaker. Label it 4. Add all reactants except H2O2 according to the table. The amounts of H2O2 and Kl are the same as in 2, but the H+ concentration is 10 times higher. Record exact volumes and concentrations on data sheet. Beaker 4: 100mL H2O; 25mL 0.05M Kl; 5mL 0.05M Na2S2O3; 5mL 0.1% starch; 30.0 PH 4.7 Buffer; 25mL of 0.30M Acetic Acid; Total Volume = 200; 10mL 0.8M H2O2 in separate beaker. 2C. Continue as in part 4A. We recorded 21.3 C end temp and 1:38 reaction time.

Part D
We will see how temperature affects the reaction.

1D. Use the same setup as Beaker 2 from Part A.

2D. Heat about 400mL of water to near boiling in a different beaker.

3D. Place about 200mL of tap water in an 800mL beaker. Add hot water while stirring until the water temp is about 20 C higher than room temperature in experiments 1-4.

4D. Put beaker containing the reaction mixture into the water bath. Stir and measure temperature. Record final temperature.

5D. Carry out reaction as in 4A. adding room temp H2O2 to the warm reaction mixture. Record the elapsed time. You can assume the addition of room temp H2O2 does not change the reaction temp. We recorded that the room temp was 22, and we heated things up to 42. The reaction time was 1L31 and the temperature of the solution went from 39.4 C to 40.6 C

Questions to Ponder:
1) For each reaction calculate the initial concentration of each reactant and the reaction rate. 2) Calculate the reaction order for each reactant. 3) Calculate the reaction constant for reactions 1-3, what is their average? 4) Write the rate law for $$H_2O_{2(aq)} + 3I^-_{(aq)}+2H^+_{(aq)}\rightarrow I^-_{3(aq)}+2H_2O_{(aq)} $$ 5) Calculate the reaction rate constant for reaction 5 (there is no reaction 5 on the sheet so he probably means beaker 5) 6) Calculate the activation energy for $$H_2O_{2(aq)} + 3I^-_{(aq)}+2H^+_{(aq)}\rightarrow I^-_{3(aq)}+2H_2O_{(aq)} $$