Record the following lab data in the table below for trial 4. Data Analysis Fill

Question

Record the following lab data in the table below for trial 4.

Data Analysis

Fill in the following table. Look below the table for instructions on how to calculate the values for each row of the table.

Calculate the final volume of the reaction mixture after the contents of beaker B are added to beaker A. Report your answer in liters.

All S2O32- is consumed at the end of the reaction. Therefore, the moles of S2O32- consumed can be calculated using the equation below.
(moles S2O32-)consumed = Mstock x (Vstock)

The I3- produced in reaction 1 reacts with S2O32- in reaction 2 as shown.
I3- (aq) + 2S2O32- (aq) ? 3I- (aq) + S4O62- (aq)

Therefore, for every mole of I3- produced, 2 moles of S2O32- have reacted.
(moles I3-)produced = (moles S2O32-)consumed / 2

[S2O32-]consumed = (moles S2O32-)consumed / (Vtotal)

Calculation is the same as d, but using I3- instead of S2O32-.

Fill in the following table. Look below the table for instructions on how to calculate the values for each column of the table.

Data can be obtained from your lab notes. ?t is the time to color change.

The rate ?[I3-] / ?t = ([I3-] produced / ?t; [I3-] produced was calculated in the first table under Data Analysis.

Take the log of the rate obtained in column b.

The volume for potassium iodide can be obtained from the table in the Procedures.

The concentration can be calculated as [S2O32-] was calculated in the first table under Data Analysis, row b.

Take the log of the concentration obtained in column e.

Follow instructions for column d, but for ammonium persulfate.

Follow instructions for column e.

Follow instructions for column f.

Conclusions

If a large amount of heat was released at the start of the reaction, what effect would this have on the rate measurements?

Determine the value of q from the graph. Explain your answer.

What is the rate law for the reaction?

For each trial, calculate the rate constant. What is average value of the rate constant?

In this experiment, you assumed that [S2O82–] >> [S2O32–]. To find out if this assumption is correct, calculate the ratio of [S2O82–] / [S2O32–] for Trial 3. (In Trial 3, the [S2O82–] was lowest, and therefore the ratio [S2O82–] / [S2O32–] is the smallest).   

In this experiment, you assumed that only a small amount of S2O82– was used during the time trial so that the concentration of this reactant did not change appreciably during the course of the reaction. Calculate how much S2O82– was used in Trial 3 and the percent remaining at the end of the reaction. Was this assumption valid?   

A student conducting the iodine clock experiment accidentally makes an S2O32- stock solution that is too concentrated. How will this affect the rate measurement and the calculated value of k?

The iodine clock experiment consists of the following three reactions, where Reaction 1 is slow relative to Reactions 2 and 3. Predict what would happen if Reaction 2 was slow relative to Reaction 1 (assume that Reaction 3 is still fast).

Reaction 1: 3I- (aq) + S2O82- (aq) ? I3- (aq) + 2SO4- (aq) slow
Reaction 2: I3- (aq) + 2S2O32- (aq) ? 3I- (aq) + S4O62- (aq) fast
Reaction 3: I3- (aq) + starch (aq) ? 3I- ---- starch (bluish black) fast

(a) volume of potassium iodide solution added to Erlenmeyer flask A in mL 12.5mL (b) volume of potassium nitrate solution added to Erlenmeyer flask A in mL 12.5mL (c) volume of starch solution added to Erlenmeyer flask A in mL 5 mL (d) volume of (NH4)2S2O8 solution added to Erlenmeyer flask B in mL 25mL (e) volume of (NH4)2SO4 solution added to Erlenmeyer flask B in mL 0 mL (f) volume of Na2S2O3 solution added to Erlenmeyer flask B in mL 10 mL (g) the initial color of the solution after adding flask B to flask A clear (h) time for color change 42 seconds (i) color of the final solution black

Explanation / Answer

ans of last question- In a reaction that occurs in various step [reaction], the reaction which is slowest is the rate determinig step.I3- formed in reaction 1 remains in the reaction mixture.Intensely colored starch triiodide complex forms since starch is present in the solution.The colour of solution is very dark blue.

Because reaction (3) is much faster than reaction (1) but depends upon the product of reaction (1). So rate of reaction(3) is equal to reaction(1).

2nd last question-In this laboratory procedure, blue color appears when all the thiosulfate is consumed.when concentration of S2O32- stock solution is too high it takes much more time to be consumed. To determine the initial rate of the reaction for any trial,the the initial concentration of thiosulfate,which is known & the final concentration of thiosulfate which is zero we know.The initial time is zero and the final time is just how long it took the solution to turn blue which becomes very long due to concentration of S2O32.

reaction rate= -1/2 delta[ S2O32]/dt

rate of reaction becomes lower than actual reaction rate.

ans-the rate law for the reaction-

rate = k[I-]^6[BrO3-][H+]^6

ans- A large amount of heat was released at the start of the reaction,the reaction is exothermic . Low temperature is favourable for reaction. No effect would this have on the rate measurements.

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