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11 Dec 2019

Write the equation for calculating the initial rates of each of these reactions. Rate = Î[I2]/Ît The indicator changed color when all the thiosulfate ions, S2O32-, were used up. Upon the instant the solutions were mixed, you calculated a molarity of S2O32- (from Q 9). Since the concentration of thiosulfate changed from this value to zero this is the value of Î[S2O32-].

What is the value of Î[I2] at the point of the color change for each of the runs? (Show your work) Consider the balanced reaction for the clock step:

I2(aq) + 2 S2O32-(aq) â 2 I-(aq) + S4O62-(aq)

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Everytime I make the graph based off my calculations in the chart it looks like this and I have no idea if it is even close? Please help ! For the questions I just need to understand how to start them and what info is nessiary to do so. very confused.

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.

a	total volume (L)	0.065L
b	mol S₂O₃^2-_consumed	0.010 mol
c	mol I₃^-_produced	0.015M
d	[S₂O₃^2-] _consumed (M)	0.154M
e	[I₃-] _produced (M)	0.231M

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 S₂O₃^2- is consumed at the end of the reaction. Therefore, the moles of S₂O₃^2- consumed can be calculated using the equation below.
(moles S₂O₃^2-)_consumed = M_stock x (V_stock)

The I₃^- produced in reaction 1 reacts with S₂O₃^2- in reaction 2 as shown.
I₃^- (aq) 2S₂O₃^2- (aq) â 3I^- (aq) S₄O₆^2- (aq)

Therefore, for every mole of I₃^- produced, 2 moles of S₂O₃^2- have reacted.
(moles I₃^-)_produced = (moles S₂O₃^2-)_consumed / 2

[S₂O₃^2-]_consumed = (moles S₂O₃^2-)_consumed / (V_total)

Calculation is the same as d, but using I₃^- instead of S₂O₃^2-.

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

	a	b	c	d	e	f	g	h	i
Trial	Ît	Î[I₃^-] / Ît (M/s)	log(Î[I₃^-] / Ît)	V 0.2 M KI added (mL)	[I^â]₀(M)	log[I^â]₀	V of 0.2 M (NH₄)S₂O₈ (mL)	[S₂O₈]₀ (M)	log[S₂O₈]₀
1	19	.011	-1.94	25	.077	-1.35	25	.077	-1.11
2	40	.0055	-2.24	25	.077	-1.11	12.5	.038	-1.42
3	85	.0030	-1.84	25	.077	-.1.11	6.25	.019	-1.72
4	42	.0055	-2.26	12.5	.038	-1.42	25	.077	-1.11

Conclusions

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

The rate of reaction would be much slower due to its exothermic reaction properties.

Click the box underneath the graph to show the trendline. It will automatically calculate your slope and intercept. Record them below.

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 [S₂O₈^2â] >> [S₂O₃^2â]. To find out if this assumption is correct, calculate the ratio of [S₂O₈^2â] / [S₂O₃^2â] for Trial 3. (In Trial 3, the [S₂O₈^2â] was lowest, and therefore the ratio [S₂O₈^2â] / [S₂O₃^2â] is the smallest).

In this experiment, you assumed that only a small amount of S₂O₈^2â 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 S₂O₈^2â was used in Trial 3 and the percent remaining at the end of the reaction. Was this assumption valid?

Chemical Kenetics

Trial 1

Flask A

25ml of 0.2 M Potassium Iodide (ml) and Volume of 0.2% starch(ml) were added together,

Flask B

25 ml of volume 0.2M Ammonium Persulfate and volume of 0.01 M sodium Thiosulfate(ml) were added together,

After adding flask A and B together, after 20 sec the solution turned from clear to black.

Trial 2

Flask A

25ml of 0.2 M Potassium Iodide (ml) and Volume of 0.2% starch(ml) were added together,

Flask B

12.5ml of 0.2 M Ammonium Persulfate, 12.5ml of 0.2 M Ammonium Sulfate(ml), 10 ml of 0.01 M Sodium Thiosulfate(ml) were added together,

When flask A and B were added, after 42 sec, it turned from clear to black.

Trial 3

Flask A

25ml of 0.2 M Potassium Iodide (ml) and Volume of 0.2% starch(ml) were added together,

Flask B

6.25ml of 0.2 M Ammonium Persulfate (ml), 18.75ml of 0.2 M Ammonium Sulfate, and 10ml of 0.01 M sodium Thiosulfate (ml) were added together.

After flask A and B were added together, at a min and 26 sec, it turned from clear to black.

Trial 4

Flask A

12.5ml of 0.2 M Potassium Iodide (ml), 12.5ml of 0.2 M Potassium Nitrate (ml), 5 ml of 0.2% starch were added together,

Flask B

25 ml of 0.2 Ammonium Persulfate, 10ml of 0.01 M Sodium Thiosulfate (ml) were added together

after mixing both A and B together, at 42 seconds, it turned clear to black.

Trial 5

Flask A

6.25ml of 0.2 M Potassium Iodide, 18.75 ml of 0.2 M Potassium Nitrate, 5ml of 0.2% starch were added together.

Flask B

25 ml of 0.2 Ammonium Persulfate, 10ml of 0.01 M Sodium Thiosulfate (ml) were added together,

After adding flask A and B together, after 1 min and 28 sec, it turned from clear to black.

All at the temp 21.5 C

Questions,

1.What was the initial concentration of thiosulfate?

A.0.00165 M

B.0.00154 M

C.0.00077 M

D.0.00134 M

2. What was the rate of reaction in Trial 1? Choose the closest answer.

A.3.13 x 10-5 M s-1

B.3.85 x 10-5 M s-1

C.4.15 x 10-5 M s-1

D.3.99 x 10-5 M s-1

3. What was the concentration of peroxodisulfate in Trial 1?

A.0.0569 M

B.0.0769 M

C.0.0669 M

D.0.0879 M

4. What was the concentration of iodide in Trial 1?

A.0.0769 M

B. 0.0659 M

C. 0.0869 M

D.0.0571 M

5. What is the order of peroxodisulfate in the rate expression?

A.0

B.2

C.1.5

D.1

6. What is the order of iodide in the rate expression?

A.0

B.2

C.1

D.3

7. What is the rate law for this reaction?

A.rate = k[I-]3[S2O82-]

B.rate = k[I-]2[S2O82-]

C.rate = k[I-][S2O82-]

D.rate = k[I-]1.5[S2O82-]

8. What is the average rate constant? Choose the closest answer.

A.5.44 x 10-3 L/(mol s)

B.5.73 x 10-3 L/(mol s)

C.6.88 x 10-3 L/(mol s)

D.6.16 x 10-3 L/(mol s)

9. The following reaction is second order with respect to A and zero order with respect to B. Identify the correct rate expression.

2A + B -> C + 2D

A.rate = k[A]4

B.rate = k[2A]2

C.rate = k[A]2

D.rate = k[A]2[B]

10. What is the overall reaction order for a reaction with the following rate expression?

rate = k[A]2[B]

A.3

B.1

C.2

D.4

11.A plot of log(rate) versus log[A] yields the following linear equation. What order is the reaction with respect to A?

y = 1.9x + 3.1

A.5

B.2

C.3

D.1.5

12. The iodine clock experiment, which consists of the following three reactions, is set up so [S2O32-] < [S2O82-]. What would happen if [S2O32-] > [S2O82-]?

Reaction 1: 3I- (aq) + S2O82- (aq) ->I3- (aq) + 2SO42- (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.No bluish black color would form because all the I- formed in reaction 1 would be converted back into I3-.

B.The bluish black color would be darker because more S2O32- is present.

C.The bluish black color would be darker because more S2O32- is present.

D.No bluish black color would form because all the I3- formed in reaction 1 would be converted back into I-.

13. 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) + 2SO42- (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.The reaction would turn blue slower because I3- is removed at a slower rate.

B.A color change would appear immediately because I3- would be removed slowly from solution, and would therefore be able to react with starch.

C.The rate of color change would indicate the rate of reaction 3 instead of reaction 1.

D.Reaction 3 would not occur.

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

A.It will take less time for the color to change, which will decrease delta t, resulting in a higher calculated rate.

B.It will take longer for the color to change, which will increase delta t, resulting in a lower calculated rate.

C.It will take less time for the color to change, which will increase delta t, resulting in a lower calculated rate.

D.It will take longer for the color to change, which will increase delta t, resulting in a higher calculated rate.

15. A student conducting the iodine clock experiment accidentally makes an S2O82- stock solution that is too concentrated. How will this affect the rate measurement?

A.Not all of the 2S2O32- will react, which will decrease the calculated rate.

B.Not all of the 2S2O32- will react, which will increase the calculated rate.

C.It will take longer for the color to change, as the reaction takes longer when more reactant is present.

D.It will take less time for the color to change, as rate increases with concentration of reactant.

16. The following reaction is first order. If the rate at [A] = 0.1 M is 3, what will the rate be at [A] = 0.2 M?

A -> B + C

A.4

B.6

C.9

D.3

17. For a zero order reaction, if the concentration of reactant is increased 4-fold, the reaction rate

A.increases four-fold.

B.increases 8-fold.

C.increases 16-fold.

D.remains the same.

18. If 2.3 x 10-3 M product forms in 46 seconds, what is the rate of the reaction?

A.50 M/s

B.2.3 x 10-3 M/s

C.5.0 x 10-5 M/s

D.2.0 x 104 M/s

19. A plot of log(delta [I3-] / delta t) versus log [I-]0 from data in Trials 1, 4, and 5 is shown below. Determine the value of p from the graph.

A.1

B.0.5

C.2

D.1.5

I know I am only supposed to post one question per answer, but I feel they all go into one, and that is alot of repitition, Anything helps, I need calculations to help me learn it, I have many more questions but I feel if I get some, and get a understanding on how to calculate it, then I will be able to do the rest. Thank you so much.

The kinetics of the reaction below were studied. The volumes of 0.040 M KBrO3, 0.040 M KI,

0.00050 M Na2S2O3, 0.040 M KCl, H2O and a buffer solution employed in each trial are shown

in the table below. One drop of 1% aqueous starch solution was also added in each trial. The

buffer solution was a pH = 4.74 solution containing 0.50 M acetic acid and 0.50 M sodium acetate; Ka = 1.8 x 10â5 for acetic acid. Each trial was performed at 25â¦C. For each trial, the reaction mixture was initially colorless after combining all solutions; appearance of a blue solution color in the reaction mixture signaled that all of the thiosulfate ion initially present had reacted and marked the end of the timing period. The initial reaction rate for each trial was calculated from the change in molar concentration of thiosulfate ion from each trial; the equation for calculating Initial Rate is shown below, where Ît is the elapsed time between combining all solutions and appearance of the blue solution color. Each solution volume was measured to the nearest 0.1 mL. (ie 5 mL shown in the table is 5.0 mL.)

BrO3- (aq) + 6Iâ(aq) + 6H+(aq) â 3H2O(l) + Brâ(aq) + 3I2(aq)

Initial rate = - 13âS2O32-ât

Volume (mL) Used

Trial #	KBrO3	KI	Na2S2O3	KCl	H2O	Buffer	Initial Rate (M/s)
1	10	10	5	0	0	15	5.0 x 10â6
2	5	10	5	5	0	15	2.5 x 10â6
3	10	5	5	5	0	15	2.5 x 10â6
4	10	10	2.5	0	2.5	15	5.0 x 10â6

a)Which pair of trials could be used to determine the order for Iâ? Justify your choice.

b)How would you determine the order for H+? Be as specific as possible.

c)What would be the value for the initial rate if 5 mL KBrO3 solution, 5 mL KI solution, 5 mL Na2S2O3 solution, 10 mL KCl solution, 15 mL pH = 4.74 buffer solution, and 1 drop of 1% starch solution were used?

d)Why do Trials 1 and 4 have the same initial rate?

e)What time was required for the color change to occur in Trial 1?

f)What was the purpose of adding 5 mL of 0.040 M KCl in Trial 2?

g)How would you experimentally verify that KCl does not cause the color change from colorless to blue? Be as specific as possible.

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