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 S2O32-consumed 0.010 mol c mol I3-produced 0.015M d [S2O32-] 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 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) â 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.
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) [Iâ]0(M) log[Iâ]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 [S2O82â] >> [S2O32â]. To find out if this assumption is correct, calculate the ratio of [S2O82â] / [S2O32â] for Trial 3. (In Trial 3, the [S2O82â] was lowest, and therefore the ratio [S2O82â] / [S2O32â] is the smallest).
In this experiment, you assumed that only a small amount of S2O82â 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 S2O82â was used in Trial 3 and the percent remaining at the end of the reaction. Was this assumption valid?
o logldelta 13/delta t and log($208)0 log[delta 13]/delta t show trendline Slope: -0.331 Intercept: -2.077
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 S2O32-consumed | 0.010 mol |
c | mol I3-produced | 0.015M |
d | [S2O32-] 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 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) â 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.
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) | [Iâ]0(M) | log[Iâ]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 [S2O82â] >> [S2O32â]. To find out if this assumption is correct, calculate the ratio of [S2O82â] / [S2O32â] for Trial 3. (In Trial 3, the [S2O82â] was lowest, and therefore the ratio [S2O82â] / [S2O32â] is the smallest).
In this experiment, you assumed that only a small amount of S2O82â 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 S2O82â was used in Trial 3 and the percent remaining at the end of the reaction. Was this assumption valid?