PART 1
>>Trial 1
Reactant Solution A
10 ml I^-, 5.0 ml S2O3^2-
Reactant Solution B
10 ml S2O8^2-, 1 ml starch
time to react = 10 seconds
>>Trial 2
Reactant Solution A
10 ml I^-, 5.0 ml S2O3^2-
Reactant Solution B
5 ml S2O8^2-, 1 ml starch
time to react = 20 seconds
>>Trial 3
Reactant Solution A
5 ml I^-, 5.0 ml S2O3^2-
Reactant Solution B
10 ml S2O8^2-, 1 ml starch
time to react = 20 seconds
How do trial 2 and 3 in the above table differ? More than one response is possible.
A) Trial 2 has twice as much I– as Trial 3
B) Trial 3 has twice as much S2O82– as Trial 2
C) Trial 3 has half as much S2O32– as Trial 2
D) Trial 2 has half as much starch as Trial 3
I chose A and B for this question
PART 2
Which of the four proposed mechanisms below would be supported by these results? Note: there may be more than one answer.
Mechanism 1:
I^- + I^- > I2^2- (slow)
I2^2- + S2O8^2– > I^2 + SO4^2- (fast)
Mechanism 2:
I2^2- + S2O8^2– > [S2O8xI]^3- (slow)
[S2O8xI]^3- +I^- > I2 + 2SO4^2- (fast)
Mechanism 3:
I^- + S2O8^2– <> [S2O8xI]^3- (fast)
[S2O8xI]^3- + I^- > I2 + 2SO4^2- (slow)
Mechanism 4:
2I^- + S2O8^2– > I2 + 2SO4^2-
I'm really not sure how to answer this part, if anyone can help that would be greatly appreciated!!
I looked part 2 for a bit and by looking at the slow step and using that to figure out the rate law I think it might be mechanism 2, would this be right?
PART 1
To compare Trial 2 and Trial 3, we need to consider the reactant solutions used in each trial. Let's analyze the differences:
Trial 2:
- Reactant Solution A: 10 ml I^-, 5.0 ml S2O3^2-
- Reactant Solution B: 5 ml S2O8^2-, 1 ml starch
- Time to react: 20 seconds
Trial 3:
- Reactant Solution A: 5 ml I^-, 5.0 ml S2O3^2-
- Reactant Solution B: 10 ml S2O8^2-, 1 ml starch
- Time to react: 20 seconds
Now, let's analyze the statements:
A) Trial 2 has twice as much I– as Trial 3:
- In Trial 2, Reactant Solution A contains 10 ml of I^-
- In Trial 3, Reactant Solution A contains 5 ml of I^-
- Therefore, Trial 2 has twice as much I^- as Trial 3.
B) Trial 3 has twice as much S2O82– as Trial 2:
- In Trial 2, Reactant Solution B contains 5 ml of S2O8^2-
- In Trial 3, Reactant Solution B contains 10 ml of S2O8^2-
- Therefore, Trial 3 has twice as much S2O82– as Trial 2.
Therefore, both A) and B) are correct responses.
PART 2
To determine which proposed mechanisms would be supported by the given results, we need to analyze the reaction outcomes in each trial.
In Trial 1:
- Reactant Solution A: 10 ml I^-, 5.0 ml S2O3^2-
- Reactant Solution B: 10 ml S2O8^2-, 1 ml starch
- Time to react: 10 seconds
In Trial 2:
- Reactant Solution A: 10 ml I^-, 5.0 ml S2O3^2-
- Reactant Solution B: 5 ml S2O8^2-, 1 ml starch
- Time to react: 20 seconds
In Trial 3:
- Reactant Solution A: 5 ml I^-, 5.0 ml S2O3^2-
- Reactant Solution B: 10 ml S2O8^2-, 1 ml starch
- Time to react: 20 seconds
From the given information, it is evident that the reaction rate differs between trials. To analyze the mechanisms, we need to compare the reactant combinations and the observed reaction rates. Let's examine each proposed mechanism:
Mechanism 1:
This mechanism assumes that the reaction between I^- and S2O8^2- is the slow step. However, since the reaction time differs in each trial, Mechanism 1 is not supported.
Mechanism 2:
This mechanism proposes a slow step where I2^2- reacts with S2O8^2- to form [S2O8xI]^3-. Given that different reactant combinations yield different reaction rates, Mechanism 2 is not supported.
Mechanism 3:
This mechanism assumes a fast step where I^- reacts with S2O8^2- to form [S2O8xI]^3- with an equilibrium. However, if the reaction rate differs between trials, this mechanism is not supported.
Mechanism 4:
This mechanism proposes a direct reaction between 2I^- and S2O8^2- to form I2 and 2SO4^2-. Since the reaction rate seems to vary among trials, it suggests that this direct reaction occurs. Therefore, Mechanism 4 is supported by the given results.
In conclusion, Mechanism 4 is the supported mechanism based on the given experimental results.