Posted by Jordan on Saturday, March 16, 2013 at 10:44pm.
So, I go to high school in Texas and we have this AP Chem lab due online tomorrow night by midnight and I need some serious help on these questions:
Lab Temp 0.2°C
Experiment Trial Time in seconds Trial 1 Trial 2 Average Time
1 1 2 256 275 268
2 1 2 215 226 220.5
3 1 2 83 59 71
4 Temp here is 35 ° C 1 83 Unnecessary
Experiment [S203^2-] [S208^2-] [I-]
1 .0022 .0044 .2222
2 .0022 .0088 .2222
3 .0022 .0044 .4444
Initial rates from experiments 1,2, and 3 from [S2O3^2-] / ∆t:
Experiment [Na2S2O3] (M) Avg. Elapsed Time (s) ∆ Na2S203/ ∆t (M/s) ∆ S208^2-/ ∆ t (M/s)
1 .0022 268 8.2X10^-6 4.10*10^-6
2 .0022 220.5 9.97X10^-6 4.98X10^-9
3 .0022 71 3.10X10^-5 1.55X10^-5
1. Determine the reaction orders to S2O8^2- using experiment 1 and experiment 2. Hint: Use m = log (rate 1/rate 2) divided by log (concentration 1/ concentration 2).
2. Determine the reaction orders to I- using experiment 1 with experiment 3. Hint: Use n = log (rate 1/rate 3) divided by log (concentration 1/ concentration 3).
3. What is the calculated rate law for this equation/reaction? Considering most rate orders have whole integers, what is most likely the correct rate law for this reaction?
4. Using the reaction rate calculated, calculate the rate constant for experiments 1, 2, and 3, and then the average rate constant for the three experiments.
5. Calculate the rate constant for experiment 4 at the elevated temperature.
6. Why are some reactions heated in laboratories?
Chemistry - URGENT - DrBob222, Saturday, March 16, 2013 at 11:10pm
I see no one has helped and that may be because all you need to do is follow the instructions in the problem. The hints tell you what to do. I suggest you do them. I'll be glad to answer specific questions about them. Post your work if you have further questions.
Chemistry - URGENT - Jordan, Saturday, March 16, 2013 at 11:17pm
I figured out 1 and 2 but i don't understand how to do the rest
Chemistry - URGENT - DrBob222, Saturday, March 16, 2013 at 11:37pm
3. The rate law expression is
rate = k(S2O8^2-)x(I^-)y where x and Y are the orders of S2O8^2- and I^-. Plug in what you found for x and y and you have it. I don't know exactly what numbers you found but the question hints that they aren't whole numbers and suggests that you round them to whole numbers. The rate comes from the delta (...)/dT but I can't read what you have because of the spacing problem with this board.
Chemistry - URGENT - Jordan, Saturday, March 16, 2013 at 11:38pm
For 1 I got m is .282 and for 2 I got n is .53
Chemistry - URGENT - Jordan, Saturday, March 16, 2013 at 11:42pm
I had to type it in word and it formatted fine even when i emailed it to myself. So how do you calculate the rate constant?
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