the [I] will halve the rate; doubling the [I] should double the rate, etc. If the reaction is 'second order' in iodide, x = 2, the rate is proportional to [I] and halving the [I] will reduce the rate by a factor of four. Look at the data and decide what the order of reaction is with respect to iodide: x = order *** *** Now, do the same thing with the second data set, where [I] is kept constant and [S2Og] changes (i.e. Row #s 3 – 5). Look at the data and decide what the order of reaction is with respect to persulphate: y = order Thus the rate law for the reaction is: The overall order of the reaction is

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Could someone double check my results from an investigating the rates of reaction lab session I did recently? Also I'd be very grateful if someone could help me with filling out the second page and if its first, second order etc as I don't properly understand that.

Results:
Fill your results in the following table.
Average initial rate of
reaction (d[l2]init/dt =
[12]10min/10 min)
Volume of
[12] produced after
10 minutes of
Initial
Initial
thiosulphate used
(Vehio) mL
Row
[r]
[S20,1
2#
reaction
M
M
M min
mL
M
5.76
0.00288
O.000288
0.2
0.028
2.52
0-000 126
0.1
0.028
O-00126
1.35
0-000675
0-0000675
3
0.05
0.028
[12] produced after
10 minutes of
Average initial rate of
reaction (d[l2]init/dt =
[l2]10min/10 min)
Volume of
Initial
[S20s?)
Initial
thiosulphate used
(Vthio) mL
Row
[r]
reaction
#3
M
M min
M
mL
2.52
0-000126
0.1
0.028
0-00126
S.58
0-00279
0-000279
4.
0.1
0.056
10.71
0-005355
0.0005355
0.1
0.112
21
Transcribed Image Text:Results: Fill your results in the following table. Average initial rate of reaction (d[l2]init/dt = [12]10min/10 min) Volume of [12] produced after 10 minutes of Initial Initial thiosulphate used (Vehio) mL Row [r] [S20,1 2# reaction M M M min mL M 5.76 0.00288 O.000288 0.2 0.028 2.52 0-000 126 0.1 0.028 O-00126 1.35 0-000675 0-0000675 3 0.05 0.028 [12] produced after 10 minutes of Average initial rate of reaction (d[l2]init/dt = [l2]10min/10 min) Volume of Initial [S20s?) Initial thiosulphate used (Vthio) mL Row [r] reaction #3 M M min M mL 2.52 0-000126 0.1 0.028 0-00126 S.58 0-00279 0-000279 4. 0.1 0.056 10.71 0-005355 0.0005355 0.1 0.112 21
the [I] will halve the rate; doubling the [I 1 should double the rate, etc.
If the reaction is 'second order' in jodide, x = 2. the rate is proportional to [I] and halving
the [I] will reduce the rate by a factor of four.
Look at the data and decide what the order of reaction is with respect to iodide:
X =
order
著*兴
Now, do the same thing with the second data set, where [I] is kept constant and
[S2O8 ] changes (i.e. Row #s 3-5).
Look at the data and decide what the order of reaction is with respect to persulphate:
y =
order
Thus the rate law for the reaction is:
The overall order of the reaction is
Transcribed Image Text:the [I] will halve the rate; doubling the [I 1 should double the rate, etc. If the reaction is 'second order' in jodide, x = 2. the rate is proportional to [I] and halving the [I] will reduce the rate by a factor of four. Look at the data and decide what the order of reaction is with respect to iodide: X = order 著*兴 Now, do the same thing with the second data set, where [I] is kept constant and [S2O8 ] changes (i.e. Row #s 3-5). Look at the data and decide what the order of reaction is with respect to persulphate: y = order Thus the rate law for the reaction is: The overall order of the reaction is
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