entry task: nov 3 rd monday sign off on rate law ws mayhan
TRANSCRIPT
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Entry Task: Nov 3rd Monday
Sign off on Rate Law ws
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Agenda:
• Go through the answers to Rate Law ws• HW: Expression and Rate Law ws
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Clear off Desks:Calculator and handout only
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1. For 2 A + B C, we’ve determined the following experimental data:
a. Rate order for A is _____and B is______b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
1 2
Rate = k[A]1[B]2 3
1.62 x10-5 = k[0.0100M]1[0.0100M]2 1.62 x10-5 M/s = k= 16.2 M-2s-1
1.0 x10-6 M3
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2. For 2 A + B C, we’ve determined the following experimental data:
a. Rate order for A is _____and B is______b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
2 1
Rate = k[A]2[B]1 3
2.80 x10-3 M/s = k[0.026 M]2[0.015 M]1
2.80 x10-3 M/s= k= 277 M-2s-1
1.01 x10-5 M3
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3. The following data were measured for the reaction of nitric oxide with hydrogen: 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)
Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [NO] = 0.050 M and [H2] = 0.150 M.
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3. Using these data, determine (a) the rate law for the reaction
Exp. 1 vs. Exp. 2, we doubled the concentration of H2, the rate doubled as well. This means [H2]1
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3. Using these data, determine (a) the rate law for the reaction
Exp. 1 vs. Exp. 2, we doubled the concentration of H2, the rate doubled as well. This means [H2]1
Exp. 2 vs. Exp. 3, we doubled the concentration of NO and the rate quadrupled or 22 . This means that NO is 2nd order [NO]2
a) Rate = k[NO]2[H2]1
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3. Using these data, determine (a) the rate law for the reaction, (b) the rate constant, c) the rate of the reaction when [NO] = 0.050 M and [H2] = 0.150 M. a) Rate = k[NO]2[H2]1
k=Rate
[NO]2[H2]1=
4.92 x 10-3 M/s[0.20M]2[0.10M]1
= 1.2 M-2/s-1
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3. Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [NO] = 0.050 M and [H2] = 0.150 M.
a) Rate = k[NO]2[H2]1
b) k = 1.2 M2 s1
Rate = (1.2 M2 s1) (0.050 M)2(0.150M)
Rate = 4.5 x 10-4 M/s
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4.
Using this information, a. Rate order for A is _____and B is______b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
2
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4.
Using this information, a. Rate order for A is _____and B is______b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
2 1
Rate = k [A]2[B] 3
0.26 x 10-9 M/s = k [1.00 x10-3M]2[0.25 x10-3M] 1.04 M-2s-1
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5.
Using this information, a. Rate order for X is _____and Y is______b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
1
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5.
Using this information, a. Rate order for X is _____and Y is______b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
1 0
Rate = k [X] 1
6.00 x 10-3 M/s = k [1.00 x10-2 M] 0.600 s-1
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6.
Using this information, a. Rate order for A is _____, B is______ , and C is_____b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
1
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6.
Using this information, a. Rate order for A is _____, B is______ , and C is_____b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
1 2
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6.
Using this information, a. Rate order for A is _____, B is______ , and C is_____b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
1 2 0
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6.
Using this information, a. Rate order for A is _____, B is______ , and C is_____b. The rate law for this reaction is: c. The overall reaction order is_______.d. Provide the rate constant for this reaction
1 2
Rate = k [A] [B]2 3
6.25 x 10-3 M/s = k [0.0500 M] [0.0500 M]2 50 M-2 s-1
0
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1. Using this data:
Rate Law:
Calculate Rate constant:
Rate = k [A]2 [B]1
0.26 x 10-9 M/s = k [1.00 x 10-3 M]2 [0.25 x10-3M] 1.04 M-2 s-1
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2. Using this data:
Rate Law:
Calculate Rate constant:
Rate = k [X]1
6.00 x 10-3 M/s = k [1.00 x 10-2 M] 0.6 s-1
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3. Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g)
a) Determine the rate law? **Hint get ratios between concentrations and rates.
b) Calculate the average value of the rate constant for the appearance of NOBr from our four data sets.
Trial [NO] (M)
[Br2] (M)
Initial Rate M/s
1 0.10 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735
0.10 = 24_ 0.25 = 150
NO is 2nd order(0.4)m = 0.16
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3. Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g)
a) Determine the rate law? **Hint get ratios between concentrations and rates.
b) Calculate the average value of the rate constant for the appearance of NOBr from our four data sets.
Trial [NO] (M)
[Br2] (M)
Initial Rate M/s
1 0.10 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735
(0.10)m = 24_ (0.25)m = 150
NO is 2nd order(0.4)m = 0.16
(0.20)m = 24_ (0.50)m = 60
Br2 is 1st order(0.4)m = 0.4
24 M/s = k [0.10 M]2 [0.20 M]
12000 M-2 s-1
Rate = k [NO]2[Br2]
The following data are for Questions 4 - 9 and refer to the reaction: A + 2B + 3C → 2Y + Z. All data were taken at 50.0°C.
The rate law derived for the reaction from the above data is:
[A] and [B] kept constant its obvious that [C] is 1st order
(0.02)m = 10_ (0.03)m = 15
B is 1st order(0.67)m = 0.67
(0.10)m = 10_ (0.20)m = 80
(0.5)m = 0.125 This is 3rd order but we need to subtract 1 for the [C] making A an 2nd order
Rate = k[A]2[B][C]
The following data are for Questions 4 - 9 and refer to the reaction: A + 2B + 3C → 2Y + Z. All data were taken at 50.0°C.
5. The value of the specific rate constant is:
10 M/s = k [0.10]2[0.02][0.04]
__ 10 M/s_ __ = k [0.10]2[0.02][0.04]
= 1.25 x 106 M-3s-1
Rate = k[A]2[B][C]
The following data are for Questions 4 - 9 and refer to the reaction: A + 2B + 3C → 2Y + Z. All data were taken at 50.0°C.
6. The missing rate (trial 5) in units of M/s should be:
? M/s = 1.25 x 106 M-3s-1 [0.05]2[0.01][0.08]
2.5 M/s
Rate = k[A]2[B][C]
The following data are for Questions 4 - 9 and refer to the reaction: A + 2B + 3C → 2Y + Z. All data were taken at 50.0°C.
7. The rate of disappearance of C in trial 2 was (in M/s):
Let’s use -1/3 ∆C/∆t = 1/2 ∆Y/∆t
We can compare C disappearing with the rate of Y appearing through stoich!
13
[C]∆t= −1
215 M/s
= 32
15M/s22.5 M
The following data are for Questions 4 - 9 and refer to the reaction: A + 2B + 3C → 2Y + Z. All data were taken at 50.0°C.
8. Doubling [B] would change the rate of formation Y by a factor of:
C would double as well!
Rate = k[A]2[B][C]
The following data are for Questions 4 - 9 and refer to the reaction: A + 2B + 3C → 2Y + Z. All data were taken at 50.0°C.
9. The rate of formation of Z in trial 3 was (in M/s):
Let’s use -1/3 ∆C/∆t = 1/2 ∆Y/∆t
We can compare Y disappearing with the rate of Z appearing through stoich!
11
[Z]∆t= −1
280 M/s
= 12
80M/s40 M/s