Question
In: Chemistry
The following set of data was obtained by the method of initial rates for the reaction:...
The following set of data was obtained by the method of initial rates for the reaction: 2NO(g) + O2 (g) ----> 2NO2(g)
| Experiment # | [NO], M | [O2], M | Initial Rate, M/s |
| 1 | 0.0126 | 0.0125 | 1.41 x 10^-2 |
| 2 | 0.0252 | 0.0250 | 1.13 x 10^-1 |
| 3 | 0.0252 | 0.0125 | 5.64 x 10^-2 |
a) What is the rate law for the reaction?
b) If you triple the concentration of both reactants, what will happen to the rate?
c) What is the value of the rate constant?
Solutions
Expert Solution
2NO(g) + O2 (g) ----> 2NO2(g)
Rate = K[NO]^m[O2]^n
1.41*10^-2 = K(0.0126)^m(0.0125)^n --------------------> 1
1.13*10^-1 = K(0.0252)^m(0.025)^n ----------------------> 2
5.64*10^-2 = K(0.0252)^m (0.0125)^n ------------------> 3
equation 1 divide by equation 3
1.41*10^-2/5.64*10^-2 = K(0.0126)^m(0.0125)^n/K(0.0252)^m (0.0125)^n
0.25 = (0.0126/0.0252)^m
0.25 = (0.5)^m
(0.5)^2 = (0.5)^m
m = 2
equation 2 divide by equation 3
1.13*10^-1/5.64*10^-2 = K(0.0252)^m(0.025)^n/K(0.0252)^m (0.0125)^n
2 = (0.025/0.0125)^n
2^1 = (2)^n
n = 1
Rate law
Rate = K[NO]^2[O2]^1
b. Rate = K(3NO)^2(3O2)^1
Rate = 27K[NO]^2[O2]^1
Rate of reaction increases 27 times
c.
Rate = K[NO]^2[O2]^1
1.41*10^-2 = K(0.0126)^2(0.0125)^1
K = 1.41*10^-2/(0.0126)^2(0.0125)^1
= 7105 M^-2 sec^-1
queen_honey_blossom answered 1 year agoRelated Solutions
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40200
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7
43800
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8
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0
7
43800
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10
43800
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5
6
44400
15
75
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45000
8
52
3
45000
12
8
19
46200
12
52
3
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70
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