14.37 Consider the gas-phase reaction between nitric oxide and bromine at 273 C
ID: 947188 • Letter: 1
Question
14.37
Consider the gas-phase reaction between nitric oxide and bromine at 273 C
2NO(g)+Br2(g)2NOBr(g).
The following data for the initial rate of appearance of NOBr were obtained:
Part A
Determine the rate law.
Determine the rate law.
d. rate=k[NO][Br2]2
Part B
Calculate the average value of the rate constant for the appearance of NOBr from the four data sets.
Part C
How is the rate of appearance of NOBr related to the rate of disappearance of Br2?
Part D
What is the rate of disappearance of Br2 when [NO]= 9.0×102 M and [Br2]= 0.16 M ?
Experiment [NO](M) [Br2](M) Initial Rate of Appearance ofNOBr(M/s) 1 0.10 0.20 24 2 0.25 0.20 150 3 0.10 0.50 60 4 0.35 0.50 735Explanation / Answer
rate= k [NO]a[Br2]b
24= k (0.1)a(0.2)b
150= k (0.25)a(0.20)b
divide both equations, you will obtain:
24/150= (0.1)a/(0.25)a
0.16= 0.4a ----> a=2
Now do the same with experiment 1 and 3:
24/60= (0.20/0.5)b
0.4= 0.4b ----> b=1
rate law: rate= k[NO]2[Br]
Part B:
24= k (0.1)2(0.2) ------> k= 12000 M-2 s-1
150= k (0.25)2(0.2) ----> k= 12000 M-2 s-1
60= k(0.1)2(0.5) ----> k=12000 M-2 s-1
735= k (0.35)2(0.5)----->k=12000 M-2 s-1
so the average value of the raqte constant is 12000 M-2 s-1
Part C:
1/2 D[NOBr]/Dt= -D[Br2]/Dt
Part D:
rate= 12000 M-2 s-1 x (9x10-2)2 x (0.16)= 15.552 M/s
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