Question

14.37 Consider the gas-phase reaction between nitric oxide and bromine at 273 ∘C 2NO(g)+Br2(g)→2NOBr(g). The following...

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:

 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 0.50 735

Part A

Determine the rate law.

Determine the rate law.

 a. rate=k[NO]2[Br2] b. rate=k[NO][Br2] c. rate=k[NO]2[Br2]2 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]= 9.0×10−2 M and [Br2]= 0.16 M ?

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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