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:

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 ?

Answer 1

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= 0.4^a ----> a=2

Now do the same with experiment 1 and 3:

24/60= (0.20/0.5)^b

0.4= 0.4^b ----> 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