Question

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 D

What is the rate of disappearance of Br2 when [NO]= 7.6×10⁻² M and [Br2]=0.21 M ?

Please help with part d!

Answer 1

Let the rate (-r) = K[ NO]^m [Br2]ⁿ

m and n are orders with respect to [NO] and [Br2] respectively

from experiment 1

24= K[0.1]^m [0.2]ⁿ      (1)

From experiments 2

150= K[0.25]^m [0.2]ⁿ      (2)

Eq.2/ Eq.1 gives   (2.5)^m= 6.25 , m=2

From experiment 3

60= K[0.1]^m [0.5]ⁿ      (3)

Eq.3/ Eq.1 gives 60/24= (0.5/0.20)n, 2.5= 2.5ⁿ,   n=1

The rate becomes –r= K[[NO]² [Br]¹

From Eq.1 24= K[ 0.1]² [0.2]¹

K=12000 /M2.sec

So the rate becomes –r = 12000[ NO]² [Br]¹

The validity of this equation for experiments 4 is also verified.

when [NO] =7.6*10-2 M and [Br2]=0.21M

Rate -r =12000*[7.6/100)2 [0.21]=14.55 M/s