Question

Rate-study data

In a study conducted at the same temperature, the following data for reaction producing nitrosyl bromide were collected.

Trial	Initial [NO] (M)	Initial [Br2] (M)	Initial rate (M/s)
1	0.150	0.240	48
2	0.300	0.240	191
3	0.300	0.480	383

Part C

What is the rate of the reaction when the initial concentration of NO is 0.400 M and that of Br2 is 0.235 M ?

Express your answer to three significant figures and include the appropriate units.

Answer 1

The reaction taking place is :

2NO +Br2 <--->2NOBr

According to rate law, Rate k[NO]^m [Br2]^2

where krate constant

and m and n are the order of the reaction with respect to NO and Br2 respectively

We need to determine the order of the reaction with respect to NO and Br2 firstly,

Initial rate k[NO]^m [Br2]^n

Putting the values of initial concentration of NO and Br from the trials 1-3 ,

48 k(0.150)^m (0.240)^n ......(1)

191k(0.300)^m (0.240)^n ...(2)

383k(0.300)^m (0.480)^n ......(3)

Divide eqn(2) by (1),

3.979(2)^m

taking log on both sides,

log (3.979)m log 2

mlog (3.979)/log 22

m2

Now divide eqn(3) by eqn(2),

2.00(2)^n

n1

Thus, the rate law becomes,

Initial rate k[NO]^2 [Br2]

Now , from eqn(1),

48 k(0.150)^2 (0.240)

k48/(0.150)^2 (0.240)8.89*10^3

Unit of k: (M/s)/(M^2)*MM^-2 s^-1

k8.89*10^3 M^-2 s^-1

part c) [NO]0.400M

[Br2]0.235M

Initial ratek[NO]^2[Br](8.89*10^3 M^-2 s^-1 )*(0.400M)^2 (0.235M)334.264 M/s