Question

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	43
2	0.300	0.240	171
3	0.300	0.480	342

Part C

What is the rate of the reaction when the initial concentration of NO is 0.400 M and that of Br2 is 0.255 M ?

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

Answer 1

Going from trial 1 to trial 2, [NO] doubles, [Br₂] remains same and rate increases by four times. Hence reaction is second order with respect to NO.

Going from trial 2 to trial 3, [NO] remains same, [Br₂] doubles and rate also gets doubled. Hence reaction is first order with respect to Br₂.

Therefore, rate law is given by:

Rate = k[NO]²[Br₂]

Consider experiment 1:

43 = k [0.150]² [0.240]

k = 7962.96 M^-2s^-1

The rate of the reaction when the initial concentration of NO is 0.400 M and that of Br₂ is 0.255 M is given by:

Rate = k[NO]²[Br₂]

Rate = 7962.96 [0.400]² [0.255]

Rate = 324.89 M/s