Question

BF₃(g) + NH₃(g) -----> F₃BNH₃(g)

              occurring at a certain temperature:

  

Experiment	[BF3]o M	[NH3]o M	Initial rate M/s
1	0.250	0.250	0.2130
2	0.250	0.125	0.1065
3	0.200	0.100	0.0682
4	0.350	0.100	0.1193
5	0.175	0.100	0.0596

            a) (12pts) Determine the rate law.

            b) (5pts) Calculate the value of the rate constant.

            c) (4pts) What is the initial rate of reaction when [BF₃^-]_o = 0.225 M   and [NH₃]_o = 0.200 M           

Answer 1

Hi,

Now to find the rate law we assume it to be R=k[BF3]^a[NH3]^b

Thus from the table given we put in the different values and get the answer that a=b=1. We use simple linear equations to solve this matter. For eg when I determine the exponential term for NH3, I try to keep the concentration of BF3 the same for ease in solving of equations

Now for b part we use experiment 1 data and get k=3.408 or the rate constant

Now we know the value of the rate expression which is

R=3.408x[BF3][NH3]

Thus the answer is 0.225x3.408x0.2=0.15336 moles/sec for c part

Thanks