Question

BF3(g) + NH3(g) -----> F3BNH3(g)               occurring at a certain temperature:    Experiment [BF3]o M [NH3]o...

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           

Homework Answers

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

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