Question

The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g):

Experiment	[BF3](M)	[NH3](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

1. What is the rate when [BF3]= 0.200 M and [NH3]= 0.530 M ?

Answer 1

let the rate law be Rate = k [Bf3]^x [NH3]^y

Comparing rate of experiment 1 and 2

rate 1 / rate 2 = 0.2130/0.1065 = {(0.25)^x(0.25)^y} / {(0.25)^x(0.125)^y}

Hence 2 = 2^y   or y =1

Similarly comparing rate 4 and rate 5

we have 0.1193/0.0596 = {(0.35)^x(0.10)} / {(0.175)^x(0.10)}

thus 2 = 2^x   or x = 1

Hence rate law is

rate = k [BF3][NH3]

Substituting any experiment values, for example eperiment No.1 values in rate expression

rate constant k = 0.2130/ (0.250x0.250)

=3.408 L/mol.s

Now when [BF3] = 0.200M and [NH3] = 0.530M,

then rate = 3.408 x(0.200) (0.530)

= 0.36128 M/s