BF 3 (g) + NH 3 (g) -----> F 3 BNH 3 (g) occurring at a certain temperature: Exp
ID: 1090425 • Letter: B
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
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
Explanation / Answer
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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