In: Chemistry
| Experiment |
A |
B | C |
intial rate of formation of D (M/min) |
| 1 | 2.0 | 2.0 | 2.0 | 2.0 |
| 2 | 2.0 | 1.0 | 2.0 | 2.0 |
| 3 | 4.0 | 5.0 | 2.0 | 8.0 |
| 4 | 2.0 | 4.0 | 1.0 | 1.0 |
| 5 | 3.0 | 4.0 | 4.0 | ? |
a) Determine the rate law.
b) Calculate the rate constant. Specify units.
c) Calculate the rate of reaction for experiment 5.
d) How is the rate of appearance of D related to the disappearance of B?
e) In experiment 3, what is the rate of disappearance of B?
Let the rate expression be rD= K[A]m [ B]n [C]p
2.0 2.0 2.0 2.0
where m,n,p are orders with respect to A, B, C respectively and K is rate constant
from First experiments
2= K [2]m [2]n [2]p (1)
From second experiment
2= K [2]m [1]n [2]p (2)
Dividing 1 and 2 gives
1= 2n , n= 0
The rate expression becomes rD= K[A]m [C]p
from 3rd experiment 8=K [ 4]m [2]p (3)
dividing 3 and 1 gives
4= (4/2)m , m= 2
From experiment 4
2.0 4.0 1.0 1.0
1= K[ 2]2 [ 1]p (4)
Dividing 4 and 1 gives
2= [2]p , p= 1
From 1
2= K[2]2 [ 2]1 K= 0.25
The rate expression then becomes
rD= 0.25 [ A]2 [C]1
The units of rate constant can be seen as
M/sec= Rate constant* [M]3
M= molarity
Rate constant = /M2.sec
following data refer to the reaction: A + 2B + C --> D + 2E
for the last experiment where A= 3 and C= 4
rD= 0.25*[3]2*4= 0.25*9*4= 9M/sec
A + 2B + C --> D + 2E
-rB/2= rD/1
Where 2 and 1 are coefficients of B and D in the reaction
In the 3rd experiments
-rB/2= rD/1= 8
-rB= 16 M/Sec.