Using the initial rates of reaction for the following reaction: A + 3B 2C [A] (m
ID: 986643 • Letter: U
Question
Using the initial rates of reaction for the following reaction:
A + 3B 2C
[A] (mol/L)
[B] (mol/L)
Initial Rate (mol/Ls)
0.210
0.150
3.41x10-3
0.210
0.300
1.36x10-2
0.420
0.300
2.73x10-2
What is the order of the reaction with respect to A?
What is the order of the reaction with respect to B?
The overall rate expression?___________________
Calculate the rate constant including the units.
[A] (mol/L)
[B] (mol/L)
Initial Rate (mol/Ls)
0.210
0.150
3.41x10-3
0.210
0.300
1.36x10-2
0.420
0.300
2.73x10-2
Explanation / Answer
Rate law = k [ A]m [B]n
K is rate constant , m and n are the order.
Lets find order by using different experimental runs.
Exp No.
[A] (mol/L)
[B] (mol/L)
Initial Rate (mol/Ls)
1
0.210
0.150
3.41x10-3
2
0.210
0.300
1.36x10-2
3
0.420
0.300
2.73x10-2
Lets take ratio of rate of exp 3:2 , then we will get order with respect A.
2.37 E-2 / 1.36 E-2 = ( .410/0.210)m
Lets find value of m
m = 1
so the order with respect to A is 1.
To get order with respect to B and for that we use ratio with 2 : 1 .
1.36 E-2/ 3.41 E-3 = (0.300 / 0.150 )n
Lets calculate n
n = 1.99 = 2
order with respect to B = 2
so the rate law becomes
Rate= k [A][B]2
Lets use extp 1 to get k
Rate= k [A][B]2
3.41 E-3 = k (0.210)x(0.150)2
k = 0.72 L mol-1 sec-1
Exp No.
[A] (mol/L)
[B] (mol/L)
Initial Rate (mol/Ls)
1
0.210
0.150
3.41x10-3
2
0.210
0.300
1.36x10-2
3
0.420
0.300
2.73x10-2
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