Question

Using the initial rates of reaction for the following reaction:

A + 3B → 2C

[A] (mol/L)	[B] (mol/L)	Initial Rate (mol/L∙s)
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.

Answer 1

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/L∙s)
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