Question

Please show work.

A. For single reactant: observe the change in rate for a change in initial reactantconcentration.

	Reaction 1		Reaction 2		Reaction 3
[A] (M)	Initial Rate (M/s)	[C] (M)	Initial Rate (M/s)	[K] (M)	Initial Rate (M/s)
0.10	0.015	0.10	0.015	0.10	0.015
0.20	0.030	0.20	0.060	0.30	0.135
0.40	0.060	0.40	0.240	0.90	1.215

The table above shows three independent reactions (not related to each other). What are the rate laws for each of thesereactions?

B. For multiple reactants: Find the reaction order for each reactant separately.

[A] (M)	[X] (M)	Initial Rate (M/s)
0.050	0.050	0.014
0.150	0.050	0.056
0.050	0.100	0.028
0.150	0.150	0.168

For the table above, find the rate law and the rate constant. What is the overall order of the reaction?

Answer 1

From the reaction we observe that there is only one reactant in each.

Rate law :

Reaction 1 ;

Rate = k [A]^m

Here k is rate constant and m is order with respect to A.

We find m by using ratio of rate for two experiment

Lets take ratio of exp 2 to exp 1

Rate 2 / rate 1 = k [A]^m/ k [A]^m

Here k is constant and it is cancelled since reaction is same.

Lets plug concentration of A and rate value.

0.20 / 0.10 = [0.030]^m/ [0.015]^m

2 = (0.030/0.015)^m

2 = 2^m

To calculate m we need to take ln of both side

Ln 2 = m ln 2

m = 1

so rate law becomes

Rate = k [A]

Reaction : 2

Rate = k [C]ⁿ

We use same strategy to solve for n

Rate of Exp 2 to exp 1

Rate 2 / rate 1 = ( 0.060/0.015)ⁿ

0.2/0.1 = ( 0.060/0.015)ⁿ

2 = 4ⁿ

Ln 2 = n ln 4

n = ½

Rate law becomes

Rate = k [C]^1/2

Reaction 3 :

Rate law :

Rate = k [K]^l

Rate 2/ rate 1 = (0.135/0.015)^l

0.30/0.10 =(0.135/0.015)^l

3 = 9^l

Ln 3 = l ln 9

l = 1/3

Rate law

Rate = k [K]^1/3

B.

Rate law :

Rate = k [A]^a [ X]^b

Here a and b are the orders with respect to A and B respectively.

Ex. No.	[A] (M)	[X] (M)	Initial Rate (M/s)
1	0.050	0.050	0.014
2	0.150	0.050	0.056
3	0.050	0.100	0.028
4	0.150	0.150	0.168

Lets use ratio of rate 2/ rate 1

Rate 2/ Rate 1 = (A2/A1)^a x (X2/X1)^b

0.056 / 0.014 = (0.150/0.050)^a   …concentration of X is same in both experiment so it is cancelled.

4 = 3^a

We use ln

ln 4 = a ln 3

a = 1.26

Lets find for b

Rate 4/ Rate 2 = (A4/A2)^a x (X4/X2)^b

0.168/0.056 = (0.150 /0.050 )^b

3 = 3^b

b = 1

Rate law :

Rate = k [A]^1.26 [X]

Calculation of rate constant

We use experiment first

0.014 = k [0.050]^1.26 [0.050]

k = 12.27

Overall order : 1 + 1.26 = 2.26