A reaction between substances A and B has been found to give the following data:

Question

A reaction between substances A and B has been found to give the following data:

3A + 2B   ---- 2C + D

[A] (mol/L)

[B] (mol/L)

Rate of appearance of C (mol/L-hr)

1.0 x 10-2

1.0

0.3 x 10-6

1.0 x 10-2

3.0

8.1 x 10-6

2.0 x 10-2

3.0

3.24 x 10-5

2.0 x 10-2

1.0

1.20 x 10-6

3.0 x 10-2

3.0

7.30 x 10-5

Using the above data, determine the order of the reaction with respect to A and B and the rate law, and calculate the specific rate constant.

[A] (mol/L)

[B] (mol/L)

Rate of appearance of C (mol/L-hr)

1.0 x 10-2

1.0

0.3 x 10-6

1.0 x 10-2

3.0

8.1 x 10-6

2.0 x 10-2

3.0

3.24 x 10-5

2.0 x 10-2

1.0

1.20 x 10-6

3.0 x 10-2

3.0

7.30 x 10-5

Explanation / Answer

if we observe the trial1 and trial 2

rate = K[A]^m[B]^n
0.3*10^-6 = K[1*10^-2]^m [1]^n .....1
8.1*10^-6 = k[1*10^-2]^m [3]^n .....2

if we do 1/2
(3/81) = (1/3)^n
(1/3)^3 =(1/3)^n

so the order w.r.t to B is 3

if we observe the trial2 and trial 3
8.1*10^-6 = k[1*10^-2]^m [3]^n .....3
3.24*10^-5 = k[2*10^-2]^m [3]^n ......4

do 4/3

(32.4/8.1) =(2)^m
2^2 = 2^m

so the order w.r.t A is 2

so the rate law
rate = K[A]^2 [B]^3
it is 5th oroder reaction

K = 0.3*10^-6/[(1*10^-2)^2 *(1)^3]
K = 0.3*10^-2 M^-4 sec^-1

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