CBSE Class 12 Chemistry 2019 Outside Delhi Set 2 Solved Paper

Question : 14

Total: 27

The following data were obtained for the reaction:
A+2B→C

Experiment	[A]∕M	[B]∕M	‌ Initial rate of formation of C∕M min−1
1	0.2	0.3	4.2×10−2
2	0.1	0.1	6.0×10−3
3	0.4	0.3	1.68×10−1
4	0.1	0.4	2.40×10−2

(a) Find the order of reaction with respect to A and B.
(b) Write the rate law and overall order of reaction.
(c) Calculate the rate constant (k).

Solution:

Let the order of reaction with respect to A be x and with respect to B be y.
∴‌‌‌ Rate of reaction ‌=k[A]x[B]y
According to details given ,
‌4.2×10−2=k[0.2]x[0.3]y .....(1)
‌6.0×10−3=k[0.1]x[0.1]y .......(2)
‌1.68×10−1=k[0.4]x[0.3]y .......(3)
‌2.40×10−2=k[0.1]x[0.4]y ........(4)
Dividing equation (4) by (2), we get
‌

2.40×10−2

6.0×10−3

‌=‌

k[0.1]x[0.4]y

k[0.1]x[0.1]y

4‌=‌

[0.4]y

[0.1]y

(4)1‌=(4)y
y‌=1
Dividing equation (1) by (3), we get
‌‌

4.2×10−2

1.68×10−1

=‌

k[0.2]x[0.3]y

k[0.4]x[0.3]y

‌0.25=‌

[0.1]x

[0.2]x

‌(0.25)=(0.5)x
‌(0.5)2=(0.5)x‌‌x=2
(a) So the rate of reaction with respect to A is 2 and with respect to B is 1 .
(b) Rate law =k[A]2[B] Overall order of reaction is 3 .
(c) Rate constant, K‌=‌

‌ Rate ‌

[A]2[B]

=‌‌

6.0×10−3

(0.1)2(0.1)

‌=6.0mol−2L2min−1

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