For the reaction {\Cl}_{2( {\g})}+2 {\NO}_{( {\g})} \rightarrow 2 {\NOCl}_{( {\g})}, the following data was obtained: Experiment No. Initial concentration of {\Cl}_2 ( M) Initial concentration of NO ( M) Initial reaction rate ( M \/ {\min}) I 0.15 0.15 0.60 II 0.30 0.15 1.20 III 0.15 0.3 2.40 IV 0.25 0.25 2.78 Identify the order of the reaction with respect to {\Cl}_2, {\NO} and the value of Rate constant.

Order with respect to {\Cl}_2=2 Order with respect to {\NO}=1 {\k}=355.5 {\mol}^{-2} {\L}^2 {\min}^{-1}

Order with respect to {\Cl}_2=0 Order with respect to {\NO}=1 {\k}=8.0 {\min}^{-1}

Order with respect to {\Cl}_2=1 Order with respect to {\NO}=1 {\k}=26.66 {\mol}^{-1} {\Lmin}^{-1}

COMEDK 2024 Shift 1 Paper

Question : 15 of 180

Marks: +1, -0

For the reaction Cl2(g)+2NO(g)⟶2NOCl(g), the following data was obtained:

Experiment No.	Initial concentration of Cl2 ( M)	Initial concentration of NO ( M)	Initial reaction rate ( M ∕min)
I	0.15	0.15	0.60
II	0.30	0.15	1.20
III	0.15	0.3	2.40
IV	0.25	0.25	2.78

Identify the order of the reaction with respect to Cl2,NO and the value of Rate constant.

Order with respect to Cl2=2‌‌ Order with respect to NO=1‌‌k=355.5mol−2L2min−1
Order with respect to Cl2=0‌‌ Order with respect to NO=1‌‌k=8.0min−1
Order with respect to Cl2=1‌‌ Order with respect to NO=2‌‌k=177.7mol−2L2min−1
Order with respect to Cl2=1‌‌ Order with respect to NO=1‌‌k=26.66mol−1Lmin−1

Validate

Solution:

To determine the order of the reaction with respect to Cl2 and NO , we can use the method of initial rates. The general rate law for the reaction can be written as:
‌ Rate ‌=k[Cl2]m[NO]n

where:
k is the rate constant
m is the order of the reaction with respect to Cl2
n is the order of the reaction with respect to NO
Next, we'll use the data from the experiments to find the values of m and n.
First, we compare Experiments I and II to determine the order with respect to Cl2. The initial concentration of NO remains constant while the concentration of Cl2 is doubled:
From Experiment I:
‌ Rate ‌1=k[Cl2]1m[NO]1n=0.60

From Experiment II:
‌ Rate ‌2=k[Cl2]2m[NO]2n=1.20

Taking the ratio of Rate ‌2 and Rate ‌1 :
‌

‌ Rate ‌2

‌ Rate ‌

=‌

k[Cl2]2m[NO]2n

k[Cl2]1n[NO]1n

=‌

1.20

0.60

=2

‌‌

(0.30)m

(0.15)m

=2
‌(‌

0.30

0.15

)m=2
‌2m=2
‌m=1
This indicates that the reaction is first order with respect to Cl2.
Next, we compare Experiments I and III to determine the order with respect to NO. The initial concentration of Cl2 remains constant while the concentration of NO is doubled:
From Experiment I:

‌ Rate ‌1=k[Cl2]1m[NO]1n=0.60
From Experiment III:
‌ Rate ‌3=k[Cl2]3m[NO]3n=2.40
Taking the ratio of Rate ‌3 and Rate ‌1 :
‌‌

‌ Rate ‌3

‌ Rate ‌

=‌

k[Cl2]3m[NO]3n

k[Cl]1m[NO]1n

=‌

2.40

0.60

=4
‌‌

(0.30)n

(0.15)n

=4
‌(‌

0.30

0.15

)n=4
‌2n=4
‌n=2
This indicates that the reaction is second order with respect to NO.
Therefore, the rate law is:
‌ Rate ‌=k[Cl2]1[NO]2
To find the rate constant k, we can use the data from any of the experiments. Let's use Experiment I :
‌0.60=k(0.15)1(0.15)2
‌0.60=k×0.15×0.0225
‌0.60=k×0.003375
k=‌

0.60

0.003375

≈177.7mol−2L2min−1
Hence, the correct option is:
Option C
Order with respect to Cl2=1‌‌ Order with respect to NO=2‌‌k=177.7mol−2L2min−1

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