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JEE Main 18 March 2021 Shift 1 Solved Paper

Section: Chemistry

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Question : 53 of 90

Marks: +1, -0

2\;\mathrm{NO}(g) + \mathrm{Cl}_2(g) \rightleftharpoons 2\;\mathrm{NOCl}(s)

This reaction was studied at

-10^{\circ}\;\mathrm{C}

and the following data was obtained run

$[\mathrm{NO}]_0$	$[\mathrm{Cl}_2]_0$	$r_0$
(1) 0.10	0.10	0.35
(2) 0.10	0.20	0.35
(3) 0.20	0.20	1.40

[\mathrm{NO}]_0

[\mathrm{Cl}_2]_0

are the initial concentrations and

r_0

is the initial reaction rate.

The overall order of the reaction is _____

(Round off to the nearest integer).

[18 Mar 2021 Shift 1]

Your Answer:

Solution:

According to given reaction,

2\;\mathrm{NO}(g) + \mathrm{Cl}_2(g) \rightleftharpoons 2\;\mathrm{NOCl}(s)

Given data are as follows,

By rate law expression,

r = k[\mathrm{NO}]^{x}[\mathrm{Cl}_2]^{y}

\text{Exp}

(I)

0.18 = k(0.1)^{x}(0.1)^{y}

...(i)

\text{Exp}

(II)

0.35 = k(0.1)^{x}(0.2)^{y}

...(ii)

\text{Exp}

(III)

1.40 = k(0.2)^{x}(0.2)^{y}

...(iii)

Dividing equation (ii) and (iii), we get

\frac{0.35}{1.40} = \frac{k(0.1)^{x}(0.2)^{y}}{k(0.2)^{x}(0.2)^{y}}

\frac{1}{4} = \left(\frac{1}{2}\right)^{x} \Rightarrow x = 2

Similarly dividing equation (i) and (ii), we get

\frac{1}{2} = \left(\frac{1}{2}\right)^{y} \Rightarrow y = 1

Now,

x+y=

order of reactions

x+y \Rightarrow 2+1 \Rightarrow 3

Overall, order of the reaction is 3 .

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