Verified Solution • 2025–26 Session moderate

QJEEM21Ch95ab: JEE Main 2021 Chemistry - Class 12 JEEM21Ch95ab Solution: 2NO(g) + Cl 2 (g) 2NOCl(s) This reaction was studied at 10 and…

2NO(g) + Cl2(g) ⇌\rightleftharpoons 2NOCl(s)

This reaction was studied at −- 10 ∘^\circ and the following data was obtained

Run [NO]0{[NO]_0} [Cl2]0{[C{l_2}]_0} r0{r_0}
1 0.10 0.10 0.18
2 0.10 0.20 0.35
3 0.20 0.20 1.40


[NO]0{[NO]_0} and [Cl2]0{[C{l_2}]_0} are the initial concentrations and r0 is the initial reaction rate.

The overall order of the reaction is __________. (Round off to the Nearest Integer).

Step-by-Step Verified Solution

We know,

Rate=K[A]x[B]yRate = K{\left[ A \right]^x}{\left[ B \right]^y}

Here from,

Exp 1 : 0.18=K[0.1]x[0.1]y0.18 = K{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y} ..... (1)

Exp. 2 : 0.35=K[0.1]x[0.2]y0.35 = K{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y} .... (2)

Exp. 3 : 1.40=K[0.2]x[0.2]y1.40 = K{\left[ {0.2} \right]^x}{\left[ {0.2} \right]^y} .... (3)

(2) ÷\div (3)

0.351.40=K×[0.1]x[0.2]yK×[0.2]x[0.2]y{{0.35} \over {1.40}} = {{K \times {{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}} \over {K \times {{\left[ {0.2} \right]}^x}{{\left[ {0.2} \right]}^y}}}

⇒14=(12)x\Rightarrow {1 \over 4} = {\left( {{1 \over 2}} \right)^x}

⇒x=2\Rightarrow x = 2

(1) ÷\div (2)

(12)=(12)y\left( {{1 \over 2}} \right) = {\left( {{1 \over 2}} \right)^y}

⇒y=1\Rightarrow y = 1

∴\therefore x + y = 3

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