(C) The rate law is given by $r = k [NO]^{m} [Cl_{2}]^{n}$.From the data:$0.18 = k (0.1)^{m} (0.1)^{n} \quad \dots (1)$$0.35 = k (0.1)^{m} (0.2)^{n} \

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(C) The rate law is given by $r = k [NO]^{m} [Cl_{2}]^{n}$.From the data:$0.18 = k (0.1)^{m} (0.1)^{n} \quad \dots (1)$$0.35 = k (0.1)^{m} (0.2)^{n} \quad \dots (2)$$1.40 = k (0.2)^{m} (0.2)^{n} \quad \dots (3)$Dividing equation $(2)$ by $(1)$:$\frac{0.35}{0.18} \approx 2 = (\frac{0.2}{0.1})^{n} = 2^{n} \implies n = 1$.Dividing equation $(3)$ by $(2)$:$\frac{1.40}{0.35} = 4 = (\frac{0.2}{0.1})^{m} = 2^{m} \implies m = 2$.The overall order of the reaction is $m + n = 2 + 1 = 3$.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

Medium

$2 \ NO_{(g)} + Cl_{2(g)} \rightleftharpoons 2 \ NOCl_{(g)}$
This reaction was studied at $-10^{\circ} C$ and the following data was obtained:

$Run$ $[NO]_{0}$ $[Cl_{2}]_{0}$ $r_{0}$

$1$ $0.10$ $0.10$ $0.18$

$2$ $0.10$ $0.20$ $0.35$

$3$ $0.20$ $0.20$ $1.40$

$[NO]_{0}$ and $[Cl_{2}]_{0}$ 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).

JEE MAIN 2021 All JEE MAIN

A
$1$
B
$2$
C
$3$
D
$4$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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$Run$	$[NO]_{0}$	$[Cl_{2}]_{0}$	$r_{0}$
$1$	$0.10$	$0.10$	$0.18$
$2$	$0.10$	$0.20$	$0.35$
$3$	$0.20$	$0.20$	$1.40$

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The following are the rate constants of two different reactions. What is the overall order of reaction for each?
$(a)$ $2.418 \times 10^{-5} \ hr^{-1}$
$(b)$ $7.1 \times 10^{-4} \ atm \ s^{-1}$