Order and Molecularity

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JEE Main 2025

LEVELJEE Advanced

Consider a reaction $A + R \to P r o d u c t$ . The rate of this reaction is measured to be $k [A] [R]$ . At the start of the reaction, the concentration of $R$ , $[R]_{0}$ , is 10-times the concentration of $A$ , $[A]_{0}$ . The reaction can be considered to be a pseudo first order reaction with assumption that $k [R] = k^{'}$ is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40% complete, is ______. [ $k$ and $k^{'}$ represent corresponding rate constants]

JEE Advanced 2019

LEVELJEE Main

Consider the kinetic data given in the following table for the reaction $A + B + C \to Product$ . $\begin{array}{|c|c|c|c|c|} \hline \text{Experiment No.} & \text{[A]} (\text{mol dm}^{-3}) & \text{[B]} (\text{mol dm}^{-3}) & \text{[C]} (\text{mol dm}^{-3}) & \text{Rate of reaction} (\text{mol dm}^{-3}\text{s}^{-1}) \\ \hline 1 & 0.2 & 0.1 & 0.1 & 6.0 \times 10^{-5} \\ \hline 2 & 0.2 & 0.2 & 0.1 & 6.0 \times 10^{-5} \\ \hline 3 & 0.2 & 0.1 & 0.2 & 1.2 \times 10^{-4} \\ \hline 4 & 0.3 & 0.1 & 0.1 & 9.0 \times 10^{-5} \\ \hline \end{array}$ The rate of the reaction for $[A] = 0.15 mol dm^{- 3}$ , $[B] = 0.25 mol dm^{- 3}$ and $[C] = 0.15 mol dm^{- 3}$ is found to be $Y \times 1 0^{- 5} mol dm^{- 3} s^{- 1}$ . The value of $Y$ is ________.

JEE Main 2019

LEVELJEE Advanced

The decomposition reaction $2 N_{2} O_{5} (g) Δ 2 N_{2} O_{4} (g) + O_{2} (g)$ is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After $Y \times 1 0^{3} s$ , the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is $5 \times 1 0^{- 4} s^{- 1}$ , assuming ideal gas behavior, the value of Y is ______.

JEE Advanced 2018

LEVELJEE Advanced

For a first order reaction A(g) $\to$ 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t = 0) and at time t are $P_{0}$ and $P_{t}$ , respectively. Initially, only A is present with concentration $[A]_{0}$ , and $t_{1/3}$ is the time required for the partial pressure of A to reach $1/ 3^{rd}$ of its initial value. The correct option(s) is (are) :- (Assume that all these gases behave as ideal gases)

* Multiple Correct Options

Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in LIST-II. $X_{s}$ and $k$ constants having appropriate units.

List-I

(P)

rate = \frac{k [ X ]}{X _{s} + [ X ]}

under all possible initial concentration of X

(Q)

rate = \frac{k [ X ]}{X _{s} + [ X ]}

where initial concentration of X are much less than

X_{s}

(R)

rate = \frac{k [ X ]}{X _{s} + [ X ]}

where initial concentration of X are much higher than

X_{s}

(S)

rate = \frac{k [ X ] ^{2}}{X _{s} + [ X ]}

where initial concentration of X is much higher than

X_{s}

List-II

(1)

Graph of half life (

t_{1/2}

) vs initial concentration of X

(2)

Graph of half life (

t_{1/2}

) vs initial concentration of X

(3)

Graph of rate vs initial concentration

(4)

Graph of

[X]

vs time

(5)

Graph of

ln [X]

vs time

JEE Advanced 2014

LEVELJEE Main

For the elementary reaction $M \to N$ , the rate of disappearance of $M$ increases by a factor of $8$ upon doubling the concentration of $M$ . The order of the reaction with respect to $M$ is

(A)

4

(B)

3

(C)

2

(D)

1

LEVELJEE Main

A reaction involving two different reactants can never be

(A)

bimolecular reaction

(B)

second order reaction

(C)

first order reaction

(D)

unimolecular reaction

JEE Main 2021

LEVELJEE Main

For a certain first order reaction 32% of the reactant is left after 570 s. The rate constant of this reaction is ...... $\times 1 0^{- 3} s^{- 1}$ . (Round off to the nearest integer). [Given, $lo g_{10} 2 = 0.301$ , $ln 10 = 2.303$ ]

JEE Main 2021

LEVELJEE Main

A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is ......... min (Round off to the nearest integer). [Use : $ln 2 = 0.69$ , $ln 10 = 2.3$ ]

JEE Main 2020

LEVELJEE Main

Consider the following reactions $A \to P 1; B \to P 2; C \to P 3; D \to P 4$ , The order of the above reactions are $a, b, c$ and $d$ , respectively. The following graph is obtained when $lo g [rate]$ vs $lo g [conc.]$ are plotted :

(A)

D > A > B > C

(B)

A > B > C > D

(C)

C > A > B > D

(D)

D > B > A > C

JEE Main 2021

LEVELJEE Main

$PCl_{5} (g) ⟶ PCl_{3} (g) + Cl_{2} (g)$ In the above first order reaction, the concentration of $PCl_{5}$ reduces from initial concentration $50 mol L^{- 1}$ to $10 mol L^{- 1}$ in $120$ minutes at $300 K$ . The rate constant for the reaction at $300 K$ is $x \times 1 0^{- 2} min^{- 1}$ . The value of $x$ is ...... . [Given, $lo g 5 = 0.6989$ ]

JEE Main 2021

LEVELJEE Main

For the following graphs Choose from the options given below, the correct one regarding order of reaction is

(A)

(B) zero order (C) and (E) first order

(B)

(A) and (B) zero order (E) first order

(C)

(B) and (D) zero order (E) first order

(D)

(A) and (B) zero order (C) and (E) first order

JEE Main 2021

LEVELJEE Main

A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non-reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is ......... min. (Round off to the nearest integer).

JEE Main 2019

LEVELJEE Main

For the reaction, $2 A + B \to C$ , the values of initial rate at different reactant concentrations are given in the table below. \begin{array}{|c|c|c|} \hline \mathbf{[A]} \text{ (mol L}^{-1}\text{)} & \mathbf{[B]} \text{ (mol L}^{-1}\text{)} & \text{\textbf{Initial rate}} \text{ (mol L}^{-1}\text{s}^{-1}\text{)} \\ \hline 0.05 & 0.05 & 0.045 \\ 0.10 & 0.05 & 0.090 \\ 0.20 & 0.10 & 0.72 \\ \hline \end{array} The rate law for the reaction is

(A)

rate =

k [A] [B]^{2}

(B)

rate =

k [A]^{2} [B]^{2}

(C)

rate =

k [A] [B]

(D)

rate =

k [A]^{2} [B]

JEE Main 2020

LEVELJEE Main

It is true that

(A)

a second order reaction is always a multistep reaction

(B)

a zero order reaction is a multistep reaction

(C)

a first order reaction is always a single step reaction

(D)

a zero order reaction is a single step reaction

JEE Main 2019

LEVELJEE Main

The following results were obtained during kinetic studies of the reaction; $2 A + B ⟶ Products$ \begin{array}{cccc} \hline \text{Experiment} & \text{[A] (mol L}^{-1}\text{)} & \text{[B] (mol L}^{-1}\text{)} & \text{Initial rate (mol L}^{-1} \text{min}^{-1}\text{)} \\ \hline \text{I.} & 0.10 & 0.20 & 6.93 \times 10^{-3} \\ \text{II.} & 0.10 & 0.25 & 6.93 \times 10^{-3} \\ \text{III.} & 0.20 & 0.30 & 1.386 \times 10^{-2} \\ \hline \end{array} The time (in minutes) required to consume half of $A$ is

For a reaction $A + 2 B ⟶ C$ , rate is given by $+ \frac{d [ C ]}{d t} = k [A] [B]$ , hence the order of the reaction is

The following mechanism has been proposed for the reaction of $NO$ with $Br_{2}$ to form $NOBr$ $NO (g) + Br_{2} (g) ⇌ NOBr_{2} (g)$ $NOBr_{2} (g) + NO (g) ⟶ 2 NOBr (g)$ If the second step is the rate determining step, the order of the reaction with respect to $NO (g)$ is

The following data was obtained for chemical reaction given below at 975 K. $2 NO (g) + 2 H_{2} (g) \to N_{2} (g) + 2 H_{2} O (g)$ $A. B. C. [NO] mol L^{- 1} 8 \times 1 0^{- 5} 24 \times 1 0^{- 5} 24 \times 1 0^{- 5} [H_{2}] mol L^{- 1} 8 \times 1 0^{- 5} 8 \times 1 0^{- 5} 32 \times 1 0^{- 5} [Rate] mol L^{- 1} 7 \times 1 0^{- 9} 2.1 \times 1 0^{- 8} 8.4 \times 1 0^{- 8}$ The order of the reaction with respect to NO is \dots\dots . [Integer answer]

LEVELBoard

Consider following two reactions, $A ⟶ Product; - \frac{d [ A ]}{d t} = k_{1} [A]^{0}$ $B ⟶ Product; - \frac{d [ B ]}{d t} = k_{2} [B]$ $k_{1}$ and $k_{2}$ are expressed in terms of molarity ( $mol L^{- 1}$ ) and time ( $s^{- 1}$ ) as

(A)

s^{- 1}, M s^{- 1}

(B)

M s^{- 1}, M s^{- 1}

(C)

s^{- 1}, M^{- 1} s^{- 1}

(D)

M s^{- 1}, s^{- 1}

LEVELJEE Main

Consider the reaction, $2 A + B \to product$ . When concentration of $B$ alone was doubled, the half-life did not change. When the concentration of $A$ alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

(A)

L mol^{- 1} s^{- 1}

(B)

no unit

(C)

mol L^{- 1} s^{- 1}

(D)

s^{- 1}

JEE Main 2021

LEVELJEE Main

The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ...... $\times 1 0^{- 3} min^{- 1}$ . (Nearest integer) [Use: $ln 10 = 2.303$ ; $lo g_{10} 3 = 0.477$ ; property of logarithm : $lo g x^{y} = y lo g x$ ]

JEE Main 2021

LEVELJEE Advanced

For the first order reaction, $A ⟶ 2 B$ , 1 mole of reactant $A$ gives 0.2 moles of $B$ after 100 minutes. The half-life of the reaction is ............... min. (Round off to the nearest integer). [Use : $ln 2 = 0.69$ , $ln 10 = 2.3$ ; properties of logarithms : $ln x^{y} = y ln x$ ; $ln (\frac{x}{y}) = ln x - ln y$ ]

JEE Main 2019

LEVELJEE Main

The reaction, $2 X \to B$ is a zeroth order reaction. If the initial concentration of $X$ is $0.2 M$ , the half-life is $6 h$ . When the initial concentration of $X$ is $0.5 M$ , the time required to reach its final concentration of $0.2 M$ will be

(A)

7.2 h

(B)

18.0 h

(C)

12.0 h

(D)

9.0 h

LEVELJEE Main

In a first order reaction, the concentration of the reactant, decreases from $0.8 M$ to $0.4 M$ in $15 min$ . The time taken for the concentration to change from $0.1 M$ to $0.025 M$ is

(A)

30 min

(B)

15 min

(C)

7.5 min

(D)

60 min

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in LIST-II. Xs​ and k constants having appropriate units.

List-I

List-II

A reaction involving two different reactants can never be

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is ......... min (Round off to the nearest integer). [Use : ln2=0.69, ln10=2.3]

For the following graphs Choose from the options given below, the correct one regarding order of reaction is

A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non-reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is ......... min. (Round off to the nearest integer).

It is true that

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a reaction A+2B⟶C, rate is given by +dtd[C]​=k[A][B], hence the order of the reaction is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is

Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in LIST-II. $X_{s}$ and $k$ constants having appropriate units.

A reaction has a half-life of 1 min. The time required for 99.9% completion of the reaction is ......... min (Round off to the nearest integer). [Use : $ln 2 = 0.69$ , $ln 10 = 2.3$ ]

For a reaction $A + 2 B ⟶ C$ , rate is given by $+ \frac{d [ C ]}{d t} = k [A] [B]$ , hence the order of the reaction is

In a first order reaction, the concentration of the reactant, decreases from $0.8 M$ to $0.4 M$ in $15 min$ . The time taken for the concentration to change from $0.1 M$ to $0.025 M$ is