Consider the given graph showing variation of reactant concentration with time. Three different reactions were started with identical initial concentration of reactants. Which of the following statement is correct? ![JEE Main 2026 Chemistry, Chemical Kinetics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Chemical_Kinetics/69d8074881239f6567c4f3f0/question_1__q_69d8074881239f6567c4f3f0__cdn-question-pool.getmarks.app__04APRS2__q63_d1__a43f921616_final_ppt_sync.jpg)

JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Consider the given graph showing variation of reactant concentration with time.

Three different reactions were started with identical initial concentration of reactants. Which of the following statement is correct?

JEE Main 2026 Chemistry, Chemical Kinetics — question figure

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
The order of all the three reactions is same.
B
The rate constant of reaction $3$ is larger than the rate constant of reaction $2$ if the order of reaction is same for both.
C
The SI unit of rate constant of reaction $1$ is s $^{-1}$ .
D
Thermal decomposition of HI on gold surface is an example of reaction $2$ .

Did you get this right?

Solution

Analysis of the graph:

The graph shows reactant concentration $[R]$ vs time $t$ for three reactions starting from the same initial concentration.

Curve 1 is a straight line with negative slope. For a zero-order reaction, the integrated rate law is $[R] = [R]_0 - kt$ , which is a straight line in $[R]$ vs $t$ . Hence, reaction $1$ is a zero-order reaction.

Curves 2 and 3 are non-linear (exponential-type decays), so they correspond to reactions of order greater than zero (e.g., first or second order).

Evaluating the options:

$(1)$ The order of all three reactions is the same.

Reaction $1$ is zero order while reactions $2$ and $3$ are of higher order, so this statement is incorrect.

$(2)$ The rate constant of reaction $3$ is larger than that of reaction $2$ , assuming both have the same order.

Assuming both curves $2$ and $3$ correspond to the same order (say first order), the integrated rate law is $[R] = [R]_0 e^{-kt}$ . At any given time $t$ , the curve for reaction $3$ lies below that of reaction $2$ , i.e., $[R]_3 < [R]_2$ . Since both start from the same $[R]_0$ , reaction $3$ is decaying faster, which implies a larger rate constant. Hence $k_3 > k_2$ , and this statement is correct.

$(3)$ The SI unit of the rate constant of reaction $1$ is $\text{s}^{-1}$ .

For a zero-order reaction, $\text{Rate} = k[R]^0 = k$ , so the unit of $k$ is that of rate, i.e., $\text{mol L}^{-1}\text{s}^{-1}$ . The unit $\text{s}^{-1}$ corresponds to a first-order reaction, so this statement is incorrect.

$(4)$ Thermal decomposition of $HI$ on a gold surface is an example of reaction $2$ .

The thermal decomposition of $HI$ on a gold surface is a well-known zero-order reaction, which corresponds to curve $1$ , not curve $2$ . Hence this statement is incorrect.

Therefore, the correct option is $(2)$ .

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

More JEE Main Physical Chemistry questions

The wavelength of photon ' A ' is 400 nm. The frequency of photon ' B ' is $10^{16} \mathrm{~s}^{-1}$. The wave number of photon ' $C^{\prime}$ is $10^{4} \mathrm{~cm}^{-1}$. The correct order of energy of these photons is :

2026medium

Given below are two statements: Statement I: The Henry's law constant $\mathrm{K}_{\mathrm{H}}$ is constant with respect to variations in solution's concentration over the range for which the solution is ideally dilute. Statement II: $\mathrm{K}_{\mathrm{H}}$ does not differ for the same solute in different solvents. In the light of the above statements, choose the correct answer from the options given below

2026medium

For the reaction, $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}$, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is $-5.40 \mathrm{~kJ} \mathrm{~mol}^{-1}$. B. As $\Delta \mathrm{G}^{\ominus}$ in graph is positive, $\mathrm{N}_{2} \mathrm{O}_{4}$ will not dissociate into $\mathrm{NO}_{2}$ at all. C. Reverse reaction will go to completion. D. When 1 mole of $\mathrm{N}_{2} \mathrm{O}_{4}$ changes into equilibrium mixture, value of $\Delta \mathrm{G}^{\ominus}=-0.84 \mathrm{~kJ} \mathrm{~mol}^{-1}$ E. When 2 mole of $\mathrm{NO}_{2}$ changes into equilibrium mixture, $\Delta \mathrm{G}^{\ominus}$ for equilibrium mixture is $-6.24 \mathrm{~kJ} \mathrm{~mol}^{-1}$. ![JEE Main 2026 Chemistry, Chemical Equilibrium — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Chemistry/images/Chemical_Equilibrium/6983c17e8bfdbb2435639572/question_1__q_6983c17e8bfdbb2435639572__cdn-question-pool.getmarks.app__21S1_q_65_1__9992d5614b_final_ppt_sync.png) Choose the correct answer from the options given below :

2026hard

The half-life of ${ }^{65} \mathrm{Zn}$ is 245 days. After $x$ days, $75 \%$ of original activity remained. The value of $x$ in days is $\_\_\_\_$. (Nearest integer) (Given: $\log 3=0.4771$ and $\log 2=0.3010$)

2026medium

One mole each of He and $A(g)$ are taken in a $10$ L closed flask and heated to $400$ K to establish the following equilibrium. $A(g) \rightleftharpoons B(g)$. $K_c$ for this reaction at $400$ K is $4.0$. The partial pressures (in atm) of He and $B(g)$ are respectively (at equilibrium) (Assume He, $A(g)$ and $B(g)$ behave as ideal gases) (Given: $R = 0.082$ L atm K$^{-1}$ mol$^{-1}$)

2026medium

View all Physical Chemistry questions

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

Options

The order of all the three reactions is same.

The rate constant of reaction $3$ is larger than the rate constant of reaction $2$ if the order of reaction is same for both.

The SI unit of rate constant of reaction $1$ is s $^{-1}$ .

Thermal decomposition of HI on gold surface is an example of reaction $2$ .

Solution

Analysis of the graph:

The graph shows reactant concentration $[R]$ vs time $t$ for three reactions starting from the same initial concentration.

Curves 2 and 3 are non-linear (exponential-type decays), so they correspond to reactions of order greater than zero (e.g., first or second order).

Evaluating the options:

$(1)$ The order of all three reactions is the same.

Reaction $1$ is zero order while reactions $2$ and $3$ are of higher order, so this statement is incorrect.

$(2)$ The rate constant of reaction $3$ is larger than that of reaction $2$ , assuming both have the same order.

$(3)$ The SI unit of the rate constant of reaction $1$ is $\text{s}^{-1}$ .

$(4)$ Thermal decomposition of $HI$ on a gold surface is an example of reaction $2$ .

The thermal decomposition of $HI$ on a gold surface is a well-known zero-order reaction, which corresponds to curve $1$ , not curve $2$ . Hence this statement is incorrect.

Therefore, the correct option is $(2)$ .

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry