For the reaction, N2O4 leftharpoons 2NO2, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is -5.40 kJ mol-1. B. As Delta G in graph is positive, N2O4 will not dissociate into NO2 at all. C. Reverse reaction will go to completion. D. When 1 mole of N2O4 changes into equilibrium mixture, value of Delta G = -0.84 kJ mol-1. E. When 2 mole of NO2 changes into equilibrium mixture, Delta G for equilibrium mixture is -6.24 kJ mol-1. Choose the correct answer from the options given below :

Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot. A. Standard free energy change (Deltar G) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Deltar G = GB - GA. Since B is higher than A, Deltar G is positive, not -5.40 kJ mol-1. Statement A is false. B. Even if Deltar G is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, partial dissociation occurs to reach equilibrium. It is false to say it will not dissociate at all. Statement B is false. C. The minimum E is not at fraction = 0 or 1, meaning an equilibrium mixture exists. Reverse reaction does not go to completion. Statement C is false. D. From 1 mole of pure reactant (point A) to the equilibrium mixture (point E), the drop in Gibbs energy is 0.84 kJ mol-1. Thus Delta G = -0.84 kJ mol-1 is correct. Statement D is true. E. The difference from pure products (point B, equivalent to 2 moles NO2) to equilibrium (E) is the entire vertical distance. The drop from B to A is 5.40, and A to E is 0.84. So drop from B to E is - (5.40 + 0.84) = -6.24 kJ mol-1. Statement E is true. Only statements D and E are correct. The free energy of mixing creates a minimum 'dip' (Equilibrium) below both pure reactants and pure products, preventing either forward or reverse reactions from going absolutely to 100% completion. Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

For the reaction, N2O4 leftharpoons 2NO2, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is -5.40 kJ mol-1. B. As Delta G in graph is pos

Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot. A. Standard free energy change (Deltar G) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Deltar G = GB - GA. Since B is higher than A, Deltar G is positive, not -5.40 kJ mol-1. Statement A is false. B. Even if Deltar G is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, p

For the reaction, N2O4 leftharpoons 2NO2, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is -5.40 kJ mol-1. B. As Delta G in graph is positive, N2O4 will not dissociate into NO2 at all. C. Reverse reaction will go to completion. D. When 1 mole of N2O4 changes into equilibrium mixture, value of Delta G = -0.84 kJ mol-1. E. When 2 mole of NO2 changes into equilibrium mixture, Delta G for equilibrium mixture is -6.24 kJ mol-1. Choose the correct answer from the options given below :

Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot. A. Standard free energy change (Deltar G) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Deltar G = GB - GA. Since B is higher than A, Deltar G is positive, not -5.40 kJ mol-1. Statement A is false. B. Even if Deltar G is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, partial dissociation occurs to reach equilibrium. It is false to say it will not dissociate at all. Statement B is false. C. The minimum E is not at fraction = 0 or 1, meaning an equilibrium mixture exists. Reverse reaction does not go to completion. Statement C is false. D. From 1 mole of pure reactant (point A) to the equilibrium mixture (point E), the drop in Gibbs energy is 0.84 kJ mol-1. Thus Delta G = -0.84 kJ mol-1 is correct. Statement D is true. E. The difference from pure products (point B, equivalent to 2 moles NO2) to equilibrium (E) is the entire vertical distance. The drop from B to A is 5.40, and A to E is 0.84. So drop from B to E is - (5.40 + 0.84) = -6.24 kJ mol-1. Statement E is true. Only statements D and E are correct. The free energy of mixing creates a minimum 'dip' (Equilibrium) below both pure reactants and pure products, preventing either forward or reverse reactions from going absolutely to 100% completion. Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

For the reaction, N2O4 leftharpoons 2NO2, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is -5.40 kJ mol-1. B. As Delta G in graph is pos

Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot. A. Standard free energy change (Deltar G) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Deltar G = GB - GA. Since B is higher than A, Deltar G is positive, not -5.40 kJ mol-1. Statement A is false. B. Even if Deltar G is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, p

For the reaction, N2O4 leftharpoons 2NO2, graph is plotted as shown below. Identify correct statements. A. Standard…

Year	2026	2025	2024	Total
Questions	12	24	8	44

Q75 jee_main_2026_22_january_morning Gibbs Energy and Equilibrium Constant

Dissociation of a gas

A₂

takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300K.

A₂(g) leftharpoons 2A(g)

The standard Gibbs energy of formation of the involved substances has been provided below:

Substance	$Δ Gf° / kJ mol⁻¹$
$A₂$ $A_{2}$	-100.00
A	-50.832

The degree of dissociation of

A₂(g)

is given by

(x \times 10⁻²)1/2

where

x =

____. (Nearest integer). [Given:

R = 8 J mol⁻¹K⁻¹

,

2 = 0.3010

,

3 = 0.48

]

Numerical Answer. Answer: 33 to 33

Solution

Related Formula

Δ G°reaction = Σ Δ G°f(products) - Σ Δ G°f(reactants)

Δ G° = -RT ln Kₚ

Kₚ = (4α² P₀)/(1 - α²)

Core Logic

First, find standard Gibbs free energy of the reaction:

Δ G°reaction = 2 \times Δ G°f(A) - Δ G°f(A₂)

Δ G° = 2(-50.832) - (-100.00) = -101.664 + 100.00 = -1.664 kJ mol⁻¹

Δ G° = -1664 J mol⁻¹

Use this to find $Kₚ$ $K_p$:

-1664 = -8 \times 300 \times ln Kₚ

1664 = 2400 ln Kₚ

ln Kₚ = (1664)/(2400) = 0.6933

Since $ln 2 \approx 0.693$ $\ln 2 \approx 0.693$, we have: $Kₚ = 2$ $K_p = 2$

Step 1: Calculate Degree of Dissociation

For the reaction $A₂ leftharpoons 2A$ $A_2 \rightleftharpoons 2A$, with initial moles 1 and degree of dissociation $α$ $\alpha$: Moles at equilibrium: $1-α$ $1-\alpha$ (for $A₂$ $A_2$) and $2α$ $2\alpha$ (for A). Total moles = $1+α$ $1+\alpha$.

Kₚ = (PA)²PA₂ = (((2α)/(1+α) P₀)²)/((1-α)/(1+α) P₀) = (4α² P₀)/(1 - α²)

Given total pressure $P₀ = 1 bar$ $P_0 = 1\text{ bar}$ and $Kₚ = 2$ $K_p = 2$:

2 = (4α² (1))/(1 - α²)

2 - 2α² = 4α²

6α² = 2 α² = (1)/(3) α = 1\sqrt(3)

Step 2: Match to Given Format

We are given $α = (x \times 10⁻²)1/2$ $\alpha = (x \times 10^{-2})^{1/2}$. Squaring both sides:

α² = x \times 10⁻²

(1)/(3) = x \times 10⁻² x = (100)/(3) = 33.33

Step 3: Rounding

Nearest integer is 33.

Pattern Recognition

Whenever $Δ G°$ $\Delta G^{\circ}$ yields an $RT ln Kₚ$ $RT \ln K_p$ around 0.693, $Kₚ$ $K_p$ is 2. The formula $Kₚ = 4α²/(1-α²)$ $K_p = 4\alpha^2/(1-\alpha^2)$ for $A₂ leftharpoons 2A$ $A_2 \rightleftharpoons 2A$ at $P=1$ $P=1$ is standard and should be memorized.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

Q72 jee_main_2026_22_january_evening Born-Haber Cycle and Lattice Enthalpy

If the enthalpy of sublimation of Li is

155 kJ mol⁻¹

, enthalpy of dissociation of

F₂

is

150 kJ mol⁻¹

, ionization enthalpy of Li is

520 kJ mol⁻¹

, electron gain enthalpy of F is

-313 kJ mol⁻¹

, standard enthalpy of formation of LiF is

-594 kJ mol⁻¹

. The magnitude of lattice enthalpy of LiF is ____

kJ mol⁻¹

(Nearest integer).

Numerical Answer. Answer: 1031 to 1031

Solution

Related Formula

Δf H⁰ = Δsub H + IE + (1)/(2)Δbond H + Δeg H + L.E.

Core Logic

Step 1: Substitute given thermodynamic cycle values into Born-Haber equation:

-594 = 155 + 520 + (150)/(2) + (-313) + L.E.

-594 = 155 + 520 + 75 - 313 + L.E.

-594 = 437 + L.E.

L.E. = -594 - 437 = -1031 kJ mol⁻¹

Step 2: Magnitude of lattice enthalpy is $1031 kJ mol⁻¹$ $1031\text{ kJ mol}^{-1}$.

Born-Haber cycle diagram for LiF for Q72 - JEE Main 2026 Evening

Pattern Recognition

Sees: Born-Haber cycle parameters for ionic solid. Shortcut: Add sublimation, ionization, half-dissociation, and electron gain enthalpies, then subtract from formation enthalpy.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q58 jee_main_2026_23_january_morning First Law of Thermodynamics and Sign Conventions

A cup of water at

5°C

(system) is placed in a microwave oven and the oven is turned on for one minute during which, the water begins to boil. Which of the following option is true?

A. $q = +ve, w = 0, Δ U = -ve$ $q = +ve, w = 0, \Delta U = -ve$
B. $q = +ve, w = -ve, Δ U = +ve$ $q = +ve, w = -ve, \Delta U = +ve$
C. $q = -ve, w = -ve, Δ U = -ve$ $q = -ve, w = -ve, \Delta U = -ve$
D. $q = +ve, w = -ve, Δ U = -ve$ $q = +ve, w = -ve, \Delta U = -ve$

Solution

Related Formula

Δ U = q + w

where: $Δ U$ $\Delta U$ = change in internal energy $q$ $q$ = heat added to system $w$ $w$ = work done on the system

Core Logic

Analyze the state changes step-by-step applying IUPAC sign conventions for thermodynamics.

First Law of Thermodynamics and Sign Conventions diagram for Q58 - JEE Main 2026 Morning

Step 1: Heat Transfer

Since heat is supplied by the microwave oven to the water (system), the system absorbs heat. Thus, $q = +ve$ $q = +ve$.

Step 2: Work Done

As water boils, it converts from liquid to vapor, which implies a massive volume expansion. Work is done by the system against the atmosphere. Thus, work done on the system is negative: $w = -ve$ $w = -ve$.

Step 3: Internal Energy Change

The temperature of water increases from $5^° C$ $5^\circ C$ to $100^° C$ $100^\circ C$, and liquid converts to gas. The internal energy of steam at $100^° C$ $100^\circ C$ is much greater than that of liquid water at $5^° C$ $5^\circ C$. Hence, $Δ U = +ve$ $\Delta U = +ve$.

Pattern Recognition

Heating + Boiling inherently means heat is entering ( $q > 0$ $q > 0$), expanding volume pushes outward doing work ( $w < 0$ $w < 0$), and increasing thermal energy raises the internal state function ( $Δ U > 0$ $\Delta U > 0$).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q63 jee_main_2026_24_january_morning Work Done in Isothermal Processes

Match the List-I with List-II

List-I (Isothermal process for ideal gas system)	List-II Work done ( $Vf > Vᵢ$ $V_f > V_i$)
A. Reversible expansion	I. $w = 0$ $w = 0$
B. Free expansion	II. $w = -nRT ln (Vf)/(Vᵢ)$ $w = -nRT \ln \frac{V_f}{V_i}$
C. Irreversible expansion	III. $w = -pₑₓ(Vf - Vᵢ)$ $w = -p_{ex}(V_f - V_i)$
D. Irreversible compression	IV. $w = -pₑₓ(Vᵢ - Vf)$ $w = -p_{ex}(V_i - V_f)$

Choose the correct answer from the options given below :

A. $A-IV, B-I, C-III, D-II$ $\text{A-IV, B-I, C-III, D-II}$
B. $A-IV, B-II, C-III, D-I$ $\text{A-IV, B-II, C-III, D-I}$
C. $A-I, B-III, C-II, D-IV$ $\text{A-I, B-III, C-II, D-IV}$
D. $A-II, B-I, C-III, D-IV$ $\text{A-II, B-I, C-III, D-IV}$

Solution

Related Formula

WRev = -\int Pgas dV

WIrrev = -Pₑₓₜ Δ V

Core Logic

(A) Reversible isothermal expansion:

WRev = -nRT ln [ (Vf)/(Vᵢ) ]

Matches with II.

(B) Free expansion (expansion into vacuum):

Pₑₓₜ = 0

$W = 0$ $W = 0$ Matches with I.

(C) Irreversible expansion against constant external pressure:

Wirrev = -Pₑₓₜ Δ V

Wirrev = -Pₑₓₜ (Vf - Vᵢ)

Matches with III.

(D) Irreversible compression: Same core formula but since it is a compression from $Vf$ $V_f$ back to $Vᵢ$ $V_i$, the volume change term flips to $Δ V = (Vᵢ - Vf)$ $\Delta V = (V_i - V_f)$, leading to:

Wirrev = -Pₑₓₜ (Vᵢ - Vf)

Matches with IV.

Step 1: Final Conclusion

The correct matches are A-II, B-I, C-III, D-IV.

Pattern Recognition

Free expansion always implies zero work because $Pₑₓₜ = 0$ $P_{\text{ext}} = 0$. Reversible implies a continuous integration over volume resulting in the natural log formula.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q67 jee_main_2026_24_january_evening Enthalpy of Atomization and Bond Enthalpy

The heat of atomisation of methane and ethane are 'x' kJ mol

⁻¹

and 'y' kJ mol

⁻¹

respectively. The longest wavelength (

λ

) of light capable of breaking the C-C bond can be expressed in SI unit as:

A. $hc1000( y-6x4)⁻¹$
B. $NAhc250(4y-6x)$ $\frac{N_{A}hc}{250(4y-6x)}$
C. $NAhc250(y-6x)$ $\frac{N_{A}hc}{250(y-6x)}$
D. $NAhc(y-(6x)/(4))⁻¹$

Solution

Related Formula

E = (hc)/(λ) λ = (hc)/(E)

Where E is the energy required to break one bond.

Core Logic

For Methane ( $CH₄$ $\mathrm{CH}_4$): $CH₄(g) arrow C(g) + 4H(g) ΔᵣH = x kJ / mole$ The energy is used to break 4 C-H bonds: $4 \times εC-H = 1000x J / mole$

For Ethane ( $C₂H₆$ $\mathrm{C}_2\mathrm{H}_6$): $C₂H₆(g) arrow 2C(g) + 6H(g) ΔᵣH = y kJ / mole$ The energy is used to break 1 C-C bond and 6 C-H bonds: $εC-C + 6 \times εC-H = 1000y J / mole$

Step 1: Extract C-C Bond Energy

From methane: $εC-H = (1000x)/(4) = 250x J / mole$ Substitute into ethane equation: $εC-C + 6 \times (250x) = 1000y$ $εC-C = 1000y - 1500x = [y - (3x)/(2)] \times 1000 J / mole$

Step 2: Wavelength Calculation

The energy calculated above is per mole. Energy required to break one C-C bond is: $E = εC-CNA$

Now, equating this to photon energy: $εC-CNA = (hc)/(λ)$ $λ = hc \cdot NAεC-C$

Substitute $εC-C$ $\varepsilon_{\mathrm{C-C}}$: $λ = hc \cdot NA[y - (3x)/(2)] \times 1000$ $λ = hc \cdot NA(2y - 3x)/(2) \times 1000$ $λ = 2 \cdot hc \cdot NA1000(2y - 3x) = hc \cdot NA500(2y - 3x) = hc \cdot NA250(4y - 6x)$

Pattern Recognition

Always convert molar quantities to per-bond (atomic scale) quantities by dividing by Avogadro's number ( $NA$ $N_A$) when equating macroscopic enthalpy data to single photon quantum limits ( $hc/λ$ $hc/\lambda$). Don't forget to multiply $kJ$ $kJ$ to $J$ $J$ by 1000 for standard SI unit consistency.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Structure of Atom

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculate Degree of Dissociation

Step 2: Match to Given Format

Step 3: Rounding

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Heat Transfer

Step 2: Work Done

Step 3: Internal Energy Change

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Extract C-C Bond Energy

Step 2: Wavelength Calculation

Pattern Recognition

Chapter Mix

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