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

Q34 jee_main_2025_28_jan_morning Phase Equilibrium and Le Chatelier's Principle

Ice and water are placed in a closed container at a pressure of 1 atm and temperature

273.15K

. If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

A. $Volume of system increases.$ $\text{Volume of system increases.}$
B. $Liquid phase disappears completely.$
C. $The amount of ice decreases.$ $\text{The amount of ice decreases.}$
D. $The solid phase (ice) disappears completely.$

Solution

Core Logic

Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water:

Vm(ice) > Vm(water)

According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water:

Phase shift diagram for Q34 - JEE Main 2025 Morning

If the pressure is increased considerably (such as doubling it to 2 atm) at $273.15K$ $273.15\mathrm{K}$, the melting point decreases, causing the entire solid phase (ice) to disappear completely.

Pattern Recognition

Sees: Ice-water system under pressure change. Trap: Assuming that an increase in pressure always favors the solid phase. Water has an anomalous phase curve with a negative slope.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q50 jee_main_2025_28_jan_morning Bond Enthalpy Calculation

The formation enthalpies,

Δ Hf

for

H(g)

and

O(g)

are 220.0 and

250.0~kJ~mol⁻¹

, respectively, at

298.15K

, and

Δ Hf⁻

for

H₂O(g)

is

-242.0kJ mol⁻¹

at the same temperature. The average bond enthalpy of the O-H bond in water at

298.15K

is

___________________________________________________ (nearest integer).

Numerical Answer. Answer: 466 to 466

Solution

Related Formula

Reaction enthalpy based on atomization processes:

Δᵣ H = Σ Δf H(products) - Σ Δf H(reactants)

Step 1: Map the Dissociation Reaction

Consider the dissociation of gas phase water molecules into constituent gaseous atoms:

H₂O(g) arrow 2H(g) + O(g)

The total energy required corresponds to breaking exactly two $O-H$ $\mathrm{O}-\mathrm{H}$ bonds:

Δᵣ H = 2 \times B.E.(O-H)

Step 2: Calculate $Δᵣ H$ $\Delta_r H$

Using the enthalpies of formation:

Δᵣ H = [2 \times Δf H(H(g)) + Δf H(O(g))] - Δf H(H₂O(g))

Δᵣ H = [2 \times 220.0 + 250.0] - (-242.0)

Δᵣ H = [440.0 + 250.0] + 242.0 = 690.0 + 242.0 = 932.0 kJ mol⁻¹

Step 3: Solve for Single Bond Enthalpy

2 \times B.E.(O-H) = 932.0

B.E.(O-H) = (932.0)/(2) = 466 kJ mol⁻¹

Pattern Recognition

Sees: Atomization state values used to evaluate single bond metrics. Shortcut: Remember $Total Dissociation Energy = Σ Δf H(atoms) - Δf H(molecule)$ . Halving the result gives the average bond enthalpy.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q35 jee_main_2025_03_april_morning Intensive and Extensive Properties

Which of the following properties will change when system containing solution 1 will become solution 2?

System mixture comparison diagram for Q35 - JEE Main 2025 Morning — Diagram showing solution 1 (10 mol solute in 10 L water) and solution 2 (1 mol solute in 1 L water).

A. Molar heat capacity
B. Density
C. Concentration
D. Gibbs free energy

Solution

Related Formula

$G = H - TS$ $G = H - TS$

Extensive properties depend on mass/amount of matter; Intensive properties are independent of amount.

Core Logic

Both solution 1 ( $10 mol/10 L = 1 M$ ) and solution 2 ( $1 mol/1 L = 1 M$ $1\text{ mol}/1\text{ L} = 1\text{ M}$) have identical concentrations and chemical compositions.

Intensive properties (molar heat capacity, density, concentration) depend only on composition and temperature, so they remain unchanged.

Gibbs free energy ( $G$ $G$) is an extensive property, proportional to the total amount of substance present, so it changes.

Step 1: Final Conclusion

Gibbs free energy changes because it is an extensive property.

Pattern Recognition

Same concentration/composition $arrow$ $\rightarrow$ Intensive properties stay equal. Total amount changes $arrow$ $\rightarrow$ Extensive properties ( $G, H, U, S$ $G, H, U, S$) change.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q49 jee_main_2025_03_april_morning Enthalpy of Formation and Bond Enthalpies

Given:

Δ Hsub [C(graphite)] = 710 ~kJ~mol⁻¹

ΔC-H H = 414 ~kJ~mol⁻¹

ΔH-H H = 436 ~kJ~mol⁻¹

ΔC=C H = 611 ~kJ~mol⁻¹

The

Δ Hf

for

CH₂ = CH₂

is ________

kJ~mol⁻¹

(nearest integer value)

Numerical Answer. Answer: 24 to 26

Solution

Related Formula

Δ Hf(C₂H₄) = 2 Δ Hsub[C(s)] + 2 ΔH-H H - [1 ΔC=C H + 4 ΔC-H H]

Core Logic

Target formation reaction:

2C(graphite, s) + 2H₂(g) arrow CH₂=CH₂(g)

Sublimation of 2 moles of C(s) to gaseous atoms: $2 \times 710 = 1420~kJ$ $2 \times 710 = 1420\mathrm{~kJ}$
Dissociation of 2 moles of $H₂(g)$ $\mathrm{H_2(g)}$ to $4H(g)$ $4\mathrm{H(g)}$ atoms: $2 \times 436 = 872~kJ$ $2 \times 436 = 872\mathrm{~kJ}$
Formation of $1 \times C=C$ $1 \times \mathrm{C=C}$ double bond: $-611~kJ$ $-611\mathrm{~kJ}$
Formation of $4 \times C-H$ $4 \times \mathrm{C-H}$ single bonds: $-4 \times 414 = -1656~kJ$ $-4 \times 414 = -1656\mathrm{~kJ}$

Step 1: Calculation

Δ Hf = 1420 + 872 - 611 - 1656

Δ Hf = 2292 - 2267 = 25 ~kJ~mol⁻¹

Pattern Recognition

Standard Enthalpy of Formation = Atomization Enthalpies of Reactants $-$ $-$ Bond Enthalpies of Products formed.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q37 jee_main_2025_04_april_evening Thermochemistry

Consider the given data : (a)

HCl(g) + 10H₂O(l)arrow HCl.10H₂O Δ H = - 6 9. 0 1 k J m o l ^ - 1

(b)

HCl(g) + 40H₂O(l)arrow HCl.40H₂O Δ H = - 7 2. 7 9 k J m o l ^ - 1

Choose the correct statement :

A. Dissolution of gas in water is an endothermic process
B. The heat of solution depends on the amount of solvent.
C. The heat of dilution for the HCl $(HCl.10H₂O$ to $HCl.40H₂O)$ is $3.78kJ mol⁻¹$ $3.78\mathrm{kJ mol}^{-1}$.
D. The heat of formation of HCl solution is represented by both (a) and (b)

Solution

Related Formula

Δ Hdilution = Δ H₂ - Δ H₁

Core Logic

Analyzing the thermodynamic statements:

$Δ H$ $\Delta H$ values are negative, so the dissolution of $HCl(g)$ $HCl(g)$ is clearly exothermic, eliminating option (1).
Since the enthalpy release changes when the moles of water solvent shift from 10 to 40 ( $-69.01$ $-69.01$ vs $-72.79$ $-72.79$), the heat of solution depends explicitly on the amount of solvent (Statement 2 is true).
Let's check Statement 3: By subtracting equation (a) from (b):

HCl\cdot10H₂O + 30H₂O arrow HCl\cdot40H₂O

Δ H = -72.79 - (-69.01) = -3.78 ~kJ\cdot mol⁻¹

The value is negative, indicating an exothermic process, so calling it $+3.78$ $+3.78$ makes option (3) incorrect.

Pattern Recognition

The standard integral enthalpy of solution varies with solvent concentration until infinite dilution is achieved. Thus, concentration dependence is a core property of partial molar solution variables.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 5

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Map the Dissociation Reaction

Step 2: Calculate Δᵣ H$\Delta_r H$

Step 3: Solve for Single Bond Enthalpy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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Step 2: Calculate $Δᵣ H$ $\Delta_r H$