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 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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot.

Gibbs Free Energy and Equilibrium diagram for Q69 - JEE Main 2026 Morning — A plot of Gibbs free energy against the fraction of N2O4 dissociated.

A. Standard free energy change ( $Δᵣ G°$ $\Delta_r G^{\circ}$) is the difference between standard free energies of pure products (point B) and pure reactants (point A): $Δᵣ G° = GB° - GA°$ . Since B is higher than A, $Δᵣ G°$ $\Delta_r G^{\circ}$ is positive, not $-5.40 kJ mol⁻¹$ $-5.40\text{ kJ mol}^{-1}$. Statement A is false.

B. Even if $Δᵣ G°$ $\Delta_r G^{\circ}$ 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 $0.84 kJ mol⁻¹$ $0.84\text{ kJ mol}^{-1}$. Thus $Δ G = -0.84 kJ mol⁻¹$ $\Delta G = -0.84\text{ kJ mol}^{-1}$ is correct. Statement D is true.

E. The difference from pure products (point B, equivalent to 2 moles $NO₂$ $NO_2$) to equilibrium (E) is the entire vertical distance. The drop from B to A is $5.40$ $5.40$, and A to E is $0.84$ $0.84$. So drop from B to E is $- (5.40 + 0.84) = -6.24 kJ mol⁻¹$ . Statement E is true.

Step 1: Final Conclusion

Only statements D and E are correct.

Pattern Recognition

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.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

More Thermodynamics Previous-Year Questions — Page 9

Q83 jee_main_2024_30_january_evening Hess's Law of Constant Heat Summation

Two reactions are given below:

2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H° = -822 kJ/mol

C(s) + (1)/(2)O2(g) arrow CO(g), Δ H° = -110 kJ/mol

Then enthalpy change for following reaction

3C(s) + Fe₂O3(s) arrow 2Fe(s) + 3CO(g)

Numerical Answer. Answer: 492 to 492

Solution

Related Formula

According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided.

Core Logic

Let the given reactions be: (1) $2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H₁ = -822 kJ/mol$ (2) $C(s) + (1)/(2)O2(g) arrow CO(g), Δ H₂ = -110 kJ/mol$

Target Reaction (3):

3C(s) + Fe₂O3(s) arrow 2Fe(s) + 3CO(g), Δ H₃ = ?

To construct the target reaction:

We need $3 CO(g)$ $3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3.
We need $Fe₂O3(s)$ $\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and $2 Fe(s)$ $2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1).

Step 1: Calculate Net Enthalpy

Target Reaction (3) = $3 \times (2) - (1)$ $3 \times (2) - (1)$

Δ H₃ = 3 \times Δ H₂ - Δ H₁

Δ H₃ = 3(-110) - (-822)

Δ H₃ = -330 + 822 = 492 kJ/mol

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q jee_main_2024_30_jan_morning Work Done in Cyclic Process

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path

Aarrow Barrow Carrow A

as shown in the diagram. The total work done in the process is ________ J.

Work Done in Cyclic Process diagram for Q83 - JEE Main 2024 Morning — The image is a graph of Volume (dm3) vs Pressure (kPa) showing a triangular cyclic process starting from A(10,10) to B(10,30) to C(30,10) and back to A.

Numerical Answer. Answer: 200 to 200

Solution

Related Formula

Wcyclic = Area enclosed in P-V graph

Core Logic

The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph. Note that the provided graph is Volume ( $V$ $V$) on the y-axis versus Pressure ( $P$ $P$) on the x-axis. The path $A arrow B arrow C arrow A$ is traced in a clockwise direction on the $V$ $V$- $P$ $P$ graph. Clockwise on a $V$ $V$- $P$ $P$ graph corresponds to anti-clockwise on a standard $P$ $P$- $V$ $V$ graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for).

Step 1: Calculating Area

The enclosed region is a right-angled triangle. Base of triangle on P-axis = $30 - 10 = 20 kPa$ $30 - 10 = 20 \text{ kPa}$ Height of triangle on V-axis = $30 - 10 = 20 dm³$ $30 - 10 = 20 \text{ dm}^3$

Area = (1)/(2) \times base \times height

Area = (1)/(2) \times 20 \times 20 = 200 kPa ³

Step 2: Unit conversion

$1 kPa = 10³ Pa$ $1 \text{ kPa} = 10^3 \text{ Pa}$ $1 dm³ = 1 Litre = 10⁻³ m³$

W = 200 \times 10³ Pa \times 10⁻³ m³

W = 200 J

Pattern Recognition

$1 kPa \cdot 1 L = 1 Joule$ . This direct conversion saves time without converting explicitly to standard SI units ( $Pa$ $Pa$ and $m³$ $m^3$).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q90 jee_main_2024_31_jan_evening Work Done in Isothermal Reversible Expansion

5 moles

of an ideal gas expands from

10 L

to a volume of

100 L

300 K

under isothermal and reversible condition then work

w

, is

-x J

. The value of

x

is ________ (Given

R = 8.314 J K⁻¹mol⁻¹

)

Numerical Answer. Answer: 28720 to 28721

Solution

Related Formula

W = -2.303 nRT ( (V₂)/(V₁) )

Core Logic

For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: $n = 5 moles$ $n = 5\text{ moles}$ $R = 8.314 J K⁻¹mol⁻¹$ $T = 300 K$ $T = 300\text{ K}$ $V₁ = 10 L$ $V_1 = 10\text{ L}$ $V₂ = 100 L$ $V_2 = 100\text{ L}$

Step 1: Calculating Work Done

W = -2.303 \times 5 \times 8.314 \times 300 \times ( (100)/(10) )

W = -2.303 \times 5 \times 8.314 \times 300 \times (10)

W = -2.303 \times 12471 \times 1

W = -28720.713 J

Step 2: Final Formatting

The question asks for work $w = -x J$ $w = -x\text{ J}$. So $x = 28720.713$ $x = 28720.713$, which rounds to $28721$ $28721$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q88 jee_main_2024_31_jan_morning Gibbs Free Energy and Equilibrium

Consider the following reaction at 298 K.

(3)/(2)O2(g) leftharpoons O3(g). Kₚ = 2.47 \times 10⁻²⁹

ΔᵣG

for the reaction is ________ kJ. (Given

R = 8.314 J K⁻¹ mol⁻¹

)

Numerical Answer. Answer: 163 to 164

Solution

Related Formula

ΔᵣG = -RT ln Kₚ

Step 1: Calculation

ΔᵣG = -8.314 \times 10⁻³ kJ K⁻¹ mol⁻¹ \times 298 K \times ln(2.47 \times 10⁻²⁹)

= -8.314 \times 10⁻³ \times 298 \times (-65.87)

= 163.19 kJ

Step 2: Nearest Integer

Rounding $163.19$ $163.19$ to the nearest integer gives $163$ $163$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Calculate Net Enthalpy

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Area

Step 2: Unit conversion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Work Done

Step 2: Final Formatting

Chapter Mix

Solution

Related Formula

Step 1: Calculation

Step 2: Nearest Integer

Chapter Mix

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