For the reaction, N_2O_4 rightleftharpoons 2NO_2, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is -5.40text kJ mol^-1. B. As Delta G^ominus in graph is positive, N_2O_4 will not dissociate into NO_2 at all. C. Reverse reaction will go to completion. D. When 1 mole of N_2O_4 changes into equilibrium mixture, value of Delta G = -0.84text kJ mol^-1. E. When 2 mole of NO_2 changes into equilibrium mixture, Delta G for equilibrium mixture is -6.24text kJ mol^-1.
Gibbs Free Energy and Equilibrium diagram for Q69 - JEE Main 2026 Morning
A plot of Gibbs free energy against the fraction of N2O4 dissociated.
Choose the correct answer from the options given below :

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 (Delta_r G^circ) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Delta_r G^circ = G_B^circ - G_A^circ. Since B is higher than A, Delta_r G^circ is positive, not -5.40text kJ mol^-1. Statement A is false. B. Even if 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 0.84text kJ mol^-1. Thus Delta G = -0.84text kJ mol^-1 is correct. Statement D is true. E. The difference from pure products (point B, equivalent to 2 moles NO_2) 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) = -6.24text 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

Reference Study Guides

Thermodynamics

More Thermodynamics Previous-Year Questions — Page 4

Q37 jee_main_2025_04_april_evening Thermochemistry
Consider the given data : (a) mathrmHCl(g) + 10mathrmH_2mathrmO(l)rightarrow mathrmHCl.10H_2O quad Delta mathrm H = - 6 9. 0 1 mathrm k J mathrm m o l ^ - 1 (b) mathrmHCl(g) + 40mathrmH_2mathrmO(l)rightarrow mathrmHCl.40H_2O quad Delta mathrm H = - 7 2. 7 9 mathrm k J mathrm 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 (mathrmHCl.10mathrmH_2mathrmO to mathrmHCl.40mathrmH_2mathrmO) is 3.78mathrmkJ mol^-1.
  • D. The heat of formation of HCl solution is represented by both (a) and (b)

Solution

### Related Formula Delta H_textdilution = Delta H_2 - Delta H_1 ### Core Logic Analyzing the thermodynamic statements: - Delta H values are negative, so the dissolution of 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 vs -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): mathrmHClcdot10H_2O + 30mathrmH_2mathrmO rightarrow mathrmHClcdot40H_2O Delta H = -72.79 - (-69.01) = -3.78 mathrm~kJcdot mol^-1 The value is negative, indicating an exothermic process, so calling it +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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics
Q27 jee_main_2025_04_april_morning Spontaneity and Gibbs Energy
Let us consider a reversible reaction at temperature, T. In this reaction, both Delta H and Delta S were observed to have positive values. If the equilibrium temperature is T_e, then the reaction becomes spontaneous at:
  • A. T = T_e
  • B. T_e > T
  • C. T > T_e
  • D. T_e = 5T

Solution

### Related Formula Delta G = Delta H - TDelta S ### Core Logic For a reaction to be spontaneous, the change in Gibbs free energy must be negative: Delta G < 0 implies Delta H - TDelta S < 0 Given that both Delta H > 0 and Delta S > 0: Delta H < TDelta S implies T > fracDelta HDelta S At the equilibrium temperature T_e, Delta G = 0, which gives: T_e = fracDelta HDelta S Substituting this back into the inequality reveals that the reaction is spontaneous when: T > T_e ### Pattern Recognition When both Delta H and Delta S are positive, the reaction is entropy-driven and becomes spontaneous only at higher temperatures (T > T_e). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics
Q30 jee_main_2025_04_april_morning Isothermal and Reversible Expansion
One mole of an ideal gas expands isothermally and reversibly from 10mathrm~dm^3 to 20mathrm~dm^3 at 300mathrm~K. Delta U, q and work done in the process respectively are: Given: R = 8.3mathrm~J~K^-1~mol^-1, ln 10 = 2.3, log 2 = 0.30, log 3 = 0.48
  • A. 0, 21.84mathrm~kJ, -1.26mathrm~kJ
  • B. 0, -17.18mathrm~kJ, 1.718mathrm~J
  • C. 0, 21.84mathrm~kJ, 21.84mathrm~kJ
  • D. 0, 1.718mathrm~kJ, -1.718mathrm~kJ

Solution

### Related Formula Delta U = n C_v Delta T w = -n R T lnleft(fracV_2V_1right) Delta U = q + w ### Core Logic Since the expansion step is strictly **isothermal** (Delta T = 0): Delta U = 0 Now compute the work command parameter w: w = -n R T lnleft(fracV_2V_1right) = -1 cdot 8.3 cdot 300 cdot lnleft(frac2010 ight) w = -2490 cdot ln(2) = -2490 cdot (2.3 cdot log 2) w = -2490 cdot (2.3 cdot 0.30) = -2490 cdot 0.69 = -1718.1mathrm~J = -1.718mathrm~kJ Applying the first law equation constraint: q = -w = +1.718mathrm~kJ Hence, Delta U = 0, q = 1.718mathrm~kJ, w = -1.718mathrm~kJ. ### Pattern Recognition Isothermal expansion of an ideal gas ALWAYS yields Delta U = 0. Work is negative (done by system) and heat exchange q matches work magnitude inversely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics
Q28 jee_main_2025_07_april_evening Lattice Enthalpy and Born-Haber Cycle
The hydration energies of textK^+ and textCl^- are -textx and -textytext kJ/mol respectively. If lattice energy of textKCl is -textztext kJ/mol, then the heat of solution of textKCl is:
  • A. +textx - texty - textz
  • B. textx + texty + textz
  • C. textz - (textx + texty)
  • D. -textz - (textx + texty)

Solution

### Related Formula Delta H_textsol = textLattice Energy (L.E.) + Delta H_texthyd(textCation) + Delta H_texthyd(textAnion) ### Core Logic According to Hess's Law, the dissolution process can be mapped as follows:
Lattice Enthalpy and Born-Haber Cycle diagram for Q28 - JEE Main 2025 Evening
Lattice Enthalpy and Born-Haber Cycle diagram for Q28 - JEE Main 2025 Evening
Given parameters: - Lattice Energy of textKCl breaking into gaseous ions = -(-textz) = textztext kJ/mol (since lattice energy released on formation is given as -textz). - Hydration energy of textK^+ = -textxtext kJ/mol - Hydration energy of textCl^- = -textytext kJ/mol ### Step 1: Computation Substituting the values into the governing formulation: Delta H_textsol = textz + (-textx) + (-texty) Delta H_textsol = textz - textx - texty = textz - (textx + texty) ### Pattern Recognition To dissolve an ionic crystal, energy equal to the lattice energy must be supplied (endothermic step, +textz), and hydration releases energy (exothermic steps, -textx and -texty). Net heat of solution is simply the sum of these parts: textz - textx - texty. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q33 jee_main_2025_07_april_evening Standard Enthalpy of Formation
The correct statement amongst the following is:
  • A. textThe term 'standard state' implies that the temperature is 0^circtextC
  • B. textThe standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K
  • C. DeltatextftextH298^thetatext is zero for O(g)
  • D. DeltatextftextH500^thetatext is zero for O2(g)

Solution

### Related Formula DeltatextfH^theta = 0 quad textfor an element in its reference/most stable standard state ### Core Logic - Standard state conditions prescribe a pressure of 1text bar. Temperature is not fixed by definition but is explicitly specified (often reference tables use 298.15text K). - Oxygen naturally and stably exists as diatomic gas molecules (textO_2(g)) at standard thresholds. - The enthalpy of formation of an element in its reference elemental state is identically zero at any reference temperature: DeltatextfH_500^theta[textO2(g)] = 0 Conversely, atomic oxygen gas (textO(g)) is not the reference phase, so its formation enthalpy is non-zero. ### Step 1: Verification of Options Statement (4) accurately aligns with thermodynamic core definitions, while statement (1) and (2) mistakenly conflate standard ambient reference states with STP conditions (273.15text K, 1text atm). ### Pattern Recognition Standard state definitions checklist: Pressure = 1text bar. Temperature is variable/assigned independently. Elements in their most stable natural form take DeltatextfH^theta = 0 at all thermal profiles. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

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