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 3

Q39 jee_main_2025_29_jan_evening Hess's Law of Constant Heat Summation
If C(textdiamond) ightarrow C(textgraphite) + Xtext kJ mol^-1 C(textdiamond) + O_2(g) ightarrow CO_2(g) + Ytext kJ mol^-1 C(textgraphite) + O_2(g) ightarrow CO_2(g) + Ztext kJ mol^-1 At constant temperature, then the correct relationship is:
  • A. X = Y + Z
  • B. -X = Y + Z
  • C. X = -Y + Z
  • D. X = Y - Z

Solution

### Core Logic Let's treat the given parameters as terms for exothermic heats evolved on the product side: 1) C(textdiamond) ightarrow C(textgraphite), Delta H = -X 2) C(textdiamond) + O_2(g) ightarrow CO_2(g), Delta H = -Y 3) C(textgraphite) + O_2(g) ightarrow CO_2(g), Delta H = -Z By subtracting Equation (3) from Equation (2): C(textdiamond) - C(textgraphite) ightarrow 0 implies C(textdiamond) ightarrow C(textgraphite) Delta H = (-Y) - (-Z) = Z - Y Matching this to Equation (1): -X = Z - Y implies X = Y - Z ### Pattern Recognition Hess's Law states that the enthalpy change of an overall reaction is equal to the sum of the enthalpy changes of its individual steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
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 273.15mathrmK . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:
  • A. textVolume of system increases.
  • B. textLiquid phase disappears completely.
  • C. textThe amount of ice decreases.
  • D. textThe 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: V_m(textice) > V_m(textwater) 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
Phase shift diagram for Q34 - JEE Main 2025 Morning
If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15mathrmK, 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics
Q50 jee_main_2025_28_jan_morning Bond Enthalpy Calculation
The formation enthalpies, Delta mathrmH_mathrmf^ominus for mathrmH_(mathrmg) and mathrmO_(mathrmg) are 220.0 and 250.0~mathrmkJ~mol^-1 , respectively, at 298.15mathrmK , and Delta mathrmH_mathrmf^- for mathrmH_2mathrmO_(mathrmg) is -242.0mathrmkJ\,mol^-1 at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15mathrmK is \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (nearest integer).
Numerical Answer. Answer: 466 to 466

Solution

### Related Formula Reaction enthalpy based on atomization processes: Delta_r H = sum Delta_f H(textproducts) - sum Delta_f H(textreactants) ### Step 1: Map the Dissociation Reaction Consider the dissociation of gas phase water molecules into constituent gaseous atoms: mathrmH_2mathrmO_mathrm(g) rightarrow 2mathrmH_mathrm(g) + mathrmO_mathrm(g) The total energy required corresponds to breaking exactly two mathrmO-mathrmH bonds: Delta_r H = 2 times textB.E.(mathrmO-mathrmH) ### Step 2: Calculate Delta_r H Using the enthalpies of formation: Delta_r H = [2 times Delta_f H(mathrmH_mathrm(g)) + Delta_f H(mathrmO_mathrm(g))] - Delta_f H(mathrmH_2mathrmO_mathrm(g)) Delta_r H = [2 times 220.0 + 250.0] - (-242.0) Delta_r H = [440.0 + 250.0] + 242.0 = 690.0 + 242.0 = 932.0\,mathrmkJ\,mol^-1 ### Step 3: Solve for Single Bond Enthalpy 2 times textB.E.(mathrmO-mathrmH) = 932.0 textB.E.(mathrmO-mathrmH) = frac932.02 = 466\,mathrmkJ\,mol^-1 ### Pattern Recognition Sees: Atomization state values used to evaluate single bond metrics. Shortcut: Remember textTotal Dissociation Energy = sum Delta_f H(textatoms) - Delta_f H(textmolecule). Halving the result gives the average bond enthalpy. ### Evaluation Rubric / Model Answer null ### 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?
Solution state transition grid for Q35 - JEE Main 2025 Morning
The flowchart maps Solution 1 containing 10 mol solute in 10 L water transitioning to Solution 2 containing 1 mol solute in 1 L water.
  • A. Molar heat capacity
  • B. Density
  • C. Concentration
  • D. Gibbs free energy

Solution

### Core Logic Let us compute the concentration of both solutions: textConcentration of Solution 1 = frac10text mol10text L = 1text mol/L textConcentration of Solution 2 = frac1text mol1text L = 1text mol/L Since concentration is identical, both systems share matching compositions. Consequently, all **intensive properties** (independent of mass/size) like concentration, density, and molar heat capacity remain exactly equal. ### Step 1: Identifying the Variable Gibbs free energy (G) is an **extensive property** that scales directly with the amount of matter in the system. Because Solution 1 contains a larger total mass and volume than Solution 2, its overall Gibbs free energy value is different. ### Pattern Recognition Shortcut: Look for the only extensive property in the options. Density, concentration, and molar parameters are always intensive. Gibbs free energy (G) scales with total matter quantity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics
Q49 jee_main_2025_03_april_morning Bond Enthalpy and Enthalpy of Formation
Given: Delta Htextsub^ominus[C(textgraphite)] = 710text kJ mol^-1 DeltaC-HH^ominus = 414text kJ mol^-1 DeltaH-HH^ominus = 436text kJ mol^-1 DeltaC=CH^ominus = 611text kJ mol^-1 The Delta Hf^ominus for CH2=CH2 is ________ textkJ mol^-1 (nearest integer value)
Numerical Answer. Answer: 25 to 25

Solution

### Related Formula The enthalpy of formation reaction can be evaluated by balancing atomization and bond cleavage details: Delta H_f^circ = sum Delta Htextatomization (reactants) - sum Delta Htextbonds broken/formed (products) ### Core Logic The target formation reaction for ethylene (textC_2textH_4) from standard elemental states is: 2C(textgraphite) + 2H_2(g) ightarrow CH_2=CH_2(g) To construct this pathway: 1. Sublime 2 moles of solid graphite: 2 times Delta H_textsub^circ[C] 2. Dissociate 2 moles of gaseous textH-textH bonds: 2 times Delta_H-HH^circ 3. Form 1 mole of textC=textC double bonds: -1 times Delta_C=CH^circ 4. Form 4 moles of textC-textH single bonds: -4 times Delta_C-HH^circ ### Step 1: Arithmetic Calculation Delta H_f^circ = [2 times 710] + [2 times 436] - 611 - [4 times 414] Delta H_f^circ = 1420 + 872 - 611 - 1656 = 2292 - 2267 = 25text kJ mol^-1 ### Pattern Recognition Shortcut: Group energy components systematically. Reactant state atomization costs +2292text kJ. Exothermic structural bond formation releases -2267text kJ. Net difference yields a small endothermic value of +25text kJ/mol. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics

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