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 6

Q47 jee_main_2025_28_jan_evening Hess's Law / Enthalpy of Formation
Consider the following data: Heat of formation of CO_2(g) = -393.5mathrm~kJ~mol^-1 Heat of formation of H_2O(l) = -286.0mathrm~kJ~mol^-1 Heat of combustion of benzene = -3267.0mathrm~kJ~mol^-1 The heat of formation of benzene is ______ mathrmkJ~mol^-1 (Nearest integer).
Numerical Answer. Answer: 48 to 48

Solution

### Related Formula Enthalpy of reaction from enthalpy of formation data: Delta H_textreaction = sum Delta H_f(textProducts) - sum Delta H_f(textReactants) ### Core Logic Write out the balanced thermochemical equation for the combustion of benzene (C_6H_6): C_6H_6(l) + frac152O_2(g) rightarrow 6CO_2(g) + 3H_2O(l) Given parameters: - Delta H_c = -3267.0mathrm\ kJ/mol - Delta H_f[CO_2] = -393.5mathrm\ kJ/mol - Delta H_f[H_2O] = -286.0mathrm\ kJ/mol - Delta H_f[O_2] = 0mathrm\ kJ/mol ### Step 1: Applying Hess's Law Substitute these values into the reaction expression: -3267 = [6(-393.5) + 3(-286.0)] - Delta H_f[C_6H_6] -3267 = [-2361.0 - 858.0] - Delta H_f[C_6H_6] -3267 = -3219.0 - Delta H_f[C_6H_6] Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48mathrm\ kJ/mol ### Pattern Recognition Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply CO_2 by 6 and H_2O by 3) to ensure accurate bookkeeping. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q jee_main_2025_29_jan_morning First Law of Thermodynamics and Heat Capacity
500 mathrm~J of energy is transferred as heat to 0.5 mathrm~mol of Argon gas at 298 mathrm~K and 1.00 mathrmatm . The final temperature and the change in internal energy respectively are : Given: mathrmR = 8.3 \, mathrmJK^-1 mathrmmol^-1
  • A. 348mathrmK and 300mathrmJ
  • B. 378mathrmK and 300mathrmJ
  • C. 368mathrmK and 500mathrmJ
  • D. 378mathrmK and 500mathrmJ

Solution

### Formulas Used For an ideal gas undergoing a constant pressure process (1.00text atm): q_p = n cdot C_p cdot Delta T Change in internal energy (Delta U): Delta U = n cdot C_v cdot Delta T For a monoatomic gas like Argon: * C_v = frac32 R * C_p = frac52 R ### Core Logic **Step 1: Calculate the final temperature (T_f)** Heat transferred at constant pressure (q_p) = 500text J 500 = 0.5 times left(frac52 times 8.3right) times (T_f - 298) 500 = 0.5 times 20.75 times (T_f - 298) 500 = 10.375 times (T_f - 298) T_f - 298 = frac50010.375 approx 48.2text K T_f = 298 + 48.2 = 346.2text K approx 348text K --- **Step 2: Calculate the change in internal energy (Delta U)** Delta U = n cdot C_v cdot Delta T Alternatively, using the ratio of heat capacities: Delta U = left(fracC_vC_pright) times q_p = frac35 times 500text J = 300text J Thus, the final temperature is **348text K** and the change in internal energy is **300text J**. ### Pattern Recognition For a monoatomic ideal gas under constant pressure, exactly 60\% of the heat added (fracC_vC_p = frac35) goes into increasing the internal energy (Delta U), while 40\% is lost to expansion work (W). **Correct Option:** **(A)**
Q79 jee_main_2024_01_february_morning First Law of Thermodynamics
Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following:
  • A. q = 0, Delta T neq 0, w = 0
  • B. q = 0, Delta T < 0, w neq 0
  • C. q neq 0, Delta T = 0, w = 0
  • D. q = 0, Delta T = 0, w = 0

Solution

### Core Logic Free expansion means expansion against a vacuum (P_ext = 0). Work done: w = -P_ext Delta V. Since P_ext = 0, w = 0. Adiabatic condition means there is no heat exchange with the surroundings. Heat transfer: q = 0. According to the First Law of Thermodynamics, Delta U = q + w. Since q = 0 and w = 0, the change in internal energy Delta U = 0. For an ideal gas, internal energy is a function of temperature only (Delta U = nC_vDelta T). If Delta U = 0, then Delta T = 0. ### Step 1: Final Parameter Check Evaluating all parameters simultaneously: q = 0 w = 0 Delta T = 0 ### Pattern Recognition Adiabatic + Free Expansion of IDEAL gas rightarrow Nothing changes thermodynamically except volume and pressure. q = 0, w = 0, Delta U = 0, Delta T = 0, Delta H = 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q88 jee_main_2024_29_january_evening Enthalpy of Phase Transition
Standard enthalpy of vapourisation for mathrmCCl_4 is 30.5mathrmkJ mol^-1. Heat required for vapourisation of 284mathrmg of mathrmCCl_4 at constant temperature is ________ kJ. (Given molar mass in g mol-; C = 12, Cl = 35.5)
Numerical Answer. Answer: 56 to 56.25

Solution

### Related Formula Q = n times Delta H_textvap^0 quad textwhere n = fractextMass, textMolar Mass ### Core Logic First, calculate the molar mass of carbon tetrachloride (CCl_4): textMolar mass = 12 + 4(35.5) = 12 + 142 = 154text g/mol Next, calculate the total number of moles present in 284text g of the substance: n = frac284, 154 approx 1.844text moles ### Step 1: Enthalpy Calculation Calculate the total energy required for vaporization: Delta H = 1.844text mol times 30.5text kJ/mol approx 56.24text kJ Rounding to the nearest integer value gives **56**. ### Pattern Recognition Enthalpy of vaporization is an intensive property given per mole. Scale it lineary by multiplying by the total number of moles to find the total extensive heat required. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics
Q84 jee_main_2024_27_jan_morning Isothermal Expansion and Work Calculations
If three moles of an ideal gas at 300text K expand isothermally from 30text dm^3 to 45text dm^3 against a constant opposing pressure of 80text kPa, then the amount of heat transferred is textquadquad J.
Numerical Answer. Answer: 1200 to 1200

Solution

### Related Formula First law of thermodynamics framework: Delta U = Q + W For an isothermal processes involving ideal gases, internal energy change is zero: Delta U = 0 implies Q = -W Irreversible work formula expanding against constant external pressure: W = -P_textext Delta V = -P_textext(V_2 - V_1) ### Step 1: Calculate structural work values Given values: P_textext = 80text kPa = 80 times 10^3text Pa V_1 = 30text dm^3 = 30 times 10^-3text m^3 V_2 = 45text dm^3 = 45 times 10^-3text m^3 Delta V = (45 - 30) times 10^-3 = 15 times 10^-3text m^3 W = -80 times 10^3 times (15 times 10^-3) = -1200text J ### Step 2: Solve for heat magnitude $Q = -W = -(-1200text J) = 1200text J ### Pattern Recognition Constant opposing pressure indicates an irreversible process path. Use W = -P\Delta V$ instead of logarithmic integrals. ### Chapter Mix Class 11 Chemistry: Thermodynamics

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