Use the following data :
SubstancefracDelta_f H^ominus(500K)kJ mol^-1fracS^ominus(500K)J K^-1 mol^-1
AB(g)32222
A_2(g)6146
B_2(g)x280
One mole each of A_2(g) and B_2(g) are taken in a 1L closed flask and allowed to establish the equilibrium at 500K. A_2(g) + B_2(g) rightleftharpoons 2AB(g) The value of x (in textkJ mol^-1) is .... (Nearest integer) (Given: log K=2.2, R=8.314text J K^-1text mol^-1)

Numerical Answer Type:
Enter a numerical value Answer: 70 to 70 +4 marks

Solution & Explanation

### Related Formula Delta_r H^circ = sum Delta_f H^circ(textproducts) - sum Delta_f H^circ(textreactants) Delta_r S^circ = sum S^circ(textproducts) - sum S^circ(textreactants) Delta_r G^circ = Delta_r H^circ - TDelta_r S^circ Delta_r G^circ = -2.303 RT log K ### Core Logic For the reaction A_2(g) + B_2(g) rightleftharpoons 2AB(g): 1. Enthalpy change (Delta_r H^circ): Delta_r H^circ = [2 times Delta_f H^circ(AB)] - [Delta_f H^circ(A_2) + Delta_f H^circ(B_2)] Delta_r H^circ = (2 times 32) - (6 + x) = (64 - 6 - x) = (58 - x)text kJ mol^-1 2. Entropy change (Delta_r S^circ): Delta_r S^circ = [2 times S^circ(AB)] - [S^circ(A_2) + S^circ(B_2)] Delta_r S^circ = (2 times 222) - (146 + 280) = 444 - 426 = 18text J K^-1text mol^-1 3. Standard Gibbs Free Energy (Delta_r G^circ) via Equilibrium Constant: Delta_r G^circ = -2.303 times R times T log K Delta_r G^circ = -2.303 times 8.314 times 500 times 2.2 = -21063.8text J mol^-1 = -21.06text kJ mol^-1 4. Using Gibbs Equation: Delta_r G^circ = Delta_r H^circ - TDelta_r S^circ -21.06 = (58 - x) - 500 times left(frac181000right) -21.06 = 58 - x - 9 -21.06 = 49 - x x = 49 + 21.06 = 70.06text kJ mol^-1 Rounding off to nearest integer, x = 70. ### 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 2

Q41 jee_main_2025_03_april_evening Latent Heat and Phase Equilibrium
Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II: At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. In the light of the above statements, choose the correct answer from the options given below:
  • A. Statement I is true but Statement II is false
  • B. Both Statement I and Statement II are false
  • C. Both Statement I and Statement II are true
  • D. Statement I is false but Statement II is true

Solution

### Related Formula During phase transition, latent heat of fusion (L_f) is absorbed: Q = m L_f Temperature remains constant because the average kinetic energy of the molecules does not change; instead, potential energy changes during phase transition: textTemperature T propto textAverage Kinetic Energy of molecules ### Core Logic Statement I Analysis: - During a phase transition (such as ice melting at 0^circmathrmC), any added heat is utilized as latent heat of fusion. The system remains at a constant temperature of 0^circmathrmC as long as both solid and liquid phases coexist in equilibrium. Thus, Statement I is True. ### Step 1: Analyze Statement II - Statement II explains *why* this happens: The thermal energy is spent entirely to break down the highly ordered crystalline hydrogen-bonded lattice of ice into liquid water. It does not increase the translational kinetic energy of the molecules. Since kinetic energy is constant, temperature remains constant. Thus, Statement II is True. ### Step 2: Conclusion Therefore, both Statement I and Statement II are True, matching Option (3). ### Pattern Recognition For any phase transition (melting, boiling, sublimation): Temperature stays flat. The added energy goes entirely into latent heat (potential energy change to overcome intermolecular forces), meaning average kinetic energy is constant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Physics: Thermal Properties of Matter
Q47 jee_main_2025_03_april_evening Work Done in Reversible Cyclic Processes
A perfect gas (0.1mathrm~mol) having barC_v=1.50mathrm~R (independent of temperature) undergoes the transformation from point 1 to point 4 as shown in the pressure-volume diagram below. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (-) ________ J. (nearest integer)
P-V path diagram for Q47 - JEE Main 2025 Evening
P-V graph showing a thermodynamic process from point 1 to point 4 containing an isobaric step and isochoric steps.
[Given: R=0.082mathrm~L~atm~K^-1~mol^-1 = 8.314mathrm~J~K^-1~mol^-1]
Numerical Answer. Answer: 304 to 304

Solution

### Related Formula Thermodynamic work done (w) for each step: - Isochoric step (V = textconstant): w = 0 - Isobaric step (P = textconstant): w = -P Delta V ### Core Logic The process from point 1 to point 4 consists of three distinct segments: 1. Step 1 rightarrow 2: Isochoric cooling at constant volume V_1 = 1000mathrm~cm^3. Work done w_1rightarrow 2 = 0. 2. Step 2 rightarrow 3: Isobaric compression at constant pressure P = 3.00mathrm~atm from volume 2000mathrm~cm^3 to 1000mathrm~cm^3. 3. Step 3 rightarrow 4: Isochoric step at constant volume. Work done w_3rightarrow 4 = 0. ### Step 1: Calculate work done in the isobaric step (2 rightarrow 3) The volume changes from V_i = 2000mathrm~cm^3 = 2.0mathrm~L to V_f = 1000mathrm~cm^3 = 1.0mathrm~L: w_2rightarrow 3 = -P Delta V = -3.00mathrm~atm times (1.0mathrm~L - 2.0mathrm~L) = +3.00mathrm~Lcdot atm ### Step 2: Convert work to Joules and analyze direction Convert \mathrm{L\cdot atm} to Joules: w_2rightarrow 3 = 3.00 times 101.325mathrm~J = 303.975mathrm~J approx 304mathrm~J The question asks for the total work done as (-) ________ J, meaning work done *by* the system (expansion) is negative and work done *on* the system (compression) is positive. Since this is compression, work done on the gas is +304\mathrm{~J}, which is represented as -(-304)\mathrm{~J} in typical IUPAC convention where work of expansion is examined. The absolute magnitude of the work is 304\mathrm{~J}. ### Pattern Recognition During any cyclic or multi-step path on a P-V graph, work is done *only* when there is a change in volume (W = -\int P dV$). Any vertical line (constant volume) represents an isochoric step where work is exactly zero. The horizontal segment directly represents rectangular area under the path. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Physics: Thermodynamics
Q48 jee_main_2025_03_april_evening Bomb Calorimetry and Heat of Combustion
A sample of n-octane (1.14mathrm~g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5mathrm~kJ~K^-1. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is ________ mathrmkJ~mol^-1. (nearest integer)
Numerical Answer. Answer: 2500 to 2500

Solution

### Related Formula Heat released at constant volume (q_v) in a bomb calorimeter is: q_v = C_textcal cdot Delta T Molar heat of combustion at constant volume (Delta U_textcomb): Delta U_textcomb = fracq_vn_textfuel ### Core Logic Given parameters: - Mass of n-octane m = 1.14mathrm~g - Heat capacity of calorimeter C_textcal = 5mathrm~kJ/K - Temperature rise Delta T = 5mathrm~K - Formula of octane: mathrmC_8H_18 Rightarrow Molar mass = 8(12) + 18(1) = 114mathrm~g/mol ### Step 1: Calculate heat absorbed by the calorimeter (q_v) q_v = 5mathrm~kJ/K times 5mathrm~K = 25mathrm~kJ ### Step 2: Calculate moles of octane n = frac1.14mathrm~g114mathrm~g/mol = 0.01mathrm~mol ### Step 3: Calculate molar heat of combustion Delta U_textcomb = frac25mathrm~kJ0.01mathrm~mol = 2500mathrm~kJ/mol The magnitude of the heat of combustion is 2500\mathrm{~kJ~mol^{-1}}. ### Pattern Recognition A bomb calorimeter operates at rigid constant volume, meaning boundary work w = 0. Thus, by the first law of thermodynamics, the measured heat flow represents the internal energy change (\Delta U), not the enthalpy change (\Delta H$, which occurs at constant pressure). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q29 jee_main_2025_07_april_morning Enthalpy Changes
Total enthalpy change for freezing of 1 mathrm~mol of water at 10^circ mathrmC to ice at -10^circ mathrmC is (Given: Delta_mathrmfusmathrmH = mathrmx kJ/mol, mathrmC_mathrmp[mathrmH_2mathrmO(ell)] = mathrmytext J mol^-1text K^-1, mathrmC_mathrmp[mathrmH_2mathrmO(texts)] = mathrmztext J mol^-1text K^-1)
  • A. -mathrmx - 10mathrmy - 10mathrmz
  • B. -10(100mathrmx + mathrmy + mathrmz)
  • C. 10(100mathrmx + mathrmy + mathrmz)
  • D. mathrmx - 10mathrmy - 10mathrmz

Solution

### Related Formula Delta H_texttotal = n C_p(ell) Delta T_1 - nDelta H_textfusion + n C_p(s) Delta T_2 ### Core Logic We need to compute the enthalpy change for the pathway: mathrmH_2O(ell, 10^circC) rightarrow mathrmH_2O(s, -10^circC) This can be divided into three consecutive steps: 1. Cool liquid water from 10^circC to 0^circC: Delta H_1 = n cdot C_p[mathrmH_2O(ell)] cdot (0 - 10) = 1 cdot y cdot (-10) = -10y text J 2. Freeze water to ice at 0^circC: Delta H_2 = -n cdot Delta_textfusH = -1 cdot x text kJ = -1000x text J 3. Cool ice from 0^circC to -10^circC: Delta H_3 = n cdot C_p[mathrmH_2O(s)] cdot (-10 - 0) = 1 cdot z cdot (-10) = -10z text J
Thermodynamics enthalpy cycle for freezing Q29
Thermodynamics enthalpy cycle for freezing Q29
Adding these three steps yields: Delta H_texttotal = -10y - 1000x - 10z = -10(100x + y + z) text Joule ### Pattern Recognition Freezing is an exothermic process, so all three steps (cooling water, freezing, and cooling ice) must carry a negative sign. Factoring out -10 cleanly yields the expression -10(100x+y+z). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics
Q46 jee_main_2025_08_april_evening Resonance Energy
Resonance in an X_2Y molecule is represented as follows: X=X=Y longleftrightarrow X equiv X^+-Y^- The experimental enthalpy of formation for gaseous X_2Y is given by the reaction: X equiv X(g) + frac12Y=Y(g) longrightarrow X_2Y(g) quad Delta H_f(textexp) = 80 text kJ mol^-1 Calculate the magnitude of the resonance energy of X_2Y in textkJ mol^-1 (as the nearest integer value). Given bond energies: * X equiv X = 940 text kJ mol^-1 * X = X = 410 text kJ mol^-1 * Y = Y = 500 text kJ mol^-1 * X = Y = 602 text kJ mol^-1 Valence settings: X:3, Y:2.
Numerical Answer. Answer: 98 to 98

Solution

### Related Formula Resonance energy equation related to experimental and theoretical formation enthalpies: Delta H_textR.E. = Delta H_f(textexp) - Delta H_f(textTheo) Theoretical enthalpy calculation using bond energies: Delta H_f(textTheo) = sum textB.E._textreactants - sum textB.E._textproducts ### Execution Step 1: Write the chemical equation to calculate the theoretical enthalpy of formation based on the localized structure X=X=Y: X equiv X(g) + frac12Y=Y(g) longrightarrow X=X=Y(g) Step 2: Substitute the localized bond energies into the reactant-minus-product relation: Delta H_f(textTheo) = left[ textB.E._X equiv X + frac12textB.E._Y=Y right] - left[ textB.E._X=X + textB.E._X=Y right] Delta H_f(textTheo) = left[ 940 + frac12(500) right] - [410 + 602] Delta H_f(textTheo) = [940 + 250] - 1012 = 1190 - 1012 = 178 text kJ mol^-1 Step 3: Calculate the resonance energy: Delta H_textR.E. = Delta H_f(textexp) - Delta H_f(textTheo) = 80 - 178 = -98 text kJ mol^-1 Taking the magnitude as requested: |Delta H_textR.E.| = 98. ### Pattern Recognition Resonance energy is always a stabilizing factor, meaning Delta H_f(textexp) is more exothermic (or less endothermic) than the theoretical localized state. The magnitude is simply the absolute value of this difference (|80 - 178| = 98). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics Class 11 Chemistry: Chemical Bonding and Molecular Structure

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