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 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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