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 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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