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 7

Q78 jee_main_2024_29_jan_morning Spontaneity and Gibbs Free Energy
Which of the following is not correct?
  • A. Delta G text is negative for a spontaneous reaction
  • B. Delta G text is positive for a spontaneous reaction
  • C. Delta G text is zero for a reversible reaction
  • D. Delta G text is positive for a non-spontaneous reaction

Solution

### Core Logic According to the second law of thermodynamics, at constant temperature and pressure, the change in Gibbs free energy (Delta G) dictates the spontaneity of a process. - If Delta G lt 0 (negative), the process is spontaneous. - If Delta G gt 0 (positive), the process is non-spontaneous. - If Delta G = 0, the system is in equilibrium (reversible process). Therefore, the statement "Delta G is positive for a spontaneous reaction" is factually incorrect. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q83 jee_main_2024_30_january_evening Hess's Law of Constant Heat Summation
Two reactions are given below: 2mathrmFe_(s) + frac32mathrmO_2(g) rightarrow mathrmFe_2mathrmO_3(s), Delta mathrmH^circ = -822 mathrmkJ/mol mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), Delta mathrmH^circ = -110 mathrmkJ/mol Then enthalpy change for following reaction 3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g)
Numerical Answer. Answer: 492 to 492

Solution

### Related Formula According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided. ### Core Logic Let the given reactions be: (1) 2mathrmFe_(s) + frac32mathrmO_2(g) rightarrow mathrmFe_2mathrmO_3(s), quad Delta H_1 = -822 \, mathrmkJ/mol (2) mathrmC_(s) + frac12mathrmO_2(g) rightarrow mathrmCO_(g), quad Delta H_2 = -110 \, mathrmkJ/mol Target Reaction (3): 3mathrmC_(s) + mathrmFe_2mathrmO_3(s) rightarrow 2mathrmFe_(s) + 3mathrmCO_(g), quad Delta H_3 = ? To construct the target reaction: - We need 3 mathrmCO_(g) on the product side, so we multiply reaction (2) by 3. - We need mathrmFe_2mathrmO_3(s) on the reactant side and 2 mathrmFe_(s) on the product side, so we reverse reaction (1). ### Step 1: Calculate Net Enthalpy Target Reaction (3) = 3 times (2) - (1) Delta H_3 = 3 times Delta H_2 - Delta H_1 Delta H_3 = 3(-110) - (-822) Delta H_3 = -330 + 822 = 492 \, mathrmkJ/mol ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q83 jee_main_2024_30_jan_morning Work Done in Cyclic Process
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path Arightarrow Brightarrow Crightarrow A as shown in the diagram. The total work done in the process is ________ J.
Work Done in Cyclic Process diagram for Q83 - JEE Main 2024 Morning
The image is a graph of Volume (dm3) vs Pressure (kPa) showing a triangular cyclic process starting from A(10,10) to B(10,30) to C(30,10) and back to A.
Numerical Answer. Answer: 200 to 200

Solution

### Related Formula W_textcyclic = textArea enclosed in P-V graph ### Core Logic The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph. Note that the provided graph is Volume (V) on the y-axis versus Pressure (P) on the x-axis. The path A rightarrow B rightarrow C rightarrow A is traced in a clockwise direction on the V-P graph. Clockwise on a V-P graph corresponds to anti-clockwise on a standard P-V graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for). ### Step 1: Calculating Area The enclosed region is a right-angled triangle. Base of triangle on P-axis = 30 - 10 = 20 text kPa Height of triangle on V-axis = 30 - 10 = 20 text dm^3 textArea = frac12 times textbase times textheight textArea = frac12 times 20 times 20 = 200 text kPacdottextdm^3 ### Step 2: Unit conversion 1 text kPa = 10^3 text Pa 1 text dm^3 = 1 text Litre = 10^-3 text m^3 W = 200 times 10^3 text Pa times 10^-3 text m^3 W = 200 text J ### Pattern Recognition 1 text kPa cdot 1 text L = 1 text Joule. This direct conversion saves time without converting explicitly to standard SI units (Pa and m^3). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q90 jee_main_2024_31_jan_evening Work Done in Isothermal Reversible Expansion
If 5text moles of an ideal gas expands from 10text L to a volume of 100text L at 300text K under isothermal and reversible condition then work w, is -xtext J. The value of x is ________ (Given R = 8.314text J K^-1textmol^-1)
Numerical Answer. Answer: 28720 to 28721

Solution

### Related Formula W = -2.303 \, nRT log left( fracV_2V_1 right) ### Core Logic For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: n = 5text moles R = 8.314text J K^-1textmol^-1 T = 300text K V_1 = 10text L V_2 = 100text L ### Step 1: Calculating Work Done W = -2.303 times 5 times 8.314 times 300 times logleft( frac10010 right) W = -2.303 times 5 times 8.314 times 300 times log(10) W = -2.303 times 12471 times 1 W = -28720.713text J ### Step 2: Final Formatting The question asks for work w = -xtext J. So x = 28720.713, which rounds to 28721. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q88 jee_main_2024_31_jan_morning Gibbs Free Energy and Equilibrium
Consider the following reaction at 298 K. frac32O_2(g) rightleftharpoons O_3(g). quad K_p = 2.47 times 10^-29 Delta_rG^ominus for the reaction is ________ kJ. (Given R = 8.314 text J K^-1 mol^-1)
Numerical Answer. Answer: 163 to 164

Solution

### Related Formula Delta_rG^ominus = -RT ln K_p ### Step 1: Calculation Delta_rG^ominus = -8.314 times 10^-3 text kJ K^-1 mol^-1 times 298 text K times ln(2.47 times 10^-29) = -8.314 times 10^-3 times 298 times (-65.87) = 163.19 text kJ ### Step 2: Nearest Integer Rounding 163.19 to the nearest integer gives 163. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

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