Question
Consider the following data for the reaction X_2(g) + Y_2(g) 2XY(g) at 600 K . The _r G^ (in kJ mol^ -1 ) for the reaction is : Compound _f H^ _ 600K (kJ mol^ -1 ) S^ _ 600K (J mol^ -1 K^ -1 ) XY(g) 42 200 X_2(g) 8 140 Y_2(g) 80 250
JEE Main
Chemistry
Thermodynamics (C)
2026
JEE Main 2026 (05 April Shift 2)
JEE Main Chemistry Question (2026) — Solution
Options
- A. -21000
- B. -10
- C. -1000
- D. -9.012
Answer
B. -10
Step-by-step solution
The given reaction is X_2(g) + Y_2(g) 2XY(g). The standard enthalpy of reaction, _r H^ , is calculated as: _r H^ = _f H^ ( products ) - _f H^ ( reactants ) _r H^ = 2 _f H^ (XY) - [ _f H^ (X_2) + _f H^ (Y_2)] _r H^ = 2(42) - (8 + 80) = 84 - 88 = -4 kJ mol ^ -1 The standard entropy of reaction, _r S^ , is calculated as: _r S^ = S^ ( products ) - S^ ( reactants ) _r S^ = 2 S^ (XY) - [S^ (X_2) + S^ (Y_2)] _r S^ = 2(200) - (140 + 250) = 400 - 390 = 10 J K ^ -1 mol ^ -1 Converting _r S^ to kJ K ^ -1 mol ^ -1 : _r S^ = 10 1000 = 0.01 kJ K ^ -1 mol ^ -1 Now, using the Gibbs free energy equation: _r G^ = _r H^ - T _r S^ _r G^ = -4 - 600 0.01 _r G^ = -4 - 6 = -10 kJ mol ^ -1 Answer: -10
Related: Chemistry — Thermodynamics (C) · All PYQ Banks