JEE Main202621 January 2026Morning ShiftChemistryThermodynamics (C)Actual
Use the following data : ( array |c|c|c| Substance & _f H ^ (500 ~K ) kJmol ⁻¹ & ~S ^ (500 ~K ) JK ⁻¹ ~mol ⁻¹ AB ( ~g ) & 32 & 222 ~A ₂( g ) & 6 & 146 ~B ₂( g ) & x & 280 array ) One mole each of A ₂( ~g ) and B ₂( ~g ) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K. A ₂( ~g )+ B ₂( ~g ) 2 AB ( ~g ) The value of x ( in kJ mol ⁻¹ ) is _ _ _ _ . (Nearest integer) (Given : K =2.2 R =8.
Correct answer
0
Step-by-step solution
Reaction: A₂(g) + B₂(g) 2AB(g) G° = -2.303RT K = -2.303 8.3 500 2.2 = -21026 J/mol = -21.03 kJ/mol S° = 2(222) - [146 + 280] = 444 - 426 = 18 J K⁻¹ mol⁻¹ H° = 2(32) - [6 + x] = (58 - x) kJ/mol Using G° = H° - T S° : -21.03 = (58 - x) - (500 0.018) -21.03 = 58 - x - 9 = 49 - x x = 49 + 21.03 = 70.03 70 kJ/mol