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 estab

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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 =8.3 ~J ~K ^ -1 ~mol ^ -1 )

Quantrex Academy · Accepted Answer

Correct answer: A. A

Reaction: A_2(g) + B_2(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

JEE Main Chemistry Question (2026) — Solution

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