JEE Main202523 Jan 2025Evening ShiftChemistryThermodynamics (C)Actual
The bond dissociation enthalpy of X ₂ H _ bond calculated from the given data is kJ mol ⁻¹ . (Nearest integer) aligned & M ⁺ X ⁻( s ) M ⁺( g )+ X ⁻( g ) H _ lattice ^*=800 ~kJ ~mol ⁻¹ & M ( ~s ) M ( ~g ) H _ sub ^ =100 ~kJ ~mol ⁻¹ aligned M ( ~g ) M ⁺( g )+ e ⁻( g ) H _ i =500 ~kJ ~mol ⁻¹ X ( ~g )+ e ⁻( g ) X ⁻( g ) H _ eg ^*=-300 ~kJ ~mol ⁻¹ M ( ~s )+ 1 2 X ₂( ~g ) M ⁺ X ⁻( s ) H _f^ =-400 ~kJ ~mol ⁻¹ [Given : M ⁺ X
Correct answer
0
Step-by-step solution
aligned & array l H _ f ( MX )= H _ sub ( M )+ I.E. ( M )+ 1 2 [ B.E. ( X - X )] + EG ( X )+ L.E. ( MX ) array & -400=(100)+(500)+ 1 2 ( B.E. )+(-300)+(-800) aligned aligned & B.E. =200 ~kJ ~mole ⁻¹ aligned