The bond dissociation enthalpy of X₂ Δ H° bond calculated from the given data is _ _ _ kJ mol⁻¹. (Nearest integer) M⁺X^-(s) → M⁺(g) + X^-(g) Δ H° lattice = 800 kJ mol⁻¹; M(s) → M(g) Δ H° sub = 100 kJ mol⁻¹; M(g) → M⁺(g) + e^-(g) Δ H° i = 500 kJ mol⁻¹; X(g) + e^-(g) → X^-(g) Δ H° eg = -300 kJ mol⁻¹; M(s) + 1/2X₂(g) → M⁺X^-(s) Δ H° f = -400 kJ mol⁻¹. [Given: M⁺X⁻ⁱs a pure ionic compound and X forms a diatomic molecule X₂ in gaseous state]

Chemical Thermodynamics: JEE Main 2025 Chemistry question

JEE Main 2025 · 23 Jan, Evening shift · Q74

JEE Main 202523 Jan, Evening shiftChemistryHardChemical Thermodynamics84× since 2002

The bond dissociation enthalpy of

\mathrm{X_2}

\Delta H^\circ_{bond}

calculated from the given data is ______

\mathrm{kJ\,mol^{-1}}

. (Nearest integer)

\mathrm{M^+X^-(s) \rightarrow M^+(g) + X^-(g)}

\Delta H^\circ_{lattice} = 800\ \mathrm{kJ\,mol^{-1}}

;

\mathrm{M(s) \rightarrow M(g)}

\Delta H^\circ_{sub} = 100\ \mathrm{kJ\,mol^{-1}}

;

\mathrm{M(g) \rightarrow M^+(g) + e^-(g)}

\Delta H^\circ_i = 500\ \mathrm{kJ\,mol^{-1}}

;

\mathrm{X(g) + e^-(g) \rightarrow X^-(g)}

\Delta H^\circ_{eg} = -300\ \mathrm{kJ\,mol^{-1}}

;

\mathrm{M(s) + \frac{1}{2}X_2(g) \rightarrow M^+X^-(s)}

\Delta H^\circ_f = -400\ \mathrm{kJ\,mol^{-1}}

. [Given:

\mathrm{M^+X^-}

is a pure ionic compound and X forms a diatomic molecule

\mathrm{X_2}

in gaseous state]

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2025 Chemical Thermodynamics Question with Solution (23 Jan Evening shift, Q74) | JEE Arc