JEE Main 2025 Jan 23 Shift 2, Chemistry Q24: Hess’s Law and Bond Energy

JEE Main2025Jan 23, Shift 2Chemistry

The bond dissociation enthalpy of $X_{2}$ , $Δ H_{b o n d}^{o}$ , calculated from the given data is _______ kJ mol $^{- 1}$ . (Nearest integer)

$M^{+} X^{-} (s) \to M^{+} (g) + X^{-} (g) Δ H_{l a tt i ce}^{\circ} = 800$ kJ mol $^{- 1}$

$M (s) \to M (g) Δ H_{s u b}^{\circ} = 100$ kJ mol $^{- 1}$

$M (g) \to M^{+} (g) + e^{-} (g) Δ H_{i}^{\circ} = 500$ kJ mol $^{- 1}$

$X (g) + e^{-} (g) \to X^{-} (g) Δ H_{e g}^{\circ} = - 300$ kJ mol $^{- 1}$

$M (s) + \frac{1}{2} X_{2} (g) \to M^{+} X^{-} (s) Δ H_{f}^{\circ} = - 400$ kJ mol $^{- 1}$

[Given : $M^{+} X^{-}$ is a pure ionic compound and X forms a diatomic molecule $X_{2}$ in gaseous state]

Your answer

Solution

Apply the Born $-$ Haber cycle for $M (s) + \frac{1}{2} X_{2} (g) \to M^{+} X^{-} (s)$ . The lattice-formation step has enthalpy $- Δ H_{l a tt i ce}^{\circ} = - 800$ kJ mol $^{- 1}$ .

$Δ H_{f}^{\circ} = Δ H_{s u b}^{\circ} + Δ H_{i}^{\circ} + \frac{1}{2} Δ H_{b o n d}^{\circ} + Δ H_{e g}^{\circ} + (- Δ H_{l a tt i ce}^{\circ})$

Substituting:

$- 400 = 100 + 500 + \frac{1}{2} Δ H_{b o n d}^{\circ} + (- 300) + (- 800)$

$- 400 = - 500 + \frac{1}{2} Δ H_{b o n d}^{\circ}$

$\frac{1}{2} Δ H_{b o n d}^{\circ} = 100 \Rightarrow Δ H_{b o n d}^{\circ} = 200$ kJ mol $^{- 1}$ .

Concept behind this question

Hess’s Law and Bond EnergyNotes, formulas and examples →

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 23 Shift 2, Chemistry Q24: Hess’s Law and Bond Energy

Concept behind this question

More previous year questions on Hess’s Law and Bond Energy

Practice more Chemistry