Question

The bond dissociation enthalpy of $X_2 \Delta H_{bond}$ calculated from the given data is _______ $kJmol^{-1}$. (Nearest integer) $M^+X^-(s) \longrightarrow M^+(g) + X^-(g) \Delta H^*_{lattice} = 800 \, kJ \, mol^{-1}$ $M(s) \longrightarrow M(g) \Delta H_{sub}^{\circ} = 100 \, kJ \, mol^{-1}$ $M(g) \longrightarrow M^+(g) + e^-(g) \Delta H_{i} = 500 \, kJ \, mol^{-1}$ $X(g) + e^-(g) \longrightarrow X^-(g) \Delta H_{eg}^* = -300 \, kJ \, mol^{-1}$ $M(s) + \frac{1}{2}X_2(g) \longrightarrow M^+X^-(s) \Delta H_{f}^{\circ} = -400 \, kJ \, mol^{-1}$

[Given: $M^+X^-$ is a pure ionic compound and $X$ forms a diatomic molecule $X_2$ in gaseous state]

Answer 1

200

Answer 2

1. Write the Born‐Haber cycle for the formation of MX:

   $$\begin{aligned} \text{M(s)} &\rightarrow \text{M(g)} \quad (\Delta H_{sub} = +100)\\[8pt] \text{M(g)} &\rightarrow \text{M}^+(g)+e^- \quad (\Delta H_i = +500)\\[8pt] \frac{1}{2}\text{X}_2(g) &\rightarrow \text{X(g)} \quad \left(\Delta H = \frac{1}{2}\Delta H_{bond}\right)\\[8pt] \text{X(g)} + e^- &\rightarrow \text{X}^-(g) \quad (\Delta H^*_{eg} = -300)\\[8pt] \text{M}^+(g) + \text{X}^-(g) &\rightarrow \text{MX(s)} \quad (\Delta H^*_{lattice} = -800)\\[8pt] \end{aligned}$$

2. The overall reaction is:

   $$\text{M(s)} + \frac{1}{2}\text{X}_2(g) \rightarrow \text{MX(s)} \quad (\Delta H_f = -400)$$

3. Adding the steps:

   $$100 + 500 + \frac{1}{2}\Delta H_{bond} - 300 - 800 = -400$$

   Simplify:

   $$(100 + 500 - 300 - 800) + \frac{1}{2}\Delta H_{bond} = -400 \\ (-500) + \frac{1}{2}\Delta H_{bond} = -400$$

4. Solve for $\Delta H_{bond}$:

   $$\frac{1}{2}\Delta H_{bond} = -400 + 500 = 100 \quad \Rightarrow \quad \Delta H_{bond} = 200 \, \text{kJ/mol}$$