Q. The bond dissociation enthalpy of $\mathrm{X_2}$ $\Delta H_{bond}$ calculated from the given data is ______ $\mathrm{kJ\,mol^{-1}}$. (Nearest integer) $M^+X^-(s)\to M^+(g)+X^-(g)$, $\Delta H^\circ_{lattice}=800$ kJ/mol; $M(s)\to M(g)$, $\Delta H^\circ_{sub}=100$ kJ/mol; $M(g)\to M^+(g)+e^-(g)$, $\Delta H_i=500$ kJ/mol; $X(g)+e^-(g)\to X^-(g)$, $\Delta H^\circ_{eg}=-300$ kJ/mol; $M(s)+\frac12X_2(g)\to M^+X^-(s)$, $\Delta H^\circ_f=-400$ kJ/mol [Given : $M^+X^-$ is a pure ionic compound and X forms a diatomic molecule $X_2$ in gaseous state]

Numerical Question

This is a numerical-answer question. The correct answer is shown below.

Explanation

[Solution by Dr. Abubacker] Born–Haber cycle: $\Delta H_f=\Delta H_{sub}+\Delta H_i+\frac12\Delta H_{bond}+\Delta H_{eg}-\Delta H_{lattice}$. $-400=100+500+\frac12\Delta H_{bond}-300-800\Rightarrow\frac12\Delta H_{bond}=100$, so $\Delta H_{bond}=200$ kJ/mol. Answer: 200.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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