JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
The bond dissociation enthalpy of \(\mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}\) calculated from the given data is _______ \(\mathrm{kJ} \mathrm{mol}^{-1}\). (Nearest integer)
\(\begin{aligned}
& \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^*=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\)
\(\mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
\(\mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{eg}}^*=-300 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
\(\mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}_f^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
[Given : \(\mathrm{M}^{+} \mathrm{X}^{-}\)is a pure ionic compound and X forms a diatomic molecule \(\mathrm{X}_2\) in gaseous state]
- A 250
- B 100
- C 200
- D 105
Answer & Solution
Correct Answer
(C) 200
Step-by-step Solution
Detailed explanation
\(\therefore \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{MX})= \) \( \Delta \mathrm{H}_{\text {sub }}(\mathrm{M})+\text { I.E. }(\mathrm{M})+\frac{1}{2}[\text { B.E. }(\mathrm{X}-\mathrm{X})] +\mathrm{EG}(\mathrm{X})+\text { L.E. }(\mathrm{MX})\)…
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