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JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry

दिए गए आँकड़ों से गणना की गई \(\mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}\) की बंध वियोजन एन्थैल्पी _______ \(\mathrm{kJ} \mathrm{mol}^{-1}\) है। (निकटतम पूर्णांक)
\(\begin{aligned}
& \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^*=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\)
\(\mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
\(\mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{eg}}^*=-300 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
\(\mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}_f^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
[दिया गया है : \(\mathrm{M}^{+} \mathrm{X}^{-}\) एक शुद्ध आयनिक यौगिक है और X गैसीय अवस्था में एक द्विपरमाणुक अणु \(\mathrm{X}_2\) बनाता है]

  1. A 250
  2. B 100
  3. C 200
  4. D 105
Verified Solution

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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