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NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Hard
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Application +4 | -1
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below
12Cl2(g)→12ΔdissH0Cl(g)→ΔEAH0Cl−(g)→ΔhydH0 Cl−(aq)\frac{1}{2}C{l_2}\left( \mathrm{g} \right)\xrightarrow[{}]{{^{\frac{1}{2}\Delta }dis{s^{{H^0}}}}}Cl\left( \mathrm{g} \right)\xrightarrow[{}]{{^\Delta E{A^{{H^0}}}}}C{l^{^ - }}\left( \mathrm{g} \right)\xrightarrow[{}]{{^\Delta hy{d^{{H^0}}}}}\,C{l^ - }\left( {aq} \right)
The energy involved in the conversion of 12Cl2(g)\frac{1}{2}\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) to Cl−(aq)\mathrm{C}{\mathrm{l}^ - }\left( {aq} \right) (Using the data)
ΔdissH∘Cl2=240  kJ  mol−1ΔEAH∘Cl=−349  kJmol−1ΔhydH∘Cl=−381  kJ  mol−1\begin{array}{l} \Delta _{\mathrm{dis}{\mathrm{s}^{H^\circ }}\mathrm{C}{\mathrm{l}_2}}^{} = 240\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}\\ \Delta _{\mathrm{E}{\mathrm{A}^{H^\circ }}\mathrm{C}{\mathrm{l}_{}}}^{} = - 349\,\,\mathrm{kJ mo}{\mathrm{l}^{ - 1}}\\ \Delta _{\mathrm{hy}{\mathrm{d}^{H^\circ }}}^{}\mathrm{Cl} = - 381\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}} \end{array} will be
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Correct option: B
−610  kJ  mol−1- 610\,\,\mathrm{kJ}\,\,\mathrm{mo}{\mathrm{l}^{ - 1}}
Option B (−610  kJ mol−1- 610\,\,\mathrm{kJ}\,\mathrm{mo}{\mathrm{l}^{ - 1}}) is correct because it represents the sum of the enthalpy changes for the given steps.

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