${C_{{\rm{graphite}}}} + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right);$ ∆H= -94.05 kcal ${\rm{mo}}{{\rm{l}}^{ - 1}}$ ${C_{{\rm{diamond}}}} + {{\rm{O}}_2}\left( {\rm{g}} \right) \to {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right);$ ∆H= -94.05 kcal ${\rm{mo}}{{\rm{l}}^{ - 1}}$ therfore :

${C_{{\rm{diamond}}}} \to {C_{{\rm{graphite}}}};\Delta H_{298{\rm{K}}}^0 = + 450{\rm{cal mo}}{{\rm{l}}^{ - 1}}$

${C_{{\rm{graphite}}}} \to {C_{{\rm{diamond}}}};\Delta H_{298{\rm{K}}}^0 = - 450{\rm{cal mo}}{{\rm{l}}^{ - 1}}$

Diamond is harder than graphite

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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Analyzing +4 | -1

{C_{\mathrm{graphite}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);

∆H= -94.05 kcal

\mathrm{mo}{\mathrm{l}^{ - 1}}

{C_{\mathrm{diamond}}} + {\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);

∆H= -94.05 kcal

\mathrm{mo}{\mathrm{l}^{ - 1}}

therfore :

View answer and explanation

Correct option: D

Graphite is the stabler allotrope

The correct answer is Graphite is the stabler allotrope. Since the combustion of graphite releases slightly more heat than the combustion of diamond, graphite is at a lower energy state and therefore more stable.

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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