The value of Delta H_{transition} for C(text{ | vedclass

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(A) $1$. Given $\Delta H_{transition} = 1.9 \ kJ/mol$ for $C(	ext{graphite}) ightarrow C(	ext{diamond})$. Since $\Delta H > 0$,the process is endo

Vedclass · Accepted Answer

$C(	ext{diamond})$ is thermodynamically more stable than $C(	ext{graphite})$ at $25^{\circ}C$

Vedclass · Answer

(A) $1$. Given $\Delta H_{transition} = 1.9 \ kJ/mol$ for $C(	ext{graphite}) ightarrow C(	ext{diamond})$. Since $\Delta H > 0$,the process is endothermic,meaning graphite is more stable than diamond at $25^{\circ}C$. Thus,option $A$ is incorrect.$2$. Entropy of graphite $(S_g)$ > entropy of diamond $(S_d)$,so $\Delta S = S_d - S_g $3$. For $C(	ext{diamond}) ightarrow C(	ext{graphite})$,$\Delta H = -1.9 \ kJ/mol$ and $\Delta S > 0$. Since $\Delta G = \Delta H - T\Delta S$,$\Delta G$ will be negative at all temperatures,making the conversion of diamond to graphite spontaneous. Thus,option $D$ is correct.$4$. Combustion of diamond releases more energy than graphite because diamond is at a higher energy state than graphite. Thus,option $C$ is correct.$5$. Thermal conductivity is a physical property independent of thermodynamic stability in this context,but option $A$ is clearly the incorrect statement regarding thermodynamic stability.

The value of $\Delta H_{transition}$ for $C(\text{graphite}) \rightarrow C(\text{diamond})$ is $1.9 \ kJ/mol$ at $25^{\circ}C$. The entropy of graphite is higher than the entropy of diamond. This implies that which of the following is incorrect $:-$

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