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.

List-$I$ Thermodynamic Process	List-$II$ Magnitude in $kJ$
$A$. Work done in reversible, isothermal expansion of $2 \ mol$ of ideal gas from $2 \ dm^3$ to $20 \ dm^3$ at $300 \ K$.	$I$. $4$
$B$. Work done in irreversible isothermal expansion of $1 \ mol$ ideal gas from $1 \ m^3$ to $3 \ m^3$ at $300 \ K$ against a constant pressure of $3 \ kPa$.	$II$. $11.5$
$C$. Change in internal energy for adiabatic expansion of a $1 \ mol$ ideal gas with change of temperature $= 320 \ K$ and $\overline{C}_V = \frac{3}{2} R$.	$III$. $6$
$D$. Change in enthalpy at constant pressure of $1 \ mole$ ideal gas with change of temperature $= 337 \ K$ and $\overline{C}_P = \frac{5}{2} R$.	$IV$. $7$

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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