The standard sate Gibbs free energies of formation of $$C$$ (graphite) and $$C$$ (diamond) at $$T=298 K$$ are
$$\eqalign{
& {\Delta _f}{G^0}\left[ {C\left( {graphite} \right)} \right] = 0\,kJ\,mo{l^{ - 1}} \cr
& {\Delta _f}{G^0}\left[ {C\left( {diamond} \right)} \right] = 2.9\,kJ\,mo{l^{ - 1}} \cr} $$
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [ $$C$$ (graphite) ] to diamond [ $$C$$ (diamond) ] reduces its volume by $$2 \times {10^{ - 6}}{m^3}mo{l^{ - 1}}.$$ If $$C$$ (graphite) is converted to $$C$$ (diamond) isothermally at $$T=298K,$$ the pressure at which $$C$$ (graphite) is in equilibrium with $$C$$ (diamond), is
[ Useful information: $${1J = 1kg\,{m^2}{s^{ - 2}};}$$ $${1Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};}$$ $${1\,bar = {{10}^5}Pa}$$ ]
Releted MCQ Question on Physical Chemistry >> Chemical Thermodynamics
Releted Question 1
The difference between heats of reaction at constant pressure and constant volume for the reaction : $$2{C_6}{H_6}\left( l \right) + 15{O_{2\left( g \right)}} \to $$ $$12C{O_2}\left( g \right) + 6{H_2}O\left( l \right)$$ at $${25^ \circ }C$$ in $$kJ$$ is
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$