The standard state Gibb's free energies of formation of  $\mathrm{C}$ (graphite) and $\mathrm{C}$ (diamond) at $\mathrm{T}=298\mathrm{ }\mathrm{K}$ are
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\mathrm{o}}\left[\mathrm{C} (\mathrm{graphite}\right]=0\mathrm{ }\mathrm{kJ}\mathrm{ }{\mathrm{mol}}^{-1}$
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\mathrm{o}}\left[\mathrm{C} \left(\mathrm{diamond}\right)\right]=2.9\mathrm{ }\mathrm{kJ}\mathrm{ }{\mathrm{mol}}^{-1}$
The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [$\mathrm{C}$ (graphite)] to diamond [$\mathrm{C}$ (diamond)] reduces its volume by $2\times {10}^{-6}\mathrm{ }{\mathrm{m}}^3 {\mathrm{mol}}^{-1}$ . If $\mathrm{C}$ (graphite) is converted to $\mathrm{C}$ (diamond) isothermally at $\mathrm{T}=298 \mathrm{K}$, the pressure at which $\mathrm{C}$ (graphite) is in equilibrium with $\mathrm{C}$ (diamond), is
[Useful information: $1\mathrm{ }\mathrm{J}=1\mathrm{ }\mathrm{kg}\mathrm{ }{\mathrm{m}}^2 {\mathrm{s}}^{-2},\mathrm{ }1\mathrm{ }\mathrm{Pa}=1 \mathrm{kg}\mathrm{ }{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}; 1\mathrm{bar}={10}^5\mathrm{Pa}$]

• Correct answer
  A

  $14501$ bar
• B

  $29001$ bar
• C

  $58001 \mathrm{bar}$
• D

  $1450 \mathrm{bar}$

Hence, the correct answer is option: 'A'.