Step-by-step Derivation:

1. Identify the chemical process and relevant thermodynamic data. The problem describes the phase transition of carbon from graphite to diamond: $C(\text{graphite}) \rightleftharpoons C(\text{diamond})$ The given data at $T=298K$ are:

   • Standard Gibbs free energy of formation of graphite: ${\Delta _f}{G^0}$ [$C$(graphite)] $= 0\,kJ\,mo{l^{ - 1}}$
   • Standard Gibbs free energy of formation of diamond: ${\Delta _f}{G^0}$ [$C$(diamond)] $= 2.9\,kJ\,mo{l^{ - 1}}$
   • Change in volume for the conversion: $\Delta V = V_{\text{diamond}} - V_{\text{graphite}} = -2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}$
   • The standard state pressure is $P^0 = 1$ bar.

2. Calculate the standard Gibbs free energy change for the reaction (${\Delta _r}{G^0}$). The standard Gibbs free energy change for the reaction is the difference between the standard Gibbs free energies of formation of the products and reactants. ${\Delta _r}{G^0} = {\Delta _f}{G^0}[C(\text{diamond})] - {\Delta _f}{G^0}[C(\text{graphite})]$ ${\Delta _r}{G^0} = 2.9\,kJ\,mo{l^{ - 1}} - 0\,kJ\,mo{l^{ - 1}} = 2.9\,kJ\,mo{l^{ - 1}}$ For consistency in units, we convert this to Joules per mole: ${\Delta _r}{G^0} = 2.9 \times 1000\,J\,mo{l^{ - 1}} = 2900\,J\,mo{l^{ - 1}}$

3. Relate Gibbs free energy to pressure. The change in Gibbs free energy ($dG$) with a change in pressure ($dP$) at constant temperature is given by the fundamental thermodynamic relation: $dG = VdP$ For a chemical reaction, this becomes: $d(\Delta_r G) = (\Delta_r V)dP$ where $\Delta_r V$ is the change in volume for the reaction. Assuming $\Delta_r V$ is constant with pressure (a valid assumption for solids), we can integrate this equation from the standard state (pressure $P^0$) to the equilibrium state (pressure $P$): $\int_{{\Delta _r}{G^0}}^{{\Delta _r}G} d(\Delta_r G) = \int_{P^0}^{P} (\Delta_r V)dP$ ${\Delta _r}G - {\Delta _r}{G^0} = (\Delta_r V)(P - P^0)$ This equation gives the Gibbs free energy change for the reaction at any pressure $P$.

4. Apply the equilibrium condition. At equilibrium, the Gibbs free energy change for the reaction is zero, i.e., ${\Delta _r}G = 0$. Substituting this into the equation from Step 3: $0 - {\Delta _r}{G^0} = (\Delta_r V)(P - P^0)$ $-{\Delta _r}{G^0} = (\Delta_r V)(P - P^0)$

5. Solve for the equilibrium pressure ($P$). Rearranging the equation to solve for the equilibrium pressure $P$: $P - P^0 = -\frac{{{\Delta _r}{G^0}}}{{\Delta_r V}}$ $P = P^0 - \frac{{{\Delta _r}{G^0}}}{{\Delta_r V}}$ Now, we substitute the known values:

   • ${\Delta _r}{G^0} = 2900\,J\,mo{l^{ - 1}}$
   • $\Delta_r V = -2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}$
   • $P^0 = 1\,bar = 10^5\,Pa$

   $P - P^0 = -\frac{2900\,J\,mo{l^{ - 1}}}{-2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}} = \frac{2900}{2 \times 10^{-6}} \frac{J}{m^3}$ $P - P^0 = 1450 \times 10^6 \,J/m^3 = 1.45 \times 10^9 \,J/m^3$ Since $1\,Pa = 1\,N/m^2 = 1\,J/m^3$, the pressure difference is: $P - P^0 = 1.45 \times 10^9\,Pa$ To express this in bars ($1\,bar = 10^5\,Pa$): $P - P^0 = \frac{1.45 \times 10^9\,Pa}{10^5\,Pa/bar} = 1.45 \times 10^4\,bar = 14500\,bar$ Finally, we find the equilibrium pressure $P$: $P = P^0 + 14500\,bar = 1\,bar + 14500\,bar = 14501\,bar$

6. Conclusion The pressure at which graphite is in equilibrium with diamond is $14501$ bar. This corresponds to option A.