The standard state Gibbs free energies of formation of ) C(graphite and C(diamond) at T = 298 K are
`Delta_(f)G^(@)["C(graphite")]=0kJ mol^(-1)`
`Delta_(f)G^(@)["C(diamond")]=2.9kJ 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 [ ) C(graphite ] to diamond [C(diamond)] reduces its volume by `2xx10^(-6)m^(3) mol^(-1).` If ) C(graphite is converted to C(diamond) isothermally at T = 298 K, the pressure at which ) C(graphite is in equilibrium with C(diamond), is
`["Useful information:"1J=1kg m^(2)s^(-2),1Pa=1kgm^(-1)s^(-2),1"bar"=10^(5)Pa]`

Options

A. 14501 bar
B. 29001 bar
C. 1450 bar
D. 58001 bar

Explanation

To solve the problem, we need to find the pressure at which carbon graphite is in equilibrium with carbon diamond during the isothermal conversion at 298 K. We will use the Gibbs free energy change and the volume change associated with the conversion.

### Step-by-step Solution:

1. **Identify the Gibbs Free Energy Change (ΔG)**:
- The standard Gibbs free energy of formation for C(graphite) is given as:
\[ \Delta_f G^\circ [C(graphite)] = 0 \, \text{kJ/mol} \]
- The standard Gibbs free energy of formation for C(diamond) is given as:
\[ \Delta_f G^\circ [C(diamond)] = 2.9 \, \text{kJ/mol} \]
- Therefore, the change in Gibbs free energy (ΔG) for the conversion from graphite to diamond is:
\[ \Delta G = \Delta_f G^\circ [C(diamond)] - \Delta_f G^\circ [C(graphite)] = 2.9 \, \text{kJ/mol} - 0 \, \text{kJ/mol} = 2.9 \, \text{kJ/mol} \]

2. **Convert ΔG from kJ to J**:
- Since 1 kJ = 1000 J, we convert ΔG:
\[ \Delta G = 2.9 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 2900 \, \text{J/mol} \]

3. **Identify the Volume Change (ΔV)**:
- The volume change during the conversion is given as:
\[ \Delta V = -2 \times 10^{-6} \, \text{m}^3/\text{mol} \]

4. **Use the Gibbs Free Energy Equation**:
- For an isothermal process, the relationship between ΔG, pressure (P), and volume change (ΔV) is given by:
\[ \Delta G = P \Delta V \]
- Rearranging for pressure gives:
\[ P = \frac{\Delta G}{\Delta V} \]

5. **Substitute the values into the equation**:
- Substitute ΔG and ΔV into the equation:
\[ P = \frac{2900 \, \text{J/mol}}{-2 \times 10^{-6} \, \text{m}^3/\text{mol}} = \frac{2900}{-2 \times 10^{-6}} \]

6. **Calculate the pressure**:
- Performing the calculation:
\[ P = -1.45 \times 10^9 \, \text{Pa} \]
- Since pressure cannot be negative in this context, we take the absolute value:
\[ P = 1.45 \times 10^9 \, \text{Pa} \]

7. **Convert Pressure from Pascals to Bar**:
- To convert from Pascals to Bar, use the conversion factor \(1 \, \text{bar} = 10^5 \, \text{Pa}\):
\[ P = \frac{1.45 \times 10^9 \, \text{Pa}}{10^5 \, \text{Pa/bar}} = 14500 \, \text{bar} \]

### Final Answer:
The pressure at which C(graphite) is in equilibrium with C(diamond) is approximately **14500 bar**.