To solve the problem of calculating the equilibrium pressure at which graphite is converted to diamond at \(25^\circ C\), we will follow these steps:
### Step 1: Calculate the Molar Volume of Graphite
Given the density of graphite, \(\rho_{\text{graphite}} = 2 \, \text{g/cm}^3\), we can find the molar volume (\(V_g\)) using the formula:
\[
V_g = \frac{\text{Molar mass}}{\text{Density}} = \frac{12 \, \text{g/mol}}{2 \, \text{g/cm}^3} = 6 \, \text{cm}^3/\text{mol}
\]
### Step 2: Calculate the Molar Volume of Diamond
Similarly, for diamond, given the density \(\rho_{\text{diamond}} = 3 \, \text{g/cm}^3\):
\[
V_d = \frac{\text{Molar mass}}{\text{Density}} = \frac{12 \, \text{g/mol}}{3 \, \text{g/cm}^3} = 4 \, \text{cm}^3/\text{mol}
\]
### Step 3: Calculate the Change in Molar Volume
Now, we can calculate the change in molar volume (\(\Delta V\)) when graphite is converted to diamond:
\[
\Delta V = V_g - V_d = 6 \, \text{cm}^3/\text{mol} - 4 \, \text{cm}^3/\text{mol} = 2 \, \text{cm}^3/\text{mol}
\]
### Step 4: Convert \(\Delta G_f\) to Appropriate Units
Given \(\Delta G_f^\circ\) for diamonds from graphite is \(3 \, \text{kJ/mol}\), we convert this to liters-atmosphere:
\[
\Delta G_f^\circ = 3 \, \text{kJ/mol} = 3000 \, \text{J/mol} = 3000 \, \text{J/mol} \times 0.00987 \, \text{L atm/J} = 29.61 \, \text{L atm/mol}
\]
### Step 5: Use the Gibbs Free Energy Equation
We can use the relationship between Gibbs free energy, change in volume, and pressure:
\[
\Delta G_f^\circ = \Delta V \cdot (P_2 - P_1)
\]
Where \(P_1\) is the atmospheric pressure (1 atm). Rearranging gives us:
\[
P_2 = \frac{\Delta G_f^\circ}{\Delta V} + P_1
\]
### Step 6: Substitute Values to Find \(P_2\)
Substituting the values we have:
\[
P_2 = \frac{29.61 \, \text{L atm/mol}}{2 \, \text{cm}^3/\text{mol} \times 0.001 \, \text{L/cm}^3} + 1 \, \text{atm}
\]
Calculating this gives:
\[
P_2 = \frac{29.61}{0.002} + 1 \approx 14805 \, \text{atm}
\]
### Final Answer
Thus, the equilibrium pressure at which graphite is converted to diamond at \(25^\circ C\) is approximately:
\[
\boxed{14805 \, \text{atm}}
\]
