The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
fG° [C(graphite)] = 0 kJ mol–1
fG° [C(diamond)] = 2.9 kJ 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 2 × 10–6 m3 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 : 1 J = 1 kg m2 s–2; 1 Pa = 1 kg m–1 s–2; 1 bar = 105 Pa]
2.9 × 103 J mol-1 = P × 2 × 10-6 m3 mol-1
P = 1.45 × 109 Pa
P = 1.45 × 109 × 10-5 bar
P = 1.45 × 104 bar
P = 14500 bar
This question involves determining the equilibrium pressure for the phase transition from graphite to diamond at 298 K using thermodynamic principles. The key concept is that at equilibrium, the Gibbs free energy change (ΔG) for the reaction is zero. We are given the standard Gibbs free energies of formation and the volume change.
Step 1: Write the reaction and the condition for equilibrium
The reaction is: C(graphite) → C(diamond)
At equilibrium, ΔG = 0 for the conversion.
Step 2: Express ΔG as a function of pressure
The Gibbs free energy change at any pressure P is related to the standard Gibbs free energy change (ΔG°) by the equation:
Since ΔV (the change in volume) is constant (given as -2 × 10-6 m³ mol⁻¹), the integral simplifies to ΔV × (P - P°), where P° is the standard pressure (1 bar). Thus:
Step 3: Apply the equilibrium condition
At equilibrium, ΔG = 0:
Rearranging for P:
Step 4: Substitute the given values
ΔG° for the reaction C(graphite) → C(diamond) is ΔG°_f(diamond) - ΔG°_f(graphite) = 2.9 kJ mol⁻¹ - 0 kJ mol⁻¹ = 2.9 kJ mol⁻¹ = 2900 J mol⁻¹ (since 1 kJ = 1000 J).
ΔV = -2 × 10⁻⁶ m³ mol⁻¹ (negative because volume decreases).
P° = 1 bar.
Substitute into the equation:
Since 1450000000 is much larger than 1, P ≈ 1450000000 bar? But wait, we must ensure unit consistency. The volume change is in m³ mol⁻¹, and energy is in J mol⁻¹ (where J = kg m² s⁻²), and pressure should be in bar. However, the useful information provides conversions: 1 Pa = 1 kg m⁻¹ s⁻², and 1 bar = 10⁵ Pa. So, we should calculate pressure in Pa first and then convert to bar.
Step 5: Calculate in SI units and convert to bar
From earlier:
ΔG° = 2900 J mol⁻¹
ΔV = -2 × 10⁻⁶ m³ mol⁻¹
So,
Since P° = 1 bar = 10⁵ Pa, we have:
Now convert to bar (since 1 bar = 10⁵ Pa):
Since P° is 1 bar, which is negligible compared to 14500 bar, the equilibrium pressure is approximately 14500 bar. Looking at the options, 14501 bar is the closest (likely due to rounding).
Final Answer: 14501 bar
Gibbs Free Energy and Pressure Dependence: The Gibbs free energy of a substance depends on pressure. For a process, ΔG = ΔG° + ∫ΔV dP. If ΔV is constant, ΔG = ΔG° + ΔV(P - P°).
Phase Equilibrium: At equilibrium between two phases, the Gibbs free energy change is zero. This principle is used to find the pressure or temperature at which two phases coexist.
Unit Conversion: Careful attention to units (J, Pa, bar) is crucial in thermodynamic calculations involving pressure-volume work.