The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm⁻³, respectively. If the standard free en

Question

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm⁻³, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol⁻¹, the pressure at which graphite will be transformed into diamond at 298 K is:

Accepted Answer

At constant temperature, the change in Gibbs free energy with pressure is given by the relation: dG = VdP.
For a macroscopic change, ΔG_P - ΔG_1 = ΔV(P - 1).
At equilibrium, the free energy of graphite and diamond must be equal, so ΔG_P = 0.
Given, ΔGº = 1895 J mol⁻¹.
The molar volume of graphite, V_graphite = Mass / Density = 12 / 2.25 = 5.333 cm³ mol⁻¹.
The molar volume of diamond, V_diamond = Mass / Density = 12 / 3.31 = 3.625 cm³ mol⁻¹.
Change in molar volume, ΔV = V_diamond - V_graphite = 3.625 - 5.333 = -1.708 cm³ mol⁻¹ = -1.708 × 10⁻⁶ m³ mol⁻¹.
Now, substituting the values into the equation:
0 - 1895 = (-1.708 × 10⁻⁶) × (P - 10⁵)
P - 10⁵ = -1895 / (-1.708 × 10⁻⁶) = 1109.48 × 10⁶ Pa = 11.09 × 10⁸ Pa.
Since 10⁵ Pa is negligible compared to 11.09 × 10⁸ Pa, the required pressure P ≈ 11.08 × 10⁸ Pa.

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