The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
Δ f Gº[C(graphite)] = 0 kJ mol –1
Δ f Gº[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 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: 1 J = 1 kg m 2 s –2 , 1 Pa = 1 kg m –1 s –2 ; 1 bar = 10 5 Pa]
Text Solution
Verified by ExpertsC
G = VdP – SdT
At 298 K, SdT = 0
∴ dG = VdP
∴ G – G° = V (P–1) [ Solids involved ∴ V almost constant]
∴ Δ r G = [G° diamond + V d (P – 1)] – [G° graphite + V g (P – 1)]
0 = 2.9 × 10 3 + (P – 1)10 5 (–2 × 10 –6 )
∴ P = 14501 bar
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems