Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are 

Δ_ƒG° [C(graphite)] = 0 kJ mol^–1 

ΔƒG° [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^–6m³mol^–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 = 1kg m²s^–2; 1Pa = 1kg m^–1 s^–2; 1 bar = 10⁵Pa] 

(A) 14501bar 

(B) 58001bar 

(C) 1450bar 

(D) 29001bar

Answer 1

The correct answer is

Comments

Can you explain delta g= dp(delta v)