Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
Δ G° f [C(graphite)] = 0 kJ mol-1 Δ G° f [C(diamond)] = 2.9 kJ mol-1The 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 x 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 : (1J = 1 kgm2
S – 2, 1 Pa = 1 Kgm –1S–2 , 1bar = 105
Pa)

Answer 1

Explanation:

C

graphite

→C

diamand

ΔG

r

o

=ΔG

diamond

o

−ΔG

graphite

o

2.9KJmol

−1

−0KJmol

−1

=2.9KJmol

−1

ΔG

r

o

=ΔH

r

o

−TΔS

r

o

ΔS

r

o

≈0

TΔS

r

o

≈0

ΔH

r

o

=ΔE

r

o

+Δ(PV)

Δ

r

o

=0 because of isothermal process ΔT=0

ΔH

r

o

=2.9×10

3

=Δ(PV)

At equilibrium pressure is constant

2.9×10

3

Jmol

−1

=P(V

2

−V

1

)

P=

2×10

−6

m

3

mol

−1

2.9×10

3

Jmol

−1

=1450×10

6

pascal

=14500bar

Total pressure = equilibrium pressure + initial pressure

=14501bar

Hence, the correct option is A.