Question

The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at $T =298 K$ are
$\Delta_{f} G^{0}[ C (\text { graphite })]=0\, kJ\,mol ^{-1}$
$\Delta_{f} G^{0}[ C (\text { diamond })]=2.9\, kJ\,mol ^{-1}$
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 $2 \times 10^{-6} m ^{3} mol ^{-1}$. 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:
$\left.1 J =1\, kg\, m ^{2} s ^{-2} ; 1\, Pa =1\, kg\, m ^{-1} s ^{-2} ; 1 \,bar =10^{5} Pa \right]$

• A. 14501 bar
• B. 58001 bar
• C. 1450 bar
• D. 29001 bar

Answer 1

Answer: (A)

$\Delta G = PdV$
$\left[\Delta_{ f } G ^{\circ}\right.$ (diamond) $-\Delta_{f} G ^{\circ}$ (graphite) $]= PdV$
$2.9 \times 10^{3} J mol ^{-1}= P \times 2 \times 10^{-6} m ^{3} mol ^{-1}$
$P =1.45 \times 10^{9} Pa$
$P =1.45 \times 10^{9} \times 10^{-5} bar$
$P =1.45 \times 10^{4} bar$
$P =14500\, bar$