Question

2. Consider the following phase transition: C (graphite) - C(diamond). (a) (6 points) The temperature dependence of reaction Gibbs energy. AGA, for the above reaction is given by AG (mol) = 1895 +3.3637 at 1 atm. Calculate AH, and AS, at 298.15K and 1 atm. (b) [9 points) The density of diamond is 3.52 g/ml and that of graphite is 2.25 g/ml. The formation free energy (AG) of diamond is 2.90 kJ/mol and that of graphite is zero at 298.15 K. Find the pressure at which diamond and graphite coexist in equilibrium at 298.15 K.

Answer 1

Part (a).

According to the Gibbs-Helmholtz equation:

$\Delta$G = $\Delta$H - T.$\Delta$S

Compare the above equation with the given equation in the problem.

Then $\Delta$H = 1895 J/mol

And $\Delta$S = -3.363 J/mol.K

Part (b).

$\Delta$G^o_reaction = P.$\Delta$V

i.e. (2.9 - 0) kJ/mol = P.{12 g/(2.25 g/mL) - 12 g/(3.52 g/mL)}

i.e. 2900 J = P * 1.924 mL

i.e. (2900/101.325) L.atm = P * 1.924*10^-3 L

Therefore, the required pressure (P) = 14874 atm or 15071 bar

Note: 1 mole of C = 12 g and density = mass/volume, i.e. volume = mass/density