S1. Use information from the appendices to predict whether diamond or graphite is the stable form of carbon at 1 atm pressure. If the density of graphite is 2.26 g/cm3 and that of diamond is 3.51 g/cm3, to what value must you change the pressure to make diamond the stable form of carbon at room temperature

a) Graphite is slightly more stable than diamond at atmospheric conditions.

b) P = 14,998 atm

Explanation:

Given:

- The gibb's free energy of graphite formation ΔG_g = 0 KJ/mol

- The gibb's free energy of a diamond formation ΔG_d = 2.90 KJ/mol

- Density of graphite is p_g = 2.26 g/cm3

- Density of diamond is p_d = 3.51 g/cm3

- Molecular Mass of diamond M_d = 12 g / mol

- Molecular Mass of graphite M_g = 12 g / mol

Find:

Q) predict whether diamond or graphite is the stable form of carbon at 1 atm pressure.

Q) to what value must you change the pressure to make diamond the stable form of carbon at room temperature

Solution:

- The stability of Diamond at atmospheric conditions can be checked by looking at the Gibb's free energy of diamond at atmospheric conditions:

ΔG_d = 2.90 KJ/mol > 0

While that of graphite is:

ΔG_g = 0 KJ/mol

- Since graphite requires no extra energy at atmospheric conditions, it means its stable in its form. Whereas, diamond requires 2.90 KJ/mol amount of energy for formation. Hence, Graphite is slightly more stable allotrope of carbon.

- Finding the minimum pressure where the ΔG of formation for diamond is equal to zero. ΔG (P) _d = 0. We can use the ideal gas law to find this pressure as follows:

ΔG (P_2) _d - ΔG (P_1) _d = Δ V (P_2 - P_1)

[ (ΔG (P_2) _d - ΔG (P_1) _d) / Δ V ] = (P_2 - P_1)

P_2 = [ (ΔG (P_2) _d - ΔG (P_1) _d) / Δ V ] + P_1

- For the expression above calculate Δ V, change in Volume for diamond and stable form graphite:

Δ V = V_diamond - V_graphite

Δ V = M_diamond / p_d - M_graphite / p_g

Plug in the values:

Δ V = (12 / 10^6*3.51) - (12 / 10^6*2.26)

Δ V = - 1.92*10^-6 m^3 / mol

- Use the Δ V change in volume and compute the pressure P_2:

P_2 = [ (0 - 2.90) / (-1.92*10^-6) ] + 1 atm

P_2 = (1.51*10^9) / (1.01*10^5) + 1 atm

P_2 = 14,998 atm

- The minimum amount of pressure required for diamond to be stable at room temperature is 15,000 times the atmospheric pressure.