Respuesta :
Answer:
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.
