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9 December, 00:05

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

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  1. 9 December, 00:26
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    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.
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