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Equilibrium pressure of calcium carbonate

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Calcium carbonate primarily occurs as two crystalline forms, calcite and aragonite. The value of∆!° for the transition
    CaCO3(calcite) ⇌ CaCO3(aragonite)
    is +1.04 kJ·mol-1 at 25°C. At that temperature the density for calcite is 2.710 g·cm-3, and that of aragonite is 2.930 g·cm-3. At what pressure will the two crystalline phases be at equilibrium at 25°C?


    2. Relevant equations
    ln(ai)= ((molar V)/RT) * (P-1)
    ΔG = -RT * ln(Kp
    Kp = aaragonite/acalcite
    3. The attempt at a solution
    So I started by finding the molar volume of each by dividing the MW by the individual density and got
    Vara = 34.157 cm^3/mol and
    Vcal = 36.930 cm^3/ mol
    From here I used the activity equation to get that

    ai = e^(((molar V)/RT) * (P-1)))

    And since

    Kp = aaragonite/acalcite and
    ΔG = -RT * ln(Kp

    I can write that

    ΔG = -RT * ln(aaragonite/acalcite)

    This leads to

    ΔG = -RT * (((molar Vara/RT) * (P-1) - ((molar Vcal/RT)*(P-1))
    Or
    ΔG = - molar Vara*(P-1)+(molar Vcal*(P-1))

    Continuing
    1040 J/mol =(-34.157 cm^3/mol ) * P + 34.157 cm^3/mol +(36.930 cm^3/mol) * P - 39.630 cm^3/mol

    Solving for P I get 376 J/cm^3.
    I am given that the answer is 3850.9 but no units and I wasn't able to convert my units to get that number...
    I'm pretty sure my math is correct but I think I messed up on units somewhere. Any help would be greatly appreciated!! Thanks!
     
  2. jcsd
  3. Dec 4, 2016 #2

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    Step one: pick CGS or MKS; do not attempt to mix them when you're not comfortable with either; i.e., densities are expressed as kg/m3 in MKS, and as g/cm3 in CGS; pressures in force per unit area, not energy per unit volume ....
     
  4. Dec 4, 2016 #3
    But can't you use energy per unit volume for pressure? Similar to how atm is J/L. Plus I'm only given ΔG in kJ/mol and I don't know how to convert that into CGS
     
  5. Dec 5, 2016 #4
    Atm is not J/L; 1 J/L = 1000 Pa; 1 atm = 101325 Pa
     
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