- #1

cheme2019

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## Homework Statement

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?

## Homework Equations

ln(a

_{i})= ((molar V)/RT) * (P-1)

ΔG = -RT * ln(K

_{p}

K

_{p}= a

_{aragonite}/a

_{calcite}

## 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

V

_{ara}= 34.157 cm^3/mol and

V

_{cal}= 36.930 cm^3/ mol

From here I used the activity equation to get that

a

_{i}= e^(((molar V)/RT) * (P-1)))

And since

K

_{p}= a

_{aragonite}/a

_{calcite}and

ΔG = -RT * ln(K

_{p}

I can write that

ΔG = -RT * ln(a

_{aragonite}/a

_{calcite})

This leads to

ΔG = -RT * (((molar V

_{ara}/RT) * (P-1) - ((molar V

_{cal}/RT)*(P-1))

Or

ΔG = - molar V

_{ara}*(P-1)+(molar V

_{cal}*(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!