The shells of marine organisms contain calcium carbonate CaCO3, largely in a crystalline form known as calcite. There is a second crystalline form of calcium carbonate known as aragonite. Physical and thermodynamic properties of calcite and aragonite are given below.

Properties
(T=298K, P=1atm) Calcite Aragonite
ΔH°f (kJ/mole) -1206.87 -1207.04
ΔG°f (kJ/mole) -1128.76 -1127.71
S° (J/mole K) 92.88 88.70
C°P (J/mole K) 81.88 81.25
Density (gm/mL) 2.710 2.930
A) Based on the thermodynamic data given, would you expect an isolated sample of
calcite at T=298K and P=1 atm to convert to aragonite, given sufficient time.
Explain.

B) What pressure must be achieved to induce the conversion of calcite to aragonite at
T=298K. Assume both calcite and aragonite are incompressible at T=298K.

C) Suppose the pressure is P=1.00 atm and the temperature is increased from
T=298K to T=400K. Based on this calculation, can isolated calcite be converted
to aragonite at P=1 atm if the temperature is increased? Explain.

D) Suppose the pressure is increased from P=1.00 atm to P=1000 atm and the
temperature is increased from T=298K to T=400K. Which form of CaCO3 is more
stable under these conditions?

Expert Answer

Check the explanation

Explanation:

A. Calcite > Argonite

At 298 K and 1 atm,

\DeltaGrxn = \DeltaGarg - \DeltaGcalc

= -1127.71-(-1128.76) = 1.05 kJ/mol

\DeltaGrxn is positive, so the reaction is not spontaneous.

Thus, the reaction would not happen even if enough time is given.

B. Molar mass of CaCO3 = 100 g/mol

Volume = mass/density

Vcalc = 100/2.710 = 36.900 ml/mol = 36.9*10-3 L = 36.9*10-6 m3

Varg = 100/2.930 = 34.130 ml/mol = 34.13*10-3 L = 34.13*10-6 m3

\DeltaGrxn = 1.05 kJ/mol = 1.05*103 J/mol

\DeltaG = -\DeltaP\DeltaV

\DeltaP = -\DeltaG/\DeltaV = -1.05*103/(34.13*10-6 - 36.9*10-6) = 0.379*109 Pa = 3.79*108 Pa

The pressure should be more than 3.79*108 Pa.

C. For reaction, \DeltaSrxn = Sarg - Scalc

= 88.70-92.88 = -4.18 J/mol.K

\DeltaHrxn = -1207.04-(-1206.87) = -0.17kJ/mol

At 400 K, \DeltaG = \DeltaH-T\DeltaS

= -0.17-(400*-4.18)

= +ve

So, the reaction is not favorable on increasing the temperature to 400 K.

D. \DeltaT = 400-298 = 102 K

\DeltaP = 1000-1 = 999.0 atm = 999*106 Pa

\DeltaG = \DeltaS\DeltaT - \DeltaV\DeltaP

= (-4.18*102)-(-2.77*10-6*999*106)

= -426.36+2767.23 = 2340.87 J/mol

\DeltaG = +ve

calcite would be more stable.

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explanation:

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