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Q.

The internal energy change in the conversion of 1.0 mol of the calcite form of CaCO3 to the aragonite form is + 0.21 kJ. Calculate the enthalpy changewhen the pressure is 1.0 bar. Given that the densities of the solids are 2.71 g cm-3 and 2.93 g cm-3 respectively.

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answer is 0.20972 KJ.

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Detailed Solution

ΔH=ΔE+PΔV Given ΔE=+0.21kJmole−1=0.21×103mole−1P=1bar=1×105PaΔV=V(aragonite) −V(Calcite) =1002.93−1002.71cm3mol−1 of CaCO3=−2.77cm3=−2.77×10−6m3ΔH=0.21×103−1×105×2.77×10−6=209.72J=0.20972kJmol−1

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