The internal energy change in the conversion of `1.0` mole of the calcite form of `CaCO_3` to the aragonite form is `+0.21 KJ`. Calculate the enthalpy change when the pressure is `1.0 `bar, given the densities of the solids are `2.71 gcm^(-3)` and `2.93 gcm^(-3)` respectively.

Explanation

`DeltaH=DeltaU+PDeltaV`
Given, `DeltaU=+0.21 kJ mol^(-1)`
`=0.21xx10^(3)H mol^(-1)`
`P=1"bar"=1.0xx10^(5) Pa`
`DeltaV=V_(("aragonite"))-V_(("calcite"))`
(mol. Wt. of `CaCO_(3)=100`)
`=(100/2.93-100/2.71)cm^(3) mol^(-1) of CaCO_(3)`
`=-2.77 cm^(3)=-2.77xx10^(-6) m^(3)`
`DeltaH=DeltaU+PDeltaV`
`DeltaH=0.21xx10^(3)-1xx10^(5)xx2.77xx10^(-6)`
`=209.72J mol^(-1)`
`=0.20972 kJ mol^(-1)`