The internal enegry change in the conversion of `1.0 mol` of the calcite form of `CaCO_(3)` to the aragonite from is `+0.2kJ`. Calculate the enthalpy change when the pressure is `10bar`, given that the densities of the solids are `2.71` and `2.93 gcm^(-3)`, respectively.

Explanation

`DeltaH = DeltaU +P DeltaV`
Given, `DeltaU = +0.21 kJ mol^(-1) = 0.21 xx 10^(3) mol^(-1)`
`P = 1 bar = 1.0 xx 10^(5) Pa`
`DeltaV = V_(("aragonite")) - V_(("Calcite"))`
`= ((100)/(2.93)-(100)/(2.71)) cm^(3) mol^(-1)of CaCO_(3)`
`=- 2.77 cm^(3) =- 2.77 xx 10^(-6) m^(3)`
`:.DeltaH = 0.21 xx 10^(3) - 1 xx 10^(5) xx 2.77 xx 10^(-6)`
`=- 209.72J = 0.20972 kJ mol^(-1)`