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The internal enegry change in the conver...

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 1 bar, given that the densities of the solids are `2.71` and `2.93 gcm^(-3)`, respectively.

Text Solution

Verified by Experts

`DeltaH = DeltaE + PDeltaV`
Given `DeltaE = +0.21 kJ mol^(-1) = 0.21 xx 10^(3) J mol^(-1)`
P = 1 bar = `1.0 xx 10^(5)` Pa
`DeltaV = V_("argonite") - 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) = -2.77 xx 10^(-6) m^(3)`
`DeltaH = 0.21 xx 10^(3) - 1 xx 10^(5) xx 2.77 xx 10^(-6)`
`=2.0972 J mol^(-1)`
`=0.20972 kJ mol^(-1)`
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