CaCO₃ exists in two forms calcite and aragonite. The conversion of 1 mole of calcite to aragonite is accompanied by internal energy change equal to +0.21kJ. Given that the densities of calcite and aragonite are 2.71 3 cm^-3 and 2.73 cm^-3 respectively. The enthalpy change at the pressure of 1.0 bar will be

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$200 .72 J {mol}^{- 1}$

$309.72 J {mol}^{- 1}$

$209 .72 J {mol}^{- 1}$

$315.00 J {mol}^{- 1}$

answer is C.

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Detailed Solution

$\begin{array}{l} ΔH = ΔU + pΔV \\ Here, \begin{aligned} ΔU & = + 0 .21 kJ {mol}^{- 1} \\ = 0 .21 \times 10^{3} {Jmol}^{- 1} \\ = 210 J {mol}^{- 1} \\ p & = 1 bar = 10^{5} Pa \end{aligned} \end{array}$
$ΔV = Molar volume of aragonite$
-Molar volume of calcite
$= \frac{100}{2 .93} - \frac{100}{2 .71} {cm}^{3} {mol}^{- 1}$
$(∵ Molar mass of {CaCO}_{3} = 100 g {mol}^{- 1})$
$= 100 (\frac{1}{2.93} - \frac{1}{2.71})$
$\begin{array}{l} = 100 \times (\frac{0 .22}{2 .93 \times 2 .71}) \\ = 2 .77 {cm}^{3} {mol}^{- 1} \\ = - 2 .77 \times 10^{- 6} m^{3} \\ ∴ ΔH = 210 \times 10^{5} (- 2 .77 \times 10^{- 6}) \\ = 210 - 0 .277 J \\ = 209 .72 J {mol}^{- 1} \end{array}$