The internal energy change in the conversion of 1 mole of the calcite form of to the aragonite form — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Molar mass of $\text{CaCO}_3 = 100\text{ g/mol}$. $V_{\text{calcite}} = 100 / 2.7 = 37.037\text{ cm}^3$. $V_{\text{aragonite}} = 100 / 3.0 = 33.333\text{ cm}^3$. $\Delta V = 33.333 - 37.037 = -3.704\text{ cm}^3 = -3.704 \times 10^{-6}\text{ m}^3$. $P\Delta V = 2.7 \times 10^5 \times (-3.704 \times 10^{-6}) = -1.0\text{ J}$. $\Delta H = \Delta U + P\Delta V = 210\text{ J} - 1.0\text{ J} = 209\text{ J}$. Format as 0209.