At 500 kbar and , the densities of graphite and diamond are 2.0 and , respectively. The value of for — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For $C(\text{graphite}) \to C(\text{diamond})$, $\Delta H - \Delta U = P \Delta V$. Molar mass $= 12 \text{ g/mol}$. $V_{graphite} = 12 / 2.0 = 6 \text{ cm}^3 = 6 \times 10^{-6} \text{ m}^3$. $V_{diamond} = 12 / 3.0 = 4 \text{ cm}^3 = 4 \times 10^{-6} \text{ m}^3$. $\Delta V = (4 - 6) \times 10^{-6} = -2 \times 10^{-6} \text{ m}^3$. Therefore, $P \Delta V = 500 \times 10^3 \text{ bar} \times 10^5 \text{ Pa/bar} \times (-2 \times 10^{-6}) \text{ m}^3 = -10 \times 10^4 \text{ J} = -100 \text{ kJ}$.