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Solved: For reactions in condensed phases (liquids and
Chapter 6, Problem 145P(choose chapter or problem)
For reactions in condensed phases (liquids and solids), the difference between \(\Delta H\) and \(\Delta U\) is usually quite small. This statement holds for reactions carried out under atmospheric conditions. For certain geochemical processes, however, the external pressure may be so great that \(\Delta H\) and \(\Delta U\) can differ by a significant amount. A well-known example is the slow conversion of graphite to diamond under Earth's surface. Calculate \((\Delta H-\Delta U)\) for the conversion of 1 mole of graphite to 1 mole of diamond at a pressure of 50,000 atm. The densities of graphite and diamond are \(2.25\ \mathrm{g} / \mathrm{cm}^{3}\) and \(3.52\ \mathrm{g} / \mathrm{cm}^{3}\), respectively.
Questions & Answers
QUESTION:
For reactions in condensed phases (liquids and solids), the difference between \(\Delta H\) and \(\Delta U\) is usually quite small. This statement holds for reactions carried out under atmospheric conditions. For certain geochemical processes, however, the external pressure may be so great that \(\Delta H\) and \(\Delta U\) can differ by a significant amount. A well-known example is the slow conversion of graphite to diamond under Earth's surface. Calculate \((\Delta H-\Delta U)\) for the conversion of 1 mole of graphite to 1 mole of diamond at a pressure of 50,000 atm. The densities of graphite and diamond are \(2.25\ \mathrm{g} / \mathrm{cm}^{3}\) and \(3.52\ \mathrm{g} / \mathrm{cm}^{3}\), respectively.
ANSWER:Step 1 of 7
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