The value of delta_f H° of U3O8(s) is -853.5 kJ mol^-1. δ H° for the reaction: 3UO2(s) + (g) -> U3O8 — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For the reaction: 3UO2(s) + $O_2$(g) -> U3O8(s), δ H° = delta_f H°(U3O8, s) - [3 * delta_f H°(UO2, s) + delta_f H°($O_2$, g)]. Given δ H° = -76.00 kJ, delta_f H°(U3O8, s) = -853.5 kJ/mol, and delta_f H°($O_2$, g) = 0: -76.00 = -853.5 - 3 * delta_f H°(UO2, s) => 3 * delta_f H°(UO2, s) = -853.5 + 76.00 = -777.5 => delta_f H°(UO2, s) = -777.5 / 3 = -259.17 kJ/mol.