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Question

The value of ΔfH° of U3O8(s) is –853.5 kJ mol–1. ΔH° for the reaction

3UO2(s) + O2(g) → U3O8(s)

is −76.00 kJ. The value of ΔfH° of UO2(s) is

Options

A.−259.17 kJ/mol
B.−310.17 kJ/mol
C.+259.17 kJ/mol
D.930.51 kJ/mol

Solution

$\Delta_{r}H^{o} = \sum_{}^{}{\Delta_{f}H_{\text{Products}}^{o}} - \sum_{}^{}{\Delta_{f}H_{\text{Reactants}}^{o}}$

$= \left\lbrack \Delta_{f}H_{U_{3}O_{8}(s)}^{o} \right\rbrack - \left\lbrack 3 \times \Delta_{f}H_{UO_{2}(s)}^{o} + \Delta_{f}H_{O_{2}(g)}^{o} \right\rbrack $$${\text{Or, } - 76.0 = ( - 853.5) - \left\lbrack 3 \times \Delta_{f}H_{UO_{2}(s)}^{o} + 0 \right\rbrack }{\therefore\Delta_{f}H_{UO_{2}(s)}^{o} = - 259.17\text{ kJ/mol}}$$

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