The standard heat of formation of U3O8 is -853.5 kcal/mol and the standard heat of the reaction3UO2+O2→ U3O8 is -76.01 kcal.The standard heat of formation of UO2 is

3UO2+O2→ U3O8 ∆H0=∆Hf0U3O8 -∆Hf0O2-3∆Hf0 -76.01 = -853.5 -0 -3∆Hf (UO2) or ∆Hf0UO2 = -76.01 +-853.5-03 =-259.17 =-1083.33 kJ/mol

The standard heat of formation of U3O8 is -853.5 kcal/mol and the standard heat of the reaction3UO2+O2→ U3O8 is -76.01 kcal.The standard heat of formation of UO2 is

3UO2+O2→ U3O8 ∆H0=∆Hf0U3O8 -∆Hf0O2-3∆Hf0 -76.01 = -853.5 -0 -3∆Hf (UO2) or ∆Hf0UO2 = -76.01 +-853.5-03 =-259.17 =-1083.33 kJ/mol

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The standard heat of formation of $U_{3} O_{8} is - 853.5 kcal / mol$ and the standard heat of the reaction
$3 {UO}_{2} + O_{2} \overset{}{\to} U_{3} O_{8} is - 76.01 kcal .$
The standard heat of formation of ${UO}_{2} is$

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$- 10.83 kJ {mol}^{- 1}$

$- 1.083 kJ {mol}^{- 1}$

$- 1083 kJ {mol}^{- 1}$

$- 108.3 kJ {mol}^{- 1}$

answer is A.

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Detailed Solution

$3 {UO}_{2} + O_{2} \overset{}{\to} U_{3} O_{8} ∆ H^{0} = ∆ H_{f}^{0} (U_{3} O_{8}) - ∆ H_{f}^{0} (O_{2}) - 3 ∆ H_{f}^{0} - 76.01 = - 853.5 - 0 - 3 ∆ H_{f ({UO}_{2})} or ∆ H_{f}^{0} ({UO}_{2}) = \frac{- (76.01) + (- 853.5) - 0}{3} = - 259.17 = - 1083.33 kJ / mol$