Calculate Delta H_f^o for UBr_4 from the Delt | vedclass

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Question

(A) The reaction is: $U_{(s)} + 2Br_2(\ell) 	o UBr_{4(s)}$.First,calculate the change in entropy for the reaction,$\Delta S^o_{rxn}$:$\Delta S^o_{rxn

Vedclass · Accepted Answer

$-822.1 \ kJ/mol$

Vedclass · Answer

(A) The reaction is: $U_{(s)} + 2Br_2(\ell) 	o UBr_{4(s)}$.First,calculate the change in entropy for the reaction,$\Delta S^o_{rxn}$:$\Delta S^o_{rxn} = \sum S^o_{products} - \sum S^o_{reactants}$$\Delta S^o_{rxn} = S^o(UBr_{4(s)}) - [S^o(U_{(s)}) + 2 	imes S^o(Br_2(\ell))]$$\Delta S^o_{rxn} = 242.6 - [50.3 + 2 	imes 152.3] \ J/K \cdot mol$$\Delta S^o_{rxn} = 242.6 - [50.3 + 304.6] = 242.6 - 354.9 = -112.3 \ J/K \cdot mol = -0.1123 \ kJ/K \cdot mol$.Using the Gibbs-Helmholtz equation: $\Delta G^o = \Delta H^o - T\Delta S^o$.Assuming standard temperature $T = 298 \ K$:$-788.6 \ kJ = \Delta H^o - (298 \ K 	imes -0.1123 \ kJ/K \cdot mol)$$-788.6 = \Delta H^o + 33.4654$$\Delta H^o = -788.6 - 33.4654 = -822.0654 \ kJ/mol \approx -822.1 \ kJ/mol$.

Calculate $\Delta H_f^o$ for $UBr_4$ from the $\Delta G^o$ of reaction and the $S^o$ values: $U_{(s)} + 2Br_2(\ell) \to UBr_{4(s)}$; $\Delta G^o = -788.6 \ kJ$; $S^o \ (J/K \cdot mol): U_{(s)} = 50.3, Br_2(\ell) = 152.3, UBr_{4(s)} = 242.6$.

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