Question

Calculate $\Delta H_f^0$ for $UB{r}_4$ from $\Delta G^0$ of reaction and the $S^0$ values.$U\left(s\right)+2B{r}_2\left(l\right)\to UB{r}_4\left(s\right);\Delta G^0=-788.6KJ;S^0(J/K-mol)50.3,152.3,242.6$

• A. -822.1 KJ/mol
• B. -841.2 KJ/mol
• C. -775.6 KJ/mol
• D. -804.3 KJ/mol

Answer 1

The correct option is C -775.6 KJ/mol

$\Delta S^o=(S^o{)}_{products}-(S^o{)}_{reactants}$

$⟹\Delta S^o=\left(242.6\right)-(1\times 50.3+2\times 152.3)$

$⟹\Delta S^o=-112.3\times {10}^{-3}$ $kJ/k-mol$

$T=298$ $K$ (Room Temperature)

$\because \Delta G=\Delta H-T\Delta S$

$⟹\Delta H=\Delta G+T\Delta S$

$⟹\Delta H=-788.6+\left(298\right)(-112.3\times {10}^{-3})$

$⟹\Delta H=-755.1$ $kJ/mol$

$\Delta H=(\Delta H{)}_{product}-(\Delta H{)}_{reactant}$

$\Delta H=\Delta H_f^o\left(UB{r}_4\right)-0$

$\therefore \Delta H_f^o=-755.1$ $kJ/mol$