Explanation
$$\Delta S^o=(S^o)_{products}-(S^o)_{reactants}$$
$$\implies \Delta S^o=(242.6)-(1\times 50.3+2\times 152.3)$$
$$\implies \Delta S^o=-112.3\times 10^{-3}$$ $$kJ/k-mol$$
$$T=298$$ $$K$$ (Room Temperature)
$$\because \Delta G=\Delta H-T\Delta S$$
$$\implies \Delta H=\Delta G+T\Delta S$$
$$\implies \Delta H=-788.6+(298)(-112.3\times 10^{-3})$$
$$\implies \Delta H=-755.1$$ $$kJ/mol$$
$$\Delta H=(\Delta H)_{product}-(\Delta H)_{reactant}$$
$$\Delta H=\Delta H_f^o(UBr_4)-0$$
$$\therefore \Delta H_f^o=-755.1$$ $$kJ/mol$$
Hint: Elements that are stable as single atoms are known as monatomic or monoatomic elements.
It is given in the question, $$P=V$$ and $$\frac{P}{V}=1$$,
Thus,
$$dg={{c}_{v}}dt+Pdv$$
For $$1$$ mole of gas $$PV=Rt$$,
$$Pdv+VdP=RdT$$
$$2PdV=RdT$$
$$PdV=\frac{RdT}{2}$$
$$dg={{C}_{r}}dt+\frac{RdT}{2}$$
$$=\frac{3}{2}R+\frac{R}{2}$$
$$=\frac{4R}{2}$$
Correct Option: $$A$$
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