Average $C-H$ bond energy is $416 \ kJ \ mol^{-1}$. Which of the following is correct?

"The quantity of heat which must be supplied to decompose a compound into its elements is equal to the heat evolved during the formation of that compound from the elements." This statement is known as

The standard enthalpy of formation of $CO_{(g)}$, $CO_{2(g)}$, $N_2O_{(g)}$ and $N_2O_{4(g)}$ are respectively $-110$, $-393$, $81$ and $-10 \ kJ \ mol^{-1}$. The enthalpy change (in $kJ$) of the following reaction is $N_2O_{4(g)} + 3CO_{(g)} \longrightarrow N_2O_{(g)} + 3CO_{2(g)}$

The correct order of $\Delta_{f} H^{\circ}$ values of Diamond $(I)$, Graphite $(II)$, and Fullerene $(III)$ is:

The enthalpy of combustion of benzene from the following data will be
$(i) \ 6C_{(s)} + 3H_{2(g)} \to C_6H_{6(l)} ; \Delta H = +45.9 \ kJ$
$(ii) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)} ; \Delta H = -285.9 \ kJ$
$(iii) \ C_{(s)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -393.5 \ kJ$
.....$kJ$

Calculate $\Delta H^{\circ}$ for the reaction,$Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$ given the following:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$