Choose the reaction$(s)$ from the following options,for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.
$(1)$ $\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$
$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$

For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is:

Calculate the amount of methane formed by the liberation of $149.6 \ kJ$ of heat using the following equation:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ/mol$ (in $g$)

For the reaction,$C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,$\Delta U$ is the heat of reaction at constant volume. Then the heat of reaction at constant pressure is:

For the reaction,$3 C_2 H_{2(g)} \longrightarrow C_6 H_{6(g)}$,calculate the standard enthalpy change. The values of $\Delta H_f$ for $C_2 H_2$ and $C_6 H_6$ respectively are $250 \ kJ \ mol^{-1}$ and $90 \ kJ \ mol^{-1}$.

Given:
$(i) \, C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = x \, kJ \, mol^{-1}$
$(ii) \, C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\ominus = y \, kJ \, mol^{-1}$
$(iii) \, CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = z \, kJ \, mol^{-1}$
Based on the above thermochemical equations,find out which one of the following algebraic relationships is correct?