The enthalpy of formation of $NH_3$ is $-46 \ kJ \ mol^{-1}$. The enthalpy change for the reaction $2 NH_{3(g)} \longrightarrow N_{2(g)} + 3 H_{2(g)}$ is

Given: $C + O_2 \rightarrow CO_2$ : $\Delta H = -395 \ kJ$,$S + O_2 \rightarrow SO_2$ : $\Delta H = -295 \ kJ$,$CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2$ : $\Delta H = -1110 \ kJ$. Calculate the heat of formation of $CS_2$ in $kJ/mol$.

The standard enthalpy of formation $(\Delta_f H^{\circ})$ of ammonia is $-46.2 \ kJ \ mol^{-1}$. What is the $\Delta_r H^{\circ}$ of the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$

The value of enthalpy of neutralization is highest for the reaction between:

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

At $298 \ K$ and constant pressure, the heat of formation of $C_2H_{2(g)}$ and $C_6H_{6(g)}$ is $230 \ kJ \ mol^{-1}$ and $85 \ kJ \ mol^{-1}$ respectively. What will be the change in heat for the reaction?
$3 \ C_2H_{2(g)} \rightarrow C_6H_{6(g)}$