Given that the bond energies of $N \equiv N$ is $946 \ kJ \ mol^{-1}$,$H-H$ is $435 \ kJ \ mol^{-1}$,$N-N$ is $159 \ kJ \ mol^{-1}$,and $N-H$ is $389 \ kJ \ mol^{-1}$,calculate the enthalpy of formation for the gas phase reaction $N_2 + 2H_2 \rightarrow N_2H_4$ in $kJ \ mol^{-1}$.

Given that,$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -x \ kJ \ mol^{-1}$ and $2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)} ; \Delta H^{\circ} = -y \ kJ \ mol^{-1}$. The enthalpy of formation of $CO$ will be:

At standard conditions,if the change in the enthalpy for the following reaction is $-109 \; kJ \ mol^{-1}$:
$H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$
Given that bond energy of $H_2$ and $Br_2$ is $435 \; kJ \ mol^{-1}$ and $192 \; kJ \ mol^{-1}$,respectively,what is the bond energy (in $kJ \ mol^{-1}$) of $HBr$?

The standard heat of formation $\left(\Delta_{f} H_{298}^{0}\right)$ of ethane in $kJ/mol$,if the heat of combustion of ethane,hydrogen and graphite are $-1560$,$-286$,and $-393.5 \; kJ/mol,$ respectively is ........... $kJ/mol$.

The equation $\frac{1}{2} H_2 + \frac{1}{2} Cl_2 \to HCl$ $(\Delta H_{298} = -22.060 \ kcal)$ means:

The standard enthalpy of formation of $NH_3$ is $-46.0 \ kJ \ mol^{-1}.$ If the enthalpy of formation of $H_2$ from its atoms is $-436 \ kJ \ mol^{-1}$ and that of $N_2$ is $-712 \ kJ \ mol^{-1},$ the average bond enthalpy of $N-H$ bond in $NH_3$ is ................ $kJ \ mol^{-1}$