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 $kJ \ mol^{-1}$.

The atomisation enthalpy of $CH_4$ is $1660 \ kJ \ mol^{-1}$. The $C-H$ bond enthalpy of each successive step in $CH_4$ $\rightarrow CH_3$ $\rightarrow CH_2$ $\rightarrow CH$ are $+15, +30$ and $+45 \ kJ \ mol^{-1}$ higher than the mean bond enthalpy of $C-H$ bonds,respectively. The bond enthalpy of the last $C-H$ unit is

The reaction for the formation of $NaCl$ is:

If the ratio of the enthalpy of formation of $CO_2$ and $SO_2$ is $4:3$ and the enthalpy of formation of $CS_2$ is $26 \ kcal/mol$,then what will be the enthalpy of formation of $SO_2(g)$ based on the following reaction?
$CS_2(l) + 3O_2(g) \to CO_2(g) + 2SO_2(g)$

Find the value of $x$ in $\text{kJ}$ using the following equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} : \Delta H = 131 \ \text{kJ}$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} : \Delta H = -282 \ \text{kJ}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} : \Delta H = -242 \ \text{kJ}$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} : \Delta H = x \ \text{kJ}$

Energy required to dissociate $16 \ g$ of $O_{2(g)}$ into free atoms is $x \ kJ$. The value of bond enthalpy of $O=O$ bond is