The atomization enthalpies of $NH_{3(g)}$ and $N_2H_{4(g)}$ are $+150 \ kJ \ mol^{-1}$ and $+310 \ kJ \ mol^{-1}$ respectively. The $\Delta H(N-N)$ bond enthalpy in $kJ \ mol^{-1}$ is:

When $2 \ mol$ of $C_2H_{6(g)}$ is completely combusted,it releases $3129 \ kJ$ of heat. What is the enthalpy of formation of $C_2H_{6(g)}$? The $\Delta H_f$ values for $CO_{2(g)}$ and $H_2O_{(l)}$ are $-395 \ kJ \ mol^{-1}$ and $-286 \ kJ \ mol^{-1}$ respectively.

If standard enthalpy of formation $(\Delta_{f} H^{\circ})$ of $CO_2, H_2 O$ and $CH_4$ are $-393, -286$ and $-74.0 \ kJ \ mol^{-1}$ respectively,the standard enthalpy of combustion of methane in $kJ \ mol^{-1}$ is

The heat of combustion of $C_xH_y$,carbon,and hydrogen are $a, b$,and $c \ cal/mole$ respectively. The heat of formation of $C_xH_y$ will be:

For the following reaction,
$C (diamond) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -97.6 \ kcal$
$C (graphite) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -94.3 \ kcal$
The heat change for the conversion of $1 \ g$ of $C (diamond) \rightarrow C (graphite)$ is (in $kcal$)

$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$