What is the enthalpy change for the reaction $2C(\text{graphite}) + 3H_{2(g)} \rightarrow C_2H_{6(g)}$ called?

Calculate $\Delta H$ for the reaction: $H_{2(g)} + O_{2(g)} \rightarrow H_2O_{2(g)}$ given the bond energies: $BE_{H-H} = 436 \ kJ/mol$,$BE_{O=O} = 499 \ kJ/mol$,$BE_{O-O} = 142 \ kJ/mol$,and $BE_{O-H} = 460 \ kJ/mol$. (in $kJ$)

Assertion $(A)$: Formation of $H_2O_{(\ell)}$ from $H_{2(g)}$ and $O_{2(g)}$ is more exothermic than formation of $H_2O_{(g)}$ from $H_{2(g)}$ and $O_{2(g)}$.
Reason $(R)$: $H_2O_{(\ell)}$ has more attraction force than $H_2O_{(g)}$ and it is more stable than $H_2O_{(g)}$.

The enthalpies of combustion of carbon and carbon monoxide are $-393.5 \, kJ$ and $-283 \, kJ$ respectively. Calculate the enthalpy of formation of carbon monoxide in $kJ$. (in $.5$)

Which of the following compounds has a standard molar enthalpy of formation equal to zero at $298 \ K$?

From the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:

$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$	$\Delta H = 42.09 \ kJ \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$	$\Delta H = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)}$	$\Delta H = 436 \ kJ \ mol^{-1}$
$O_{2(g)} \rightarrow 2 O_{(g)}$	$\Delta H = 496 \ kJ \ mol^{-1}$