The enthalpy of neutralization is about $57.3 \ kJ$ for the pair:

What will be the enthalpy of formation of benzene in $kJ$ for the reaction: $6C_{(s)} + 3H_{2(g)} \rightarrow C_6H_{6(l)}$? Given that the enthalpy of combustion of benzene is $-3268 \ kJ$, the enthalpy of formation of $CO_{2(g)}$ is $-393.5 \ kJ$, and the enthalpy of formation of $H_2O_{(l)}$ is $-285.8 \ kJ$.

Which of the following reactions defines the standard enthalpy of combustion,$\Delta H_c^ \circ$?

Given $C + O_2 \rightarrow CO_2 + 94.2 \ kcal$,$\Delta H = -94.2 \ kcal$; $H_2 + 1/2 O_2 \rightarrow H_2O + 68.3 \ kcal$,$\Delta H = -68.3 \ kcal$ and $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + 210.8 \ kcal$,$\Delta H = -210.8 \ kcal$. Calculate the heat of formation of methane $(CH_4)$ in $kcal$.

The heat of formation is the change in enthalpy accompanying the formation of a substance from its elements at $298 \ K$ and $1 \ atm$ pressure. Since the enthalpies of elements are taken to be zero,the heat of formation $(\Delta H_f)$ of compounds

$C + O_2 \to CO_2; \Delta H = X$
$CO + \frac{1}{2} O_2 \to CO_2; \Delta H = Y$
Then the heat of formation of $CO$ is