$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H$ at $298 \ K = -285.8 \ kJ$. The molar enthalpy of vaporization of water at $1 \ atm$ and $25^{\circ}C$ is $44 \ kJ$. The standard enthalpy of formation of $1 \ mole$ of water vapor at $25^{\circ}C$ is $...... \ kJ$. (in $.8$)

Determine the enthalpy of formation for $H_2O_{2(l)}$,using the listed enthalpies of reaction:
$N_2H_{4(l)} + 2H_2O_{2(l)} \to N_{2(g)} + 4H_2O_{(l)}$; $\Delta_r H_1^o = -818 \ kJ/mol$
$N_2H_{4(l)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(l)}$; $\Delta_r H_2^o = -622 \ kJ/mol$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta_r H_3^o = -285 \ kJ/mol$

"The resultant heat change in a reaction is the same whether it takes place in one or several stages." This statement is called

The bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio $1 : 1 : 0.5$. If the enthalpy of formation of $XY$ is $\Delta_fH = -200 \ kJ \ mol^{-1}$,find the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$.

Calculate the enthalpy of formation of ethylene in $kJ\,mol^{-1}$ from the following data:
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -393.5\,kJ$
$H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.2\,kJ$
$C_2H_{4(g)} + 3 O_{2(g)} \to 2 CO_{2(g)} + 2 H_2O_{(l)} ; \Delta H = -1410.8\,kJ$

Find the value $\Delta H_f^o[NH_{3(g)}]$ if $\Delta H_r$ for the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is $-\,183.6 \ kJ/mol$ of $N_{2(g)}$.