If $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = r$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}; \Delta H = s$,then the heat of formation of $CO$ is

$Fe_2O_{3(s)} + \frac{3}{2} C_{(s)} \to \frac{3}{2} CO_{2(g)} + 2Fe_{(s)}$
$\Delta H^o = +234.1 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
$\Delta H^o = -393.5 \ kJ$
Use these equations and $\Delta H^o$ values to calculate $\Delta H^o$ for this reaction:
$4Fe_{(s)} + 3O_{2(g)} \to 2Fe_2O_{3(s)}$
..... $kJ$

Based on the bond enthalpy $(B.E.)$ values given,the standard enthalpy of formation $(\Delta_fH^o)$ of $N_2H_{4(g)}$ is ...... $kJ\ mol^{-1}$.
Given: $B.E.(N-N) = 159\ kJ\ mol^{-1}$,$B.E.(H-H) = 436\ kJ\ mol^{-1}$,$B.E.(N \equiv N) = 941\ kJ\ mol^{-1}$,$B.E.(N-H) = 398\ kJ\ mol^{-1}$.

Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:
$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$

The bond enthalpies of heavy hydrogen $(D-D)$, oxygen $(O=O)$, and heavy water $(D-O)$ are $+400$, $+498$, and $+490 \ kJ \ mol^{-1}$, respectively. The $\Delta_{r} H^{\circ}$ of the reaction to produce $D_2O$ is:

What will be the heat of formation of methane,if the heat of combustion of carbon is $-x \ kJ$,heat of formation of water is $-y \ kJ$ and heat of combustion of methane is $z \ kJ$?