Enthalpy of a compound is equal to its

If the combustion of $1 \ g$ of graphite produces $20.7 \ kJ$ of heat,what will be the molar enthalpy change? Give the significance of the sign also.

Given that bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively and $\Delta H_f$ for $HCl$ is $-90 \ kJ \ mol^{-1},$ the bond enthalpy of $HCl$ is ............... $kJ \ mol^{-1}$.

The enthalpy changes at $298 \ K$ in successive breaking of $O-H$ bonds of $H_2O$ are:
$H_2O_{(g)} \to H_{(g)} + OH_{(g)}, \Delta H = 498 \ kJ \ mol^{-1}$
$OH_{(g)} \to H_{(g)} + O_{(g)}, \Delta H = 428 \ kJ \ mol^{-1}$
The bond enthalpy of the $O-H$ bond is ..... $kJ \ mol^{-1}$.

The enthalpy of neutralization of $NaOH$ by $HCl$ is $-55.84 \, kJ/mol$ and that of $NH_4OH$ by $HCl$ is $-51.34 \, kJ/mol$. The enthalpy of ionization of $NH_4OH$ is ..... $kJ/mol$.

Given the following thermochemical equations:
$(1) \ H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} ; \Delta H_1 = 100 \ kJ$
$(2) \ CO_{(g)} + \frac{1}{2}O_{2_{(g)}} \to CO_{2_{(g)}} ; \Delta H_2 = -300 \ kJ$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2_{(g)}} \to H_2O_{(g)} ; \Delta H_3 = -250 \ kJ$
Calculate the value of $x$ for the reaction:
$(4) \ C_{(s)} + O_{2_{(g)}} \to CO_{2_{(g)}} ; \Delta H_4 = -x \ kJ$