The value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$. Then,the formation of one mole of water in the gaseous state from $H_{(g)}$ and $O_{(g)}$ atoms is accompanied by:

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:
$(4) \ C_{(s)} + O_{2_{(g)}} \to CO_{2_{(g)}} ; \Delta H_4 = -x \ kJ$

Enthalpy change of the reaction $4H_{(g)} \to 2H_{2(g)}$ is $-869.6\, kJ$. The bond dissociation energy of $H-H$ bond is .....$kJ$

Enthalpies of formation of $CO_{(g)}$,$CO_{2(g)}$,$N_2O_{(g)}$ and $N_2O_{4(g)}$ are $-110$,$-393$,$81$ and $9.7 \, kJ \, mol^{-1}$ respectively. Find the value of $\Delta_r H$ for the reaction:
$N_2O_{4(g)} + 3 CO_{(g)} \rightarrow N_2O_{(g)} + 3 CO_{2(g)}$

If the enthalpy of neutralization of $HCN$ and $NaOH$ is $-12.13 \, kJ/mol$,then the enthalpy of ionization of $HCN$ will be ...... $kJ/mol$.

Consider the following data:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by: