The heats of formation of $CO_{2(g)}$,$H_2O_{(l)}$,and $CH_{4(g)}$ are $-94.0$,$-68.4$,and $-17.9 \ kcal$ respectively. The heat of combustion of methane is.....$kcal$.

From the given reaction,$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta H = -92.6 \ kJ$,the enthalpy of formation of $NH_3$ is (in $kJ$):

Given that $C + O_{2} \longrightarrow CO_{2} ; \Delta H^{\circ} = -x \ kJ$ and $2 CO + O_{2} \longrightarrow 2 CO_{2} ; \Delta H^{\circ} = -y \ kJ$. The heat of formation of carbon monoxide will be

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:

Calculate the enthalpy change for the process $CCl_{4(g)} \to C_{(g)} + 4Cl_{(g)}$ and calculate the bond enthalpy of the $C-Cl$ bond in $CCl_{4(g)}$.
$\Delta_{vap} H^{\theta}(CCl_{4}) = 30.5 \, kJ \, mol^{-1}$
$\Delta_{f} H^{\theta}(CCl_{4}) = -135.5 \, kJ \, mol^{-1}$
$\Delta_{a} H^{\theta}(C) = 715.0 \, kJ \, mol^{-1}$ (where $\Delta_{a} H^{\theta}$ is enthalpy of atomisation)
$\Delta_{a} H^{\theta}(Cl_{2}) = 242 \, kJ \, mol^{-1}$

The reaction for the formation of $NaCl$ is: