If $C + O_2 \to CO_2 + 94.2 \ kcal$,$H_2 + \frac{1}{2} O_2 \to H_2O + 68.3 \ kcal$,and $CH_4 + 2O_2 \to CO_2 + 2H_2O + 210.8 \ kcal$,then the heat of formation of methane will be $... \ kcal$.

Given the bond energies of $N \equiv N$,$H - H$ and $N - H$ bonds are $945$,$436$ and $391 \ kJ \cdot mol^{-1}$ respectively,the enthalpy of the following reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$ is ...... $kJ$.

Given:
$(A) \ 2 \ CO_{(g)} + O_{2(g)} \rightarrow 2 \ CO_{2(g)} \quad \Delta H_1^\theta = -x \ kJ \ mol^{-1}$
$(B) \ C \ (\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H_2^\theta = -y \ kJ \ mol^{-1}$
The $\Delta H^\theta$ for the reaction $C \ (\text{graphite}) + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$ is:

Standard molar enthalpy of formation,$\Delta _{f}H^{o}$ is just a special case of enthalpy of reaction,$\Delta _{r}H^{o}$. Is the $\Delta _{r}H^{o}$ for the following reaction same as $\Delta _{f}H^{o}$? Give reason for your answer. $CaO_{(s)} + CO_{2_{(g)}} \to CaCO_{3_{(s)}}$; $\Delta _{r}H^{o} = -178.3 \ kJ \ mol^{-1}$

If the $\Delta H_{fusion}$ of a substance is $'x'$ and $\Delta H_{vap}$ is $'y'$,then $\Delta H_{sublimation}$ will be:

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 ..... $kJ/mol$.