$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = -94 \ kcal$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -67.7 \ kcal$
Then the heat of formation of $CO_{(g)}$ is $..... \ kcal$

Find the value of enthalpy of formation of $PCl_5(s)$ given the following thermochemical equations:
$1) \frac{1}{2} P_{4(s)} + 3Cl_{2(g)} \to 2PCl_3(\ell) ; \Delta H = -635 \ kJ$
$2) PCl_3(\ell) + Cl_{2(g)} \to PCl_{5(s)} ; \Delta H = -137 \ kJ$

With the help of the following data,find out the change in heat content for the reaction:
$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

For the reaction,$3 C_2 H_{2(g)} \longrightarrow C_6 H_{6(g)}$,calculate the standard enthalpy change. The values of $\Delta H_f$ for $C_2 H_2$ and $C_6 H_6$ respectively are $250 \ kJ \ mol^{-1}$ and $90 \ kJ \ mol^{-1}$.

The enthalpy of $1 \ mol$ of a compound is equal to its ...... (when it is formed from its constituent elements).

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 $... \ kcal$.