The heat of formation is the change in enthalpy accompanying the formation of a substance from its elements at $298 \ K$ and $1 \ atm$ pressure. Since the enthalpies of elements are taken to be zero,the heat of formation $(\Delta H_f)$ of compounds

Hess's law is based on

Calculate the enthalpy of formation of carbon monoxide $(CO).$ Given: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -393.3 \ kJ \ mol^{-1}$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -282.2 \ kJ \ mol^{-1}.$

Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)

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 $... \ kcal$.

The standard enthalpy of combustion of $C$ (graphite),$H_{2(g)}$ and $CH_3OH_{(l)}$ respectively are $-393 \ kJ \ mol^{-1}$,$-286 \ kJ \ mol^{-1}$ and $-726 \ kJ \ mol^{-1}$. What is the standard enthalpy of formation of methanol?