Enthalpy Change Of Formation
$$\ce{C(graphite) + O2(g) -&gt; CO2(g)}$$

$$\Delta H^{\varnothing} = -393.5\ \mathrm{kJ}\,\mathrm{mol}^{-1}$$

That is, the reaction is exothermic.

Since the reactants are elements in standard state, the heat of formation for carbon dioxide is the same as the enthalpy change of the above reaction.