The equation for standard enthalpies of formation has a similar structure, except that the reactants are in their standard states and the products are in their standard states. The standard enthalpy change of any reaction can be calculated from the standard enthalpies of formation of reactants and products by substituting in their respective standard states and multiplying by their respective stoichiometric coefficients. The equation for standard enthalpy changes is as follows:
The standard enthalpy of formation of water in its most stable state is zero (0). Thus, the standard enthalpy change of (ii) reaction 1, and (iii) reaction 2, are 0. The standard enthalpy change of (iv) reaction 1, and (v) reaction 2, are -64.4 and -9.5 kJ, respectively. This means that for reaction 1, the reaction is exothermic, while for reaction 2, the reaction is endothermic.
Note that the stoichiometry coefficients are equal to 1 when the reactants are in their standard states and the products are in their standard states because the reactants are the same in both reactions.
Disclaimer: I am using this reaction because it is the first reaction in my chemistry course. I am aware it is only hypothetical, but I do want to use it to illustrate the overall process of chemical reactions, not to provide a method to calculate the standard enthalpy of formation of methane.
Also, I would be curious to know if propyne would have a positive standard enthalpy of formation, meaning it would be more stable than methane at standard conditions. Or is the enthalpy of formation always negative for carbonyl compounds?
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