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Standard enthalpy of formation
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== Use in calculation for other reactions == The [[Standard enthalpy of reaction|standard enthalpy change of any reaction]] can be calculated from the standard enthalpies of formation of reactants and products using Hess's law. A given reaction is considered as the decomposition of all reactants into elements in their standard states, followed by the formation of all products. The heat of reaction is then ''minus'' the sum of the standard enthalpies of formation of the reactants (each being multiplied by its respective stoichiometric coefficient, {{mvar|Ξ½}}) ''plus'' the sum of the standard enthalpies of formation of the products (each also multiplied by its respective stoichiometric coefficient), as shown in the equation below:<ref>{{cite web|url=http://www.science.uwaterloo.ca/~cchieh/cact/c120/heatreac.html|title=Enthalpies of Reaction|website=www.science.uwaterloo.ca|access-date=2 May 2018|url-status=live|archive-url=https://web.archive.org/web/20171025201240/http://www.science.uwaterloo.ca/~cchieh/cact/c120/heatreac.html|archive-date=25 October 2017}}</ref> :<math>\Delta_{\text{r}} H^{\ominus } = \sum \nu \Delta_{\text{f}} H^{\ominus }(\text{products}) - \sum \nu \Delta_{\text{f}} H^{\ominus}(\text{reactants}).</math> If the standard enthalpy of the products is less than the standard enthalpy of the reactants, the standard enthalpy of reaction is negative. This implies that the reaction is exothermic. The converse is also true; the standard enthalpy of reaction is positive for an endothermic reaction. This calculation has a tacit assumption of [[ideal solution]] between reactants and products where the [[enthalpy of mixing]] is zero. For example, for the combustion of methane, <chem>CH4 + 2O2 -> CO2 + 2H2O</chem>: :<math>\Delta_{\text{r}} H^{\ominus } = [\Delta_{\text{f}} H^{\ominus }(\text{CO}_2) + 2\Delta_{\text{f}} H^{\ominus } (\text{H}_2{}\text{O})] - [\Delta_{\text{f}} H^{\ominus }(\text{CH}_4) + 2\Delta_{\text{f}} H^{\ominus }(\text{O}_2)].</math> However <chem>O2</chem> is an element in its standard state, so that <math> \Delta_{\text{f}} H^{\ominus }(\text{O}_2) = 0</math>, and the heat of reaction is simplified to :<math>\Delta_{\text{r}} H^{\ominus } = [\Delta_{\text{f}} H^{\ominus }(\text{CO}_2) + 2\Delta_{\text{f}} H^{\ominus } (\text{H}_2{}\text{O})] - \Delta_{\text{f}} H^{\ominus }(\text{CH}_4),</math> which is the equation in the previous section for the enthalpy of combustion <math>\Delta_{\text{comb}}H^{\ominus }</math>.
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