Name: calculating reaction heat with hess's law
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Enthalpies of Reaction

Para uma reação que ocorra em condições padrão, é utilizada uma equação geral para calcular a alteração da entalpia padrão da reação. Esta equação é resolvida através da descoberta de a diferença entre a soma das entalpias padrão da formação dos produtos e a soma das entalpias padrão da formação dos reagentes, cada um multiplicado pelo seu coeficiente estequiométrico. Consideremos a combustão de 2 mols de gás de acetileno com 5 mols de gás de oxigênio para formar 4 mols de gás de dióxido de carbono e 2 mols de vapor de água em condições normais.A entalpia para a reação é igual à soma de quatro vezes a entalpia da formação de gás de dióxido de carbono e duas vezes a entalpia da formação do vapor de água, menos a soma de duas vezes a entalpia da formação de gás de acetileno e cinco vezes a entalpia da formação de oxigênio gasoso. A equação de entalpia é derivada da combinação de dois conceitos:entalpia padrão da formação e a lei de Hess. O primeiro termo representa os entalpia-padrão de formação dos produtos;a formação de dióxido de carbono a partir de carbono e oxigênio equação 1, e a formação de água a partir de hidrogênio e oxigênio equação 2.As entalpias de formação padrão conhecidas para o dióxido de carbono e para a água são 393, 5 kilojoules negativos e 241, 8 kilojoules negativos, respetivamente. Uma vez que a combustão produz 4 mols de dióxido de carbono, o delta H1 seria a entalpia padrão de formação de dióxido de carbono vezes 4, que seria de 1574 kilojoules negativos. A combustão também produz 2 mols de água, por isso a alteração da entalpia delta H2 seria a entalpia padrão da formação da água vezes 2, que seria 483, 6 kilojoules negativos.Isto produz 2058 kilojoules negativos como a entalpia padrão líquida da formação dos produtos. O segundo termo representa a decomposição do acetileno em carbono e hidrogênio equação 3. Este é o inverso da reação para a entalpia padrão de formação do reagente, e, portanto, o valor da sua entalpia, 227, 4 kilojoules, é precedida por um sinal negativo, que também pode ser visto na equação da entalpia.Porque a reação consome 2 mols de acetileno, o delta H3 seria o negativo da entalpia padrão de formação de acetileno vezes 2, o que equivale a 453, 4 kilojoules negativos. A entalpia de formação padrão do oxigênio é zero. Portanto, a entalpia líquida padrão de formação dos reagentes é de 453, 4 kilojoules negativos.Recordemos a Lei de Hess, que se for levada a cabo uma reação de um passo em vários passos, depois a soma das entalpias de cada passo é igual à alteração da entalpia líquida. A substituição dos valores para as entalpias de formação na equação dá a entalpia da reação como 2511 kilojoules negativos.

<ol>
	<li>Davis, Thomas W. "Um mal-entendido comum da lei de Hess." Jornal de Educação Química 28, nº 11 (1951): 584.</li>
	<li>Lee, A. L., H. L. Feldkirchner, F. C. Schora e J. J. Henry. "Calor de Reação de Hidrogênio e Carvão." Projeto e Desenvolvimento de Processos de Química Industrial e de Engenharia 7, nº 2 (1968): 244-249.</li>
</ol>

entalpias de reações

A lei de Hess pode ser usada para determinar a alteração da entalpia de qualquer reação se as correspondentes entalpiaa de formação dos reagentes e produtos estiverem disponíveis. A reação principal pode ser dividida em reações por passos : (i) decomposições dos reagentes nos seus elementos componentes, para os quais as alterações da entalpia são proporcionais ao negativo das entalpias de formação dos reagentes, −ΔHf°(reagentes), seguida de (ii) recombinações dos elementos (obtidos no passo 1) para originar os produtos, com as alterações de entalpia proporcionais às entalpias de formação dos produtos,  ΔHf° (produtos). A alteração da entalpia padrão da reação global é portanto igual a: (ii) soma das entalpias padrão de formação de todos os produtos mais (i) soma dos negativos das entalpias padrão de formação dos reagentes, tal como indicado pela equação seguinte, onde ∑ representa “a soma de” e n representa os coeficientes estequiométricos.

<img alt="Eq1" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-01.png" style="height: 40px;" />

A equação é geralmente rearranjada ligeiramente para ser escrita como se segue: 

<img alt="Eq2" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-02.png" style="height: 40px;" />

O exemplo a seguir mostra em detalhes por que esta equação é válida e como usá-la para calcular a alteração da entalpia padrão para uma reação:

<img alt="Eq3" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-03.png" style="height: 40px;" />

Aqui, utiliza-se a forma especial da lei de Hess e os valores de formação de calor para os reagentes e produtos: ΔHf° (HNO3) = −206,64 kJ/mol; ΔHf° (NO) = +90,2 kJ/mol; ΔHf° (NO2) = +33 kJ/mol; ΔHf° (H2O) = −285,8 kJ/mol.

<img alt="Eq4" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-04.png" style="height: 40px;" />

<img alt="Eq5" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-05.png" style="height: 40px;" />

<img alt="Eq6" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-06.png" style="height: 40px;" />

<img alt="Eq7" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-07.png" style="height: 40px;" />

 
Este texto é adaptado de <a href="https://openstax.org/books/chemistry-2e/pages/5-3-enthalpy">Openstax, Chemistry 2e, Section 5.3: Enthalpy.</a>

Entalpias de Reações

A lei de Hess pode ser usada para determinar a mudança de entalpia de qualquer reação se as entalpias correspondentes de formação dos reagentes e produtos ...

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calculating reaction heat with hess's law

Calculating reaction heat with Hess's law uses standard enthalpies of formation to find the enthalpy change of a reaction. When the formation values for the reactants and products are known, the overall reaction can be broken into simpler stepwise reactions. This approach helps show how the heat change for the full reaction is built from smaller parts.

First, the reactants are thought of as being decomposed into their component elements. The enthalpy change for this step is proportional to the negative of the standard enthalpies of formation of the reactants, written as -ΔHf°(reactants). Next, those elements are recombined to form the products, and this step has an enthalpy change proportional to ΔHf°(products).

The standard enthalpy change of the overall reaction is found by adding the sum of the standard enthalpies of formation of all the products and the sum of the negatives of the standard enthalpies of formation of the reactants. In the equation, Σ means &quot;the sum of&quot; and n is the stoichiometric coefficient. The formula is usually rearranged into a simpler form for calculation.

A worked example shows why the equation works and how to use it to calculate the standard enthalpy change of a reaction. In that example, the heat of formation values used are ΔHf°(HNO3) = -206.64 kJ/mol, ΔHf°(NO) = +90.2 kJ/mol, ΔHf°(NO2) = +33 kJ/mol, and ΔHf°(H2O) = -285.8 kJ/mol. This method follows the special form of Hess's law described in OpenStax Chemistry 2e, Section 5.3: Enthalpy.

Calculating Reaction Heat with Hess's Law

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enthalpies of reaction

Calculating reaction heat with Hess's law uses standard enthalpies of formation to find the enthalpy change of a reaction. When the formation values for the reactants and products are known, the overall reaction can be broken into simpler stepwise reactions. This approach helps show how the heat change for the full reaction is built from smaller parts. First, the reactants are thought of as being decomposed into their component elements. The enthalpy change for this step is proportional to the...

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Calculating Reaction Heat with Hess's Law

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