Name: calculating reaction heat with hess's law
Uploaded: 2020-09-03T13:38:27.000Z
Duration: 3 min 33 s
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PS.Chm.7.2

Enthalpies of Reaction

Per una reazione che si verifica in condizioni standard, viene utilizzata un'equazione generale per calcolare la variazione standard di entalpia della reazione. Questa equazione si risolve trovando la differenza tra la somma delle entalpie standard di formazione dei prodotti e la somma delle entalpie standard di formazione dei reagenti, ciascuna moltiplicata per il proprio coefficiente stechiometrico. Considerate la combustione di 2 moli di gas acetilene con 5 moli di ossigeno gassoso per formare 4 moli di anidride carbonica e 2 moli di vapore acqueo, in condizioni standard.

L'entalpia per la reazione è uguale alla somma di quattro volte l'entalpia di formazione del gas di anidride carbonica e due volte l'entalpia di formazione del vapore acqueo, meno la somma di due volte l'entalpia di formazione del gas acetilene e cinque volte l'entalpia di formazione di ossigeno gassoso. L'equazione dell'entalpia deriva dalla combinazione di due concetti:entalpia standard di formazione e legge di Hess. Il primo termine rappresenta le entalpie standard di formazione dei prodotti;la formazione di anidride carbonica da carbonio e ossigeno, equazione 1, e la formazione di acqua da idrogeno e ossigeno, equazione 2.

Le entalpie standard di formazione per anidride carbonica e acqua note sono, rispettivamente, 393, 5 kilojoule negative e 241, 8 kilojoule negative. Poiché la combustione produce 4 moli di anidride carbonica, delta H1 sarebbe l'entalpia standard di formazione di anidride carbonica per 4, che sarebbe negativa per 1574 kilojoule. La combustione produce anche 2 moli di acqua, quindi la variazione di entalpia, delta H2, sarebbe l'entalpia standard di formazione dell'acqua per 2, che sarebbe 483, 6 kilojoule negativa.

Ciò produce 2058 kilojoule negativi come entalpia standard netta di formazione dei prodotti. Il secondo termine rappresenta la decomposizione dell'acetilene in carbonio e idrogeno, equazione 3. Questo è il contrario della reazione per l'entalpia standard di formazione del reagente, e quindi il suo valore di entalpia, 227, 4 kilojoule, è preceduto da un segno negativo, che può essere visto anche nell'equazione dell'entalpia.

Poiché la reazione consuma 2 moli di acetilene, delta H3 sarebbe il negativo dell'entalpia standard di formazione dell'acetilene per 2, che equivale a 453, 4 kilojoule negativi. L'entalpia di formazione standard dell'ossigeno è zero. Pertanto, l'entalpia standard netta di formazione dei reagenti è 453, 4 kilojoule negativa.

Ricordate dalla legge di Hess, che se una reazione ad una fase viene eseguita in più fasi, la somma delle entalpie di ciascuna fase è uguale alla variazione di entalpia netta. Sostituendo i valori delle entalpie di formazione nell'equazione si ottiene l'entalpia della reazione come 2511 kilojoule negativi.

<ol>
	<li>Davis, Thomas W. "Un malinteso comune della legge di Hess". Journal of Chemical Education 28, n. 11 (1951): 584.</li>
	<li>Lee, A. L., H. L. Feldkirchner, F. C. Schora e J. J. Henry. "Calore di reazione dell'idrogeno e del carbone". Industrial & Engineering Chemistry Process Design and Development 7, n. 2 (1968): 244-249.</li>
</ol>

entalpie di reazione

La legge di Hess pu essere utilizzata per determinare la variazione di entalpia di qualsiasi reazione se sono disponibili le corrispondenti entalpie di formazione dei reagenti e dei prodotti. La reazione principale pu essere suddivisa in reazioni graduali: (i) decomposizioni dei reagenti nei loro elementi componenti, per cui le variazioni di entalpia sono proporzionali al negativo delle entalpie di formazione dei reagenti, −ΔHf°(reagenti), seguita da (ii) ricombinazioni degli elementi (ottenute nel passaggio 1) per dare i prodotti, con variazioni di entalpia proporzionali alle entalpie di formazione dei prodotti,  ΔHf° (prodotti). La variazione di entalpia standard della reazione complessiva quindi pari a: (ii) la somma delle entalpie standard di formazione di tutti i prodotti pi (i) la somma dei negativi delle entalpie standard di formazione dei reagenti, come dato dalla seguente equazione, dove ∑ rappresenta “la somma di? e n sta per i coefficienti stechiometrici.

<img alt="Eq1" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-01.png" style="height: 40px;" />

L'equazione viene solitamente riorganizzata leggermente per essere scritta come segue:

<img alt="Eq2" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-02.png" style="height: 40px;" />

L'esempio seguente mostra in dettaglio perch questa equazione valida e come utilizzarla per calcolare la variazione di entalpia standard per una reazione:

<img alt="Eq3" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-03.png" style="height: 40px;" />

Qui viene utilizzata la forma speciale della legge di Hess e i valori del calore di formazione dei reagenti e dei prodotti: ΔHf° (HNO_3) =−206.64 kJ/mol; ΔHf°(NO) = +90,2 kJ/mol; ΔHf° (NO_2) = +33 kJ/mol; ΔHf° (H_2O) = −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;" />

Questo testo adattato da <a href="https://openstax.org/books/chemistry-2e/pages/5-3-enthalpy">Openstax, Chemistry 2e, Section 5.3: Enthalpy.</a>

Entalpie di reazione

La legge di Hess può essere utilizzata per determinare la variazione di entalpia di qualsiasi reazione se le corrispondenti entalpie di formazione dei reagenti e dei prodotti ...

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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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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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