Name: calculating reaction heat with hess's law
Uploaded: 2020-09-03T13:38:27.000Z
Duration: 3 min 33 s
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Enthalpies of Reaction

eine allgemeine Gleichung verwendet. die Differenz zwischen der Summe der Standardenthalpien der Bildung der Produkte und der Summe der Standardbildungsenthalpien der Reaktanten findet, und Betrachten Sie die Verbrennung von 2 Mol Acetylengas mit 5 Mol Sauerstoffgas zur Bildung von 4 Mol Kohlendioxid-Gas und 2 Mol Wasserdampf unter Standardbedingungen. der vierfachen Bildungsenthalpie von Kohlenstoffdioxid-Gas und zwei Mal die Enthalpie der Bildung von Wasserdampf, minus der Summe der zweifachen-Bildungsenthalpie von Sauerstoffgas.Die Enthalpie-Gleichung ergibt sich aus der Kombination zweier Konzepte:der Standardbildungsenthalpie und dem Hess'schem Gesetz. der Bildung der Produkte;die Bildung von Kohlenstoffdioxid aus Kohlenstoff und Sauerstoff Gleichung 1, und die Bildung von Wasser aus Wasserstoff und Sauerstoff Gleichung 2. und Wasser sind 393, 5 Kilojoule und 241, 8 Kilojoule.Da bei der Verbrennung 4 Mol Kohlenstoffdioxid entstehen, der Bildung von Kohlendioxid mal 4, Bei der Verbrennung entstehen auch 2 Mol Wasser, die Standardenthalpie der Wasserbildung mal 2, was einem Minus von 483, 6 Kilojoule entspricht. Dies ergibt 2058 Kilojoule als Netto-Enthalpie der Bildung der Produkte. in Kohlenstoff und Wasserstoff Gleichung 3.der Bildungsenthalpie des Reaktanten und damit ist seinem Enthalpiewert von 227, 4 Kilojoule ein negatives Vorzeichen vorangestellt, was sich auch in der Enthalpiegleichung zeigt. Weil bei der Reaktion 2 Mol Acetylen verbraucht werden, ist delta H3 das Negative der Standardenthalpie der Bildung von Acetylen mal 2, was einem Negativwert von 453, 4 Kilojoule entspricht. Die Standardbildungsenthalpie von Sauerstoff ist Null.der Bildung der Reaktanten 453, 4 negative Kilojoule. Erinnern Sie sich an das Hess'sche Gesetz, wenn eine Ein-Schritt-Reaktion in die Gleichung, ergibt die Enthalpie der Reaktion 2511 negative Kilojoule.

<ol>
	<li>Davis, Thomas W. "Ein häufiges Missverständnis des Hessschen Gesetzes." Zeitschrift für chemische Bildung 28, Nr. 11 (1951): 584.</li>
	<li>Lee, A. L., H. L. Feldkirchner, F. C. Schora und J. J. Henry. "Reaktionswärme von Wasserstoff und Kohle." Industrial & Engineering Chemistry Process Design and Development 7, Nr. 2 (1968): 244-249.</li>
</ol>

reaktionsenthalpien

Mit dem Hess'schen Gesetz lässt sich die Enthalpieänderung jeder Reaktion bestimmen, solange die entsprechenden Bildungsenthalpien der Reaktanten und Produkte bekannt sind. Die Hauptreaktion kann in schrittweise Reaktionen unterteilt werden: (i) Aufteilen der Reaktanten in ihre Bestandteile wobei die Enthalpieänderungen proportional zur Stoffmenge sind und das umgekehrte Vorzeichen der Bildungsenthalpien der Reaktanten erhalten, −ΔHf°(Reaktanten), gefolgt von (ii) zusammenfassen der Bildungsenthalpien der Produkte wobei die Enthalpieänderungen wieder die Bildungsenthalpien proportional zur Stoffmenge der Produkte, ΔHf° (Produkte), sind. Die Standardenthalpieänderung der Gesamtreaktion ist daher gleich: (iii) der Summe der Standardbildungsenthalpien aller Produkte plus (i) der Summe der Negativen Standardbildungsenthalpien der Reaktanten, wie gegeben durch die folgende Gleichung, wobei ∑ „die Summe“ darstellt und n für die stöchiometrischen Koeffizienten steht. <img alt="Eq1" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-01.png" style="height: 40px;"> Die Gleichung wird normalerweise leicht umgestellt und lautet wie folgt: <img alt="Eq2" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-02.png" style="height: 40px;"> Das folgende Beispiel zeigt im Detail, warum diese Gleichung gültig ist und wie man sie zur Berechnung der Standardenthalpieänderung für eine Reaktion verwenden kann: <img alt="Eq3" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-03.png" style="height: 40px;"> Dabei werden die spezielle Form des Hess'schen Gesetzes und die Bildungswärmenwerte für die Reaktanten und Produkte verwendet: 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;"> Dieser Text wurde angepasst von <a href="https://openstax.org/books/chemistry-2e/pages/5-3-enthalpy">Openstax, Chemistry 2e, Section 5.3: Enthalpy.</a>

Reaktionsenthalpien

Das Hess-Gesetz kann verwendet werden, um die Enthalpieänderung einer Reaktion zu bestimmen, wenn die entsprechenden Bildungsenthalpien der Reaktanten und Produkte ...

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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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