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

Voor een reactie die onder standaardomstandigheden plaatsvindt, wordt een algemene vergelijking gebruikt om de standaard enthalpie-verandering van de reactie te berekenen. Deze vergelijking wordt opgelost door het verschil te vinden tussen de som van de standaard vormingsenthalpie van de producten en de som van de standaard vormingsenthalpie van de reactanten, elk vermenigvuldigd met hun stoichiometrische coëfficiënt. Overweeg de verbranding van 2 mol acetyleengas met 5 mol zuurstofgas om onder standaardomstandigheden 4 mol kooldioxidegas en 2 mol waterdamp te vormen.De enthalpie voor de reactie is gelijk aan de som van vier keer de enthalpie van de vorming van kooldioxidegas en twee keer de enthalpie van de vorming van waterdamp, minus de som van twee keer de enthalpie van de vorming van acetyleengas en vijf keer de enthalpie van de vorming van zuurstofgas. De enthalpie-vergelijking wordt afgeleid door het combineren van twee concepten:standaard vormingsenthalpie en de wet van Hess. De eerste term geeft de standaard vormingsenthalpie van de producten weer;de vorming van kooldioxide uit koolstof en zuurstof vergelijking 1, en de vorming van water uit waterstof en zuurstof vergelijking 2.De bekende standaard vormingsenthalpie voor kooldioxide en water zijn respectievelijk negatief 393, 5 kilojoule en negatief 241, 8 kilojoule. Aangezien de verbranding 4 mol kooldioxide produceert, zou delta H1 de standaard enthalpie zijn van vorming van kooldioxide maal 4, wat negatief 1574 kilojoule zou zijn. De verbranding produceert ook 2 mol water, dus de enthalpie-verandering delta H2 zou de standaard enthalpie zijn van de vorming van water maal 2, wat negatief 483, 6 kilojoule zou zijn.Dit levert een negatieve 2058 kilojoule op als de netto standaard vormingsenthalpie van de producten. De tweede term staat voor de ontleding van acetyleen in koolstof en waterstof vergelijking 3. Dit is het omgekeerde van de reactie voor de standaard-vormingsenthalpie van de reactant, en dus wordt zijn enthalpie-waarde, 227, 4 kilojoule, voorafgegaan door een negatief teken, dat ook te zien is in de enthalpie-vergelijking.Omdat de reactie 2 mol acetyleen verbruikt, zou delta H3 het negatief zijn van de standaard enthalpie van vorming van acetyleen maal 2, wat gelijk is aan negatief 453, 4 kilojoule. Standaard formatie-enthalpie van zuurstof is nul. Daarom is de netto standaard vormingsenthalpie van de reactanten min 453, 4 kilojoule.Bedenk uit de wet van Hess dat als een eenstapsreactie in meerdere stappen wordt uitgevoerd, de som van de enthalpie van elke stap gelijk is aan de netto enthalpie-verandering. Door de waarden voor de formatie-enthalpie in de vergelijking te vervangen, wordt de enthalpie van de reactie min 2511 kilojoule.

enthalpieën van reactie

De wet van Hess kan worden gebruikt om de enthalpieverandering van elke reactie te bepalen als de overeenkomstige vormingsenthalpieën van de reactanten en producten beschikbaar zijn. De hoofdreactie kan worden onderverdeeld in stapsgewijze reacties: (i) ontledingen van de reactanten in hun samenstellende elementen, waarbij de enthalpieveranderingen evenredig zijn met het negatieve van de vormingsenthalpieën van de reactanten, −ΔHf°(reactanten), gevolgd door (ii) recombinaties van de elementen (verkregen in stap 1) om de producten te geven, waarbij de enthalpieveranderingen evenredig zijn met de vormingsenthalpieën van de producten, ΔHf° (producten). De standaard enthalpieverandering van de totale reactie is daarom gelijk aan: (ii) de som van de standaard vormingsenthalpieën van alle producten plus (i) de som van de negatieven van de standaard vormingsenthalpieën van de reactanten, zoals gegeven door de volgende vergelijking, waarbij ∑ staat voor “de som van” en n staat voor de stoichiometrische coëfficiënten. <img alt="Eq1" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-01.png" style="height: 40px;"> De vergelijking wordt gewoonlijk enigszins herschikt en als volgt geschreven: <img alt="Eq2" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-02.png" style="height: 40px;"> Het volgende voorbeeld laat in detail zien waarom deze vergelijking geldig is en hoe je deze kunt gebruiken om de standaard enthalpieverandering voor een reactie te berekenen: <img alt="Eq3" src="https://www.jove.com/files/core/CDNSource/lessons/802/802_Equations-03.png" style="height: 40px;"> Hier wordt de speciale vorm van de wet van Hess en de vormingswarmtewaarden voor de reactanten en producten gebruikt: 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;"> Deze tekst is aangepast van <a href="https://openstax.org/books/chemistry-2e/pages/5-3-enthalpy">Openstax, Chemistry 2e, Section 5.3: Enthalpy.</a>

Enthalpieën van reactie

Hess&rsquo;s law can be used to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products ...

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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...

key terms and concepts

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Calculating Reaction Heat with Hess's Law

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