(Go: >> BACK << -|- >> HOME <<)

Hess's law: Difference between revisions

Content deleted Content added
m →‎Examples: {{serif}}
 
(36 intermediate revisions by 25 users not shown)
Line 1:
{{Short description|Observation that total enthalpy change of a chemical reaction is independent of the steps taken}}
[[Image:Hess cycle.svg|thumb|A representation of Hess's law (where H represents enthalpy)]]
 
'''Hess'sHess’ law of constant heat summation''', also known simply as '''Hess' law''', is a relationship in [[physical chemistry]] named after [[Germain Henri Hess|Germain Hess]], a [[Switzerland|Swiss]]-born [[Russia]]n [[chemist]] and [[physician]] who published it in 1840. The law states that the total [[enthalpy]] change during the complete course of a [[chemical reaction]] is independent of the numbersequence of steps taken.<ref name= "Aakash1">{{cite book|title=Chemistry for ISEET - Volume 1, Part A|year=2012|publisher=Varsity Education Management Limited|location=[[Hyderabad, India]]|author=Mannam Krishnamurthy|edition=2012|author2=Subba Rao Naidu|editor=Lokeswara Gupta|page=244|chapter=7}}</ref><ref>{{cite web|title=Hess' Law - Conservation of Energy|url=http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html|publisher=University of Waterloo|access-date=12 January 2014|archive-url=https://web.archive.org/web/20150109195637/http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html|archive-date=9 January 2015|url-status=dead}}</ref>
 
Hess' law is now understood as an expression of the [[principlefact of conservation of energy]], also expressed inthat the [[firstStandard lawenthalpy of thermodynamics]], and the fact that the [[reaction|enthalpy]] of a chemical process]] is independent of the path taken from the initial to the final state (i.e. enthalpy is a [[state function]]). ReactionAccording to the [[first law of thermodynamics]], the enthalpy changeschange in a system due to a reaction at [[Isobaric process|constant pressure]] is equal to the heat absorbed (or the negative of the heat released), which can be determined by [[calorimetry]] for many reactions. The values are usually stated for processesreactions with the same initial and final temperatures and pressures, although the(while conditions canare allowed to vary during the reactioncourse of the reactions). Hess' law can be used to determine the overall energy required for a chemical reaction, when itthat can be divided into synthetic steps that are individually easier to characterize. This affords the compilation of [[standard enthalpy of formation|standard enthalpies of formation]], thatwhich may be used asto apredict basisthe toenthalpy designchange in complex synthesessynthesis.
 
==Theory==
Hess'Hess’ law states that the change of enthalpy in a chemical reaction is the same regardless of whether the reaction takes place in one step or several steps, provided the initial and final states of the reactionsreactants and products are the same. Enthalpy is an [[Intensive and extensive properties|extensive property]], meaning that its value is proportional to the system size.<ref>{{cite book |last1=Engel |first1=Thomas |last2=Reid |first2=Philip |title=Physical Chemistry |date=2006 |publisher=Pearson / Benjamin Cummings |isbn=0-8053-3842-X |page=6 |quote=A variable ... proportional to the size of the system is referred to as an extensive variable.}}</ref> Because of this, the enthalpy change is proportional to the number of moles participating in a given reaction.
 
In other words, if a chemical change takes place by several different routes, the overall enthalpy change is the same, regardless of the route by which the chemical change occurs (provided the initial and final condition are the same). If this were not true, then one could violate the [[first law of thermodynamics]].
 
Hess' law allows the [[enthalpy]] change (ΔHΔ''H'') for a reaction to be calculated even when it cannot be measured directly. This is accomplished by performing basic algebraic operations based on the [[chemical equation]]s of reactions using previously determined values for the enthalpies of formation.
 
Combination of chemical equations leads to a net or overall equation. If the enthalpy changechanges isare known for eachall equation,the equations in the resultsequence, their sum will be the enthalpy change for the net equation. If the net enthalpy change is negative (ΔH<submath>\Delta H_\text{net}<0</submath> < 0), the reaction is [[Exothermic reaction|exothermic]] and is more likely to be [[Gibbs free energy|spontaneous]]; positive ΔHΔ''H'' values correspond to [[endothermic]] reactions. ([[Entropy]] also plays an important role in determining spontaneity, as some reactions with a positive enthalpy change are nevertheless spontaneous due to an entropy increase in the reaction system.)
 
==Use of enthalpies of formation==
Hess' law states that enthalpy changes are additive. Thus the ΔH for a single reaction
Hess' law states that enthalpy changes are additive. Thus the value of the [[standard enthalpy of reaction]] can be calculated from [[Standard enthalpy of formation|standard enthalpies of formation]] of products and reactants as follows:
:<math>\Delta H_{\text {reaction}}^\ominus = \sumsum_{i} a_i\Delta H_Delta_{\mathrmtext {f}} \,(\textH_{p})products}^{\ominus} - \sumsum_{i} b_i\Delta H_Delta_{\mathrmtext {f}} \,(\textH_{r})reactants}^{\ominus}</math>
whereHere, the first sum is over all products and the second over all reactants, <math>\Deltaa_i</math> and H_{f(p)}<math>b_i</math> isare the [[standard enthalpystoichiometric of formation|enthalpy of formationcoefficients]] of productproducts pand reactants respectively, <math>\DeltaDelta_{\text {f}} H_{products}^{\ominus}</math> and<math>\Delta_{\text {f(r)}} H_{reactants}^{\ominus}</math> isare the enthalpystandard enthalpies of formation of reactantproducts and reactants rrespectively, and the <sup><s>o</s></sup> superscript indicates [[standard state]] values. This may be considered as the sum of two (real or fictitious) reactions:
:Reactants → Elements (in their standard states)
:<math>\Delta HH_{\text{RE}}^{\ominus} = - \sumsum_{i} b_i\Delta H_Delta_{\mathrmtext {f}} \,(\textH_{r})reactants}^{\ominus}</math>
and Elements → Products <br>
:<math>\Delta HH_{\text{EP}}^{\ominus} = \sumsum_{i} a_i\Delta H_Delta_{\mathrmtext {f}} \,(\textH_{p})products}^{\ominus}</math>
 
==Examples==
<ol>
<li> Given:
<ol type="a">
<li>C<sub>graphite</sub> + O<sub>2</sub> → CO<sub>2</sub> ({{serif|g}}) ; ( Δ''H'' = −393.5 kJ/mol) (direct step)</li>
<li>C<sub>graphite</sub> + 1/2 O<sub>2</sub> → CO ({{serif|g}}) ; (Δ''H'' = −110.5 kJ/mol)</li>
<li>CO ({{serif|g}}) +1/2 O<sub>2</sub> → CO<sub>2</sub> ({{serif|g}}); (Δ''H'' = −283.020 kJ/mol)</li>
</ol>
Reaction (a) is the sum of reactions (b) and (c), for which the total Δ''H'' = −393.5 kJ/mol, which is equal to Δ''H'' in (a).
 
The difference in the value of Δ''H'' is 0.02 kJ/mol which is due to measurement errors.
</li>
 
<li>Given:
*B<sub>2</sub>O<sub>3</sub> ({{serif|s}}) + 3H<sub>2</sub>O ({{serif|g}}) → 3O<sub>2</sub> ({{serif|g}}) + B<sub>2</sub>H<sub>6</sub> ({{serif|g}}) (Δ''H'' = 2035 kJ/mol)
*H<sub>2</sub>O ({{serif|l}}) → H<sub>2</sub>O ({{serif|g}}) (Δ''H'' = 44 kJ/mol)
*H<sub>2</sub> ({{serif|g}}) + 1/2 O<sub>2</sub> ({{serif|g}}) → H<sub>2</sub>O ({{serif|l}}) (Δ''H'' = −286 kJ/mol)
*2B ({{serif|s}}) + 3H<sub>2</sub> ({{serif|g}}) → B<sub>2</sub>H<sub>6</sub> ({{serif|g}}) (Δ''H'' = 36 kJ/mol)
Find the Δ''H''<sub>f</sub>''H'' of:
*2B ({{serif|s}}) + 3/2 O<sub>2</sub> ({{serif|g}}) → B<sub>2</sub>O<sub>3</sub> ({{serif|s}})
 
After multiplying the equations (and their enthalpy changes) by appropriate factors and reversing the direction when necessary, the result is:
*B<sub>2</sub>H<sub>6</sub> ({{serif|g}}) + 3O<sub>2</sub> ({{serif|g}}) → B<sub>2</sub>O<sub>3</sub> ({{serif|s}}) + 3H<sub>2</sub>O ({{serif|g}}) (Δ''H'' = 2035 × (−1) = −2035 kJ/mol)
*3H<sub>2</sub>O ({{serif|g}}) → 3H<sub>2</sub>O ({{serif|l}}) (Δ''H'' = 44 × (−3) = −132 kJ/mol)
*3H<sub>2</sub>O ({{serif|l}}) → 3H<sub>2</sub> ({{serif|g}}) + (3/2) O<sub>2</sub> ({{serif|g}}) (Δ''H'' = −286 × (−3) = 858 kJ/mol)
*2B ({{serif|s}}) + 3H<sub>2</sub> ({{serif|g}}) → B<sub>2</sub>H<sub>6</sub> ({{serif|g}}) (Δ''H'' = 36 kJ/mol)
 
Adding these equations and canceling out the common terms on both sides, we obtain
*2B ({{serif|s}}) + 3/2 O<sub>2</sub> ({{serif|g}}) → B<sub>2</sub>O<sub>3</sub> ({{serif|s}}) (Δ''H'' = −1273 kJ/mol)
</li>
</ol>
 
==Extension to free energy and entropy==
The concepts of Hess' law can be expanded to include changes in [[entropy]] and in [[Gibbs free energy]], whichsince these are also [[state function]]s. The [[Bordwell thermodynamic cycle]] is an example of such an extension whichthat takes advantage of easily measured [[Chemical equilibrium|equilibria]] and [[redox potential]]s to determine experimentally inaccessible [[Gibbs free energy]] values. Combining ΔGΔ''G''<sup><s>o</s></sup> values from Bordwell thermodynamic cycles and ΔHΔ''H''<sup><s>o</s></sup> values found with Hess'Hess’ law can be helpful in determining entropy values whichthat arehave not been measured directly, and therefore mustneed to be calculated through alternative paths.
 
For the free energy:
:<math>\Delta G_{\text{reaction}}^\ominus = \sum \nu_{\text{p}} \Delta G_{\mathrm f \,(\text{p})}^{\ominus} - \sum \nu_{\text{r}} \Delta G_{\mathrm f \,(\text{r})}^{\ominus}.</math>
 
For [[entropy]], the situation is a little different. Because entropy can be measured as an absolute value, not relative to those of the elements in their reference states (as with ΔHΔ''H''<sup><s>o</s></sup> and ΔGΔ''G''<sup><s>o</s></sup>), there is no need to use the entropy of formation; one simply uses the absolute entropies for products and reactants:
:<math>\Delta S_{\text{reaction}}^\ominus = \sum \nu_{\text{p}} S_{(\text{p})}^{\ominus} - \sum \nu_{\text{r}} S_{(\text{r})}^{\ominus}.</math>
 
==Applications==
Line 67 ⟶ 68:
#Heat changes in phase transitions and [[allotropy|allotropic]] transitions.
#[[Lattice energy|Lattice energies]] of ionic substances by constructing [[Born–Haber cycle]]s if the [[electron affinity]] to form the anion is known, or
#Electron affinities using a Born–Haber cycle with a theoretical [[lattice energy]].
 
==See also==