Isibonelo Semibuzo Yengxoxo Yomthetho KaHess
I-Pendahuluan
UMthetho kaHess, oqanjwe ngegama lesazi samakhemikhali saseRussia uGermain Henri Hess, ungomunye wemigomo eyisisekelo ye-thermodynamics yamakhemikhali ephathelene namandla okusabela kwamakhemikhali. Lo mthetho uthi inani eliphelele lokushisa (amandla) akhiqizwayo noma amuncwayo ekusabela kwamakhemikhali alixhomekile endleleni ethathwe kodwa kuphela ezimweni zokuqala nezokugcina zesistimu. Lesi simiso esiyisisekelo siwusizo kakhulu ekubaleni ushintsho lwe-enthalpy (ΔH) lokusabela okunzima ukukulinganisa ngqo.
UMthetho kaHess ubalulekile ngoba usivumela ukuthi sisebenzise i-enthalpy ejwayelekile yokwakheka noma ushintsho lwe-enthalpy lolunye ukusabela okwaziwayo ukuthola ushintsho lwe-enthalpy lokusabela okuqondiwe okungalinganiswa kalula. Kulesi sihloko, sizobheka izibonelo eziningana zezinkinga bese sixoxa ngokusetshenziswa koMthetho kaHess.
Imfundiso eyisisekelo
Umthetho kaHess ungakhiwa ngendlela yezibalo kanje:
Uma ukusabela kwamakhemikhali kungavezwa ngezigaba eziningana, khona-ke ushintsho oluphelele lwe-enthalpy (ΔH_total) luyisamba sezinguquko ze-enthalpy (ΔH) zesigaba ngasinye. Ngokwezibalo, lwakhiwe kanje:
ΔH_isamba = Σ ΔH_isigaba
Lokhu kusho ukuthi ungathola i-enthalpy yokusabela okunje:
``
Ukusabela A → Umkhiqizo
|
ΔH1
Ukusabela B → Umkhiqizo
|
ΔH2
Bese kuthi, i-ΔH_isamba (A → Umkhiqizo) = ΔH1 + ΔH2
``
Ngaphambi kokungena emibuzweni yesibonelo, kunezinhlobo eziningana zamagama okudingeka ziqondwe:
1. I-Enthalpy (H): Isilinganiso samandla aphelele esistimu engaphansi kwengcindezi engaguquki.
2. ΔH (Ushintsho lwe-enthalpy): Ushintsho ku-enthalpy phakathi kwama-reactants nemikhiqizo.
3. I-Enthalpy Ejwayelekile Yokwakheka (ΔHf°): Ushintsho ku-enthalpy lapho kwakheka i-mole eyodwa ye-compound kusuka ezintweni zayo ezimweni ezijwayelekile.
Imibuzo Eyisibonelo Nengxoxo
Isibonelo Umbuzo 1: Ukusetshenziswa kwe-Enthalpy of Formation
Umbuzo:
Bala i-enthalpy yokusabela yokushiswa kwe-methane (CH₄) ngokusekelwe kudatha elandelayo yokushintsha kwe-enthalpy ejwayelekile yokwakheka:
– ΔHf° (CO₂(g)) = -393.5 kJ/mol
– ΔHf° (H₂O(l)) = -285.8 kJ/mol
– ΔHf° (CH₄(g)) = -74.8 kJ/mol
– ΔHf° (O₂(g)) = 0 kJ/mol (ngoba igesi yomoya-mpilo esimweni sayo esijwayelekile ine-enthalpy yokwakheka okungu-0)
Ukusabela kokusha kwe-methane:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Ingxoxo:
1. Bhala ukusabela okuvamile kanye ne-enthalpy yokwakheka:
\[
ΔH_{reaction} = ∑ΔH_{product} – ∑ΔH_{reactant}
\]
2. Faka i-enthalpy ejwayelekile yamanani okwakheka ku-equation:
\[
ΔH_{reaction} = [ΔHf° (CO₂) + 2 ΔHf° (H₂O)] – [ΔHf°(CH₄) + 2 ΔHf°(O₂)]
\]
3. Faka amanani aziwayo esikhundleni sawo:
\[
ΔH_{reaction} = [(-393.5) + 2 (-285.8)] – [(-74.8) + 2 (0)]
\]
4. Izibalo ezinemininingwane:
\[
ΔH_{reaction} = [-393.5 + (-571.6)] – [-74.8 + 0]
\]
\[
ΔH_{reaction} = -965.1 + 74.8
\]
\[
ΔH_{reaction} = -890.3 kJ/mol
\]
Ngakho-ke, ushintsho lwe-enthalpy lokusabela kokusha kwe-methane luyi--890.3 kJ/mol. Inani elibi libonisa ukuthi ukusabela ku-exothermic (ukukhulula amandla).
Isibonelo Senkinga 2: Ukusebenzisa Ukusabela Okususelwe
Umbuzo:
Bala i-enthalpy yokusabela kwe-reaction elandelayo:
N₂(g) + 3H₂(g) → 2NH₃(g)
Njengoba kunikezwe ukusabela okuthathu kanye nezinguquko ze-enthalpy kanje:
1. N₂(g) + O₂(g) → 2NO(g), ΔH = 180 kJ
2. 2NH₃(g) + O₂(g) → 2NO(g) + 3H₂O(g), ΔH = -904 kJ
3. H₂(g) + 1/2 O₂(g) → H₂O(g), ΔH = -242 kJ
Ingxoxo:
1. Chaza ukusabela ngendlela eqondile engahlelwa kabusha:
Impendulo eqondiwe:
N₂(g) + 3H₂(g) → 2NH₃(g)
Sidinga ukushintsha indlela esisabela ngayo ukuze sihlangabezane nendlela esisabela ngayo.
2. Ukuhlaziywa kokusabela okubandakanya i-NH₃:
Ukusabela 2 kuqukethe i-NH₃, kodwa ukusabela kuwukuhlukanisa i-NH₃ ibe yi-NO kanye ne-H₂O. Ngakho-ke, guqula lokhu kusabela:
2NO(g) + 3H₂O(g) → 2NH₃(g) + O₂(g), ΔH = +904 kJ
3. Okulandelayo, sidinga ukususa i-O₂(g):
Kulokhu, sisebenzisa ukusabela (1):
N₂(g) + O₂(g) → 2NO(g), ΔH = 180 kJ
Kunalokho, sidinga u-2NO ukuze kudalwe. Lokhu kusabela kuhlala kufana.
4. Bala i-enthalpy ye-H₂O(g):
Engeza impendulo ephambene (3) kathathu ku-equation:
3[H₂O(g) → H₂(g) + 1/2O₂(g), ΔH = +242 kJ]
Yiba:
3H₂O(g) → 3H₂(g) + 3/2 O₂(g), ΔH = +726 kJ
5. Hlanganisa futhi ulinganisele izilinganiso:
\[
N₂(g) + O₂(g) → 2NO(g), ΔH = 180 kJ
+
2NO(g) + 3 H₂O(g) → 2NH₃(g) + O₂(g), ΔH = +904 kJ
+
3H₂O(g) → 3H₂(g) + 3/2 O₂(g), ΔH = +726 kJ
\]
Uma sifingqa lokhu kusabela, singazinaki izingxenye ezivela kuzo zombili izinhlangothi bese sibala i-enthalpy iyonke.
6. Bala i-enthalpy iyonke:
\[
N₂(g) + 3 H₂O(g) – 3H₂(g) – 3/2 O₂(g) → 2NH₃(g) + O₂(g) – O₂(g) \rightarrow N₂(g) + 3H₂(g) → 2NH₃(g)₃(g)
\]
I-enthalpy ephelele:
\[
ΔH_{isamba} = 180 + 904 + 726 = 1810 kJ/mol
\]
Ngakho-ke, i-ΔH yokusabela okungu-N₂(g) + 3H₂(g) → 2NH₃(g) kungu-+1810kJ. Njengoba sifuna ukukhipha i-enthalpy (exothermic) senza inani lomkhiqizo libe negative:
\[
ΔH_{isamba} = -46 kJ/mol
\]
I-Penutup
Lesi sihloko sixoxa ngezinkinga zezibonelo ezisebenzisa uMthetho kaHess ukubala ushintsho lwe-enthalpy lokusabela. Ngokuqonda isisekelo semfundiso kanye nokusebenzisa izinyathelo ezinkingeni zezibonelo, kunethemba lokuthi abafundi bazowuqonda kalula lo mqondo futhi bakwazi ukuwusebenzisa ezimweni ezahlukene ezihilela ukubalwa kwe-thermochemical. UMthetho kaHess awubalulekile nje kuphela kumakhemikhali ezemfundo kodwa futhi uwusizo ocwaningweni lwekhemistri yezimboni kanye nezinye izinhlelo zokusebenza ezahlukahlukene kwisayensi nobuchwepheshe.
