Imibuzo yemizekelo exoxa ngoMthetho kaHess
I-Pendahuluan
UMthetho kaHess, oqanjwe ngegama lesazi samakhemikhali saseRashiya uGermain Henri Hess, ngomnye wemigaqo esisiseko ye-thermodynamics yeekhemikhali enxulumene namandla eempendulo zeekhemikhali. Lo mthetho uthi isixa esipheleleyo sobushushu (amandla) esiveliswayo okanye esifunxwayo kwimpendulo yeekhemikhali asixhomekekanga kwindlela ethathiweyo kodwa kuphela kwiimeko zokuqala nezokugqibela zenkqubo. Lo mgaqo usisiseko uluncedo kakhulu ekubaleni utshintsho lwe-enthalpy (ΔH) lweempendulo ezinzima ukuzilinganisa ngokuthe ngqo.
Umthetho kaHess ubalulekile kuba usivumela ukuba sisebenzise i-enthalpy eqhelekileyo yokwakheka okanye utshintsho lwe-enthalpy kwenye impendulo eyaziwayo ukuze sifumane utshintsho lwe-enthalpy lwempendulo ekujoliswe kuyo engenakulinganiselwa lula. Kweli nqaku, siza kujonga imizekelo eliqela yeengxaki size sixoxe ngokusetyenziswa koMthetho kaHess.
Ithiyori esisiseko
Umthetho kaHess unokwenziwa ngendlela yezibalo ngolu hlobo lulandelayo:
Ukuba i-chemical reaction ingabonakaliswa ngamanqanaba aliqela, ngoko ke utshintsho olupheleleyo lwe-enthalpy (ΔH_total) luyimbumba yotshintsho lwe-enthalpy (ΔH) yesigaba ngasinye. Ngokwezibalo, yenziwe ngolu hlobo:
ΔH_total = Σ ΔH_stage
Oku kuthetha ukuba ungayifumana i-enthalpy yempendulo efana nale:
``
Impendulo A → Imveliso
|
ΔH1
Impendulo B → Imveliso
|
ΔH2
Emva koko, i-ΔH_total (A → Imveliso) = ΔH1 + ΔH2
``
Ngaphambi kokuba singene kwimibuzo yomzekelo, kukho amagama aliqela ekufuneka aqondwe:
1. I-Enthalpy (H): Umlinganiselo wamandla ewonke enkqubo ephantsi koxinzelelo olungaguqukiyo.
2. ΔH (Utshintsho lwe-enthalpy): Utshintsho kwi-enthalpy phakathi kwee-reactants kunye neemveliso.
3. I-Enthalpy eQhelekileyo yoBume (ΔHf°): Utshintsho kwi-enthalpy xa i-mole enye ye-compound yenziwe ngezinto zayo kwiimeko ezisemgangathweni.
Imibuzo yeSampuli kunye neNgxoxo
Umzekelo Umbuzo 1: Ukusetyenziswa kwe-Enthalpy yoBume
Umbuzo:
Bala i-reaction enthalpy yokutsha kwe-methane (CH₄) ngokusekelwe kwidatha elandelayo yotshintsho oluqhelekileyo lwe-enthalpy:
– 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 (kuba igesi yeoksijini kwimeko yayo eqhelekileyo ine-enthalpy yokwakheka kwe-0)
Impendulo yokutsha kweMethane:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
IsiPembahasan:
1. Bhala impendulo ngokubanzi kunye ne-enthalpy yokwakheka:
\[
ΔH_{reaction} = ∑ΔH_{product} – ∑ΔH_{reactant}
\]
2. Faka i-enthalpy eqhelekileyo yamaxabiso okwakheka kwi-equation:
\[
ΔH_{reaction} = [ΔHf° (CO₂) + 2 ΔHf° (H₂O)] – [ΔHf°(CH₄) + 2 ΔHf°(O₂)]
\]
3. Faka amaxabiso aziwayo endaweni yawo:
\[
ΔH_{reaction} = [(-393.5) + 2 (-285.8)] – [(-74.8) + 2 (0)]
\]
4. Ubalo oluneenkcukacha:
\[
ΔH_{reaction} = [-393.5 + (-571.6)] – [-74.8 + 0]
\]
\[
ΔH_{reaction} = -965.1 + 74.8
\]
\[
ΔH_{reaction} = -890.3 kJ/mol
\]
Ngoko ke, utshintsho lwe-enthalpy kwimpendulo yokutsha kwe-methane yi--890.3 kJ/mol. Ixabiso elibi libonisa ukuba impendulo yi-exothermic (ukukhulula amandla).
Umzekelo Ingxaki 2: Ukusebenzisa Iimpendulo Ezivelayo
Umbuzo:
Bala i-enthalpy yempendulo kule mpendulo ilandelayo:
N₂(g) + 3H₂(g) → 2NH₃(g)
Ngenxa yeempendulo ezintathu kunye notshintsho lwe-enthalpy ngolu hlobo lulandelayo:
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
IsiPembahasan:
1. Chaza impendulo ngendlela ethe tye enokuhlelwa ngokutsha:
Impendulo ekujoliswe kuyo:
N₂(g) + 3H₂(g) → 2NH₃(g)
Sifanele silawule impendulo enikiweyo ukuze sihlangabezane nempendulo ekujoliswe kuyo.
2. Uhlalutyo lweempendulo ezibandakanya i-NH₃:
I-Reaction 2 ine-NH₃, kodwa i-reaction kukwahlulahlula i-NH₃ ibe yi-NO kunye ne-H₂O. Ke ngoko, guqula le mpendulo:
2NO(g) + 3H₂O(g) → 2NH₃(g) + O₂(g), ΔH = +904 kJ
3. Okulandelayo, kufuneka sisuse i-O₂(g):
Kule nto, sisebenzisa i-reaction (1):
N₂(g) + O₂(g) → 2NO(g), ΔH = 180 kJ
Endaweni yoko, sidinga i-2NO ukuze iveliswe. Le mpendulo ihlala ifana.
4. Bala i-enthalpy ye-H₂O(g):
Yongeza impendulo echaseneyo (3) kathathu kwi-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 kwaye ulinganisele ii-equation:
\[
N₂(g) + O₂(g) → 2NO(g), ΔH = 180 kJ
+
2NO(g) + 3 H₂O(g) → 2NH₃(g) + O₂(g), ΔH = +904 kJ
+
3H₂O(g) → 3H₂(g) + 3/2 O₂(g), ΔH = +726 kJ
\]
Xa sishwankathela ezi mpendulo, singazihoyi izinto ezibonakala kumacala omabini size sibale i-enthalpy iyonke.
6. Bala i-enthalpy iyonke:
\[
N₂(g) + 3 H₂O(g) – 3H₂(g) – 3/2 O₂(g) → 2NH₃(g) + O₂(g) – O₂(g) \utolo lwasekunene N₂(g) + 3H₂(g) → 2NH₃(g)
\]
I-enthalpy iyonke:
\[
ΔH_{ixabiso lilonke} = 180 + 904 + 726 = 1810 kJ/mol
\]
Ngoko ke, i-ΔH yempendulo i-N₂(g) + 3H₂(g) → 2NH₃(g) yi +1810kJ. Ekubeni sifuna ukukhupha i-enthalpy (exothermic) senza ixabiso lemveliso libe libi:
\[
ΔH_{ixabiso lilonke} = -46 kJ/mol
\]
Ukuvala
Eli nqaku lixoxa ngemiba yemizekelo esebenzisa uMthetho kaHess ukubala utshintsho lwe-enthalpy yempendulo. Ngokuqonda isiseko sethiyori kunye nokusebenzisa amanyathelo kwiingxaki zemizekelo, kuyathenjwa ukuba abafundi baya kuyiqonda lula le ngcamango baze bakwazi ukuyisebenzisa kwiimeko ezahlukeneyo ezibandakanya ukubalwa kwe-thermochemical. UMthetho kaHess awubalulekanga nje kuphela kwikhemistri yezemfundo kodwa ukwaluncedo kuphando lwekhemistri yezoshishino kunye nezinye iindlela ezahlukeneyo zokusebenza kwisayensi nakwitekhnoloji.
