Imibuzo Yesibonelo Ekhuluma Ngezinguquko Ze-Enthalpy Ezimweni Ezijwayelekile
I-Pendahuluan
Ushintsho lwe-enthalpy luwumqondo oyisisekelo ku-thermochemistry odlala indima ebalulekile ezinqubweni ezahlukene zamakhemikhali. Kulesi sihloko, sizoxoxa ngokuningiliziwe ukuthi singabala futhi siqonde kanjani ushintsho lwe-enthalpy ngaphansi kwezimo ezijwayelekile ngezinkinga eziningana zezibonelo kanye nezingxoxo ezibanzi. Lesi sihloko sizoba usizo kubafundi, abafundi basekolishi, nanoma ubani ofunda i-chemistry ukuze aqonde kangcono lesi sihloko.
Ukuqonda i-Enthalpy kanye nezinguquko zayo
I-Enthalpy (H) amandla aphelele esistimu, aqukethe amandla angaphakathi kanye namandla ahlotshaniswa nokucindezela kanye nomthamo. Endabeni yokusabela kwamakhemikhali, sivame ukuba nesithakazelo ekushintsheni kwe-enthalpy (ΔH), okubonisa ushintsho lwamandla aphelele ngesikhathi sokusabela ekucindezelekeni okungaguquki.
Ushintsho lwe-enthalpy ngaphansi kwezimo ezijwayelekile (ΔH⁰) lushintsho lwe-enthalpy lapho zonke izinto ezisabelayo kanye nemikhiqizo zisesimweni sazo esijwayelekile, okungukuthi, ekucindezelweni okungu-1 atm kanye nokushisa okuvame ukuba ngu-25°C (298 K).
Imibuzo Eyisibonelo Nengxoxo
Umbuzo 1: Ukusha kweMethane
Umbuzo: Bala ushintsho olujwayelekile lwe-enthalpy (ΔH⁰) lokushiswa kwe-mole eyi-1 ye-methane (\(CH_4\)) ngokusekelwe kulesi sibalo esilandelayo:
\[ CH_4(g) + 2O_2(g) \umcibisholo wangakwesokudla CO_2(g) + 2H_2O(l) \]
Kuyaziwa:
– ΔH⁰f \(CH_4(g)\) = -74.8 kJ/mol
– ΔH⁰f \(CO_2(g)\) = -393.5 kJ/mol
– ΔH⁰f \(H_2O(l)\) = -285.8 kJ/mol
Ingxoxo:
Ushintsho olujwayelekile lwe-enthalpy lokusabela kwamakhemikhali lungabalwa kusetshenziswa umthetho kaHess ngale ndlela elandelayo:
\[ \Delta H⁰ = ∑ ΔH⁰f(umkhiqizo) – ∑ ΔH⁰f(i-reactant) \]
Okokuqala, thola i-enthalpy ejwayelekile yokwakheka kwento ngayinye ekuphenduleni:
– \( ΔH⁰f_{CH_4(g)} = -74.8 \) kJ/mol
– \( ΔH⁰f_{CO_2(g)} = -393.5 \) kJ/mol
– \( ΔH⁰f_{H_2O(l)} = -285.8 \) kJ/mol (×2 ngama-moles amabili \(H_2O\))
Bese, bala inani lama-enthalpies okwakheka kwemikhiqizo kanye nama-reactant:
\[ ∑ ΔH⁰f(umkhiqizo) = [-393.5] + [2(-285.8)]
= -393.5 + (-571.6)
= -965.1 \umbhalo{kJ/mol} \]
\[ ∑ ΔH⁰f(i-reactant) = [-74.8] + [0] \]
(Zonke izinto eziyinhloko ezitholakala ngesimo sazo esiyisisekelo zine-enthalpy ejwayelekile yokwakheka okungu-0 kJ/mol)
Bese, bala ushintsho olujwayelekile lwe-enthalpy (ΔH⁰):
\[ ΔH⁰ = -965.1 – (-74.8)
= -965.1 + 74.8
= -890.3 \umbhalo{kJ/mol} \]
Ngakho-ke, ushintsho olujwayelekile lwe-enthalpy lokushiswa kwe-mole eyi-1 ye-methane luyi--890.3 kJ/mol.
Umbuzo 2: Ukusabela Kokwakheka Kwamanzi
Umbuzo: Bala ushintsho olujwayelekile lwe-enthalpy (ΔH⁰) lokwakheka kwamanzi avela ku-hydrogen kanye ne-oxygen ngokusekelwe kulesi sibalo esilandelayo:
\[ 2H_2(g) + O_2(g) \umcibisholo ongakwesokudla 2H_2O(l) \]
Kuyaziwa:
– ΔH⁰f \(H_2O(l)\) = -285.8 kJ/mol
Ingxoxo:
Sidinga ukuthola ushintsho olujwayelekile lwe-enthalpy lokusabela okuvela ezintweni zokuqala kuya emikhiqizweni oyifunayo. Ukusebenzisa i-enthalpy ejwayelekile yokwakheka:
\[ ΔH⁰ = ∑ ΔH⁰f(umkhiqizo) – ∑ ΔH⁰f(i-reactant) \]
Bala i-enthalpy yokwakheka kwemikhiqizo kanye nama-reactants:
\[
\begin{align }
∑ ΔH⁰f(umkhiqizo) kanye = [2(-285.8)] \\
∑ ΔH⁰f(umkhiqizo) kanye = -571.6 \text{ kJ/mol}
\end{align }
\]
\[
ΔH⁰f(H_2(g)) = 0 \umbhalo{ kJ/mol} \\
ΔH⁰f(O_2(g)) = 0 \umbhalo{ kJ/mol} \\
∑ ΔH⁰f(ama-reactants) = [2(0)] + [0] = 0 \text{ kJ/mol}
\]
Bese, bala ushintsho olujwayelekile lwe-enthalpy (ΔH⁰):
\[ ΔH⁰ = -571.6 \umbhalo{ kJ/mol} \]
Ngakho-ke, ushintsho olujwayelekile lwe-enthalpy ekwakhekeni kwamanzi luyi--571.6 kJ/mol.
Umbuzo 3: Ukubola kwe-Nitrogen Dioxide
Umbuzo: Bala ushintsho olujwayelekile lwe-enthalpy (ΔH⁰) lokubola kwe-nitrogen dioxide (\(NO_2\)) ibe yi-nitrogen monoxide gas (\(NO\)) kanye ne-oxygen gas (O₂) ngokusekelwe kulesi sibalo esilandelayo:
\[ 2NO_2(g) \umcibisholo ongakwesokudla 2NO(g) + O_2(g) \]
Kuyaziwa:
– ΔH⁰f \(NO_2(g)\) = 33.2 kJ/mol
– ΔH⁰f \(CHA(g)\) = 90.3 kJ/mol
Ingxoxo:
Izibalo ezifanayo:
\[ ΔH⁰ = ∑ ΔH⁰f(umkhiqizo) – ∑ ΔH⁰f(i-reactant) \]
Bala i-enthalpy yokwakheka komkhiqizo:
\[
\begin{align }
∑ ΔH⁰f(umkhiqizo) kanye = [2( ΔH⁰f_{CHA(g)} )] + [ ΔH⁰f_{O_2(g)}] \\
& = [2(90.3)] + [0] \\
& = 180.6 \umbhalo{kJ/mol}
\end{align }
\]
Bala i-enthalpy yokwakheka kwama-reactants:
\[
\begin{align }
∑ ΔH⁰f(i-reactant) kanye = [2( ΔH⁰f_{NO_2(g)} )] \\
& = [2(33.2)] \\
& = 66.4 \umbhalo{kJ/mol}
\end{align }
\]
Bala ushintsho olujwayelekile lwe-enthalpy (ΔH⁰):
\[ ΔH⁰ = 180.6 – 66.4 = 114.2 \umbhalo{ kJ/mol} \]
Ngakho-ke, ushintsho olujwayelekile lwe-enthalpy lokubola kwe-nitrogen dioxide luyi-114.2 kJ/mol.
Isiphetho
Ukubala izinguquko ezijwayelekile ze-enthalpy (ΔH⁰) kuyindlela ebalulekile ku-thermochemistry. Ngokuqonda ukuthi singazisebenzisa kanjani i-enthalpies ezijwayelekile zokwakheka nokusebenzisa umthetho kaHess, singakwazi ukuthola izinguquko zamandla ekusabeleni okuhlukahlukene kwamakhemikhali. Ngezinkinga zesibonelo ezingenhla, abafundi kulindeleke ukuthi bathole ukuqonda kanye nekhono lokubala izinguquko ze-enthalpy zesabeleni esihlukahlukene samakhemikhali. Lolu lwazi alubalulekile kuphela ezifundweni zemfundo kodwa futhi nasezinhlelweni ezahlukene zezimboni kanye nocwaningo lwesayensi.
