Mehlala ea Lipotso tse Buisanang ka Liphetoho tsa Enthalpy Maemong a Tloaelehileng
Pendahuluan
Phetoho ea enthalpy ke khopolo ea motheo ho thermochemistry e bapalang karolo ea bohlokoa lits'ebetsong tse fapaneng tsa lik'hemik'hale. Sehloohong sena, re tla buisana ka botlalo ka mokhoa oa ho bala le ho utloisisa phetoho ea enthalpy tlas'a maemo a tloaelehileng ka mathata a 'maloa a mehlala le lipuisano tse felletseng. Sehlooho sena se tla ba molemo ho baithuti, baithuti ba koleche, le mang kapa mang ea ithutang k'hemistri ho fumana kutloisiso e betere ea sehlooho sena.
Ho Utloisisa Enthalpy le Liphetoho tsa Eona
Enthalpy (H) ke matla ohle a sistimi, a nang le matla a ka hare le matla a amanang le khatello le bophahamo ba modumo. Tabeng ea lik'hemik'hale, hangata re thahasella phetoho ea enthalpy (ΔH), e bonts'ang phetoho eohle ea matla nakong ea karabelo khatellong e sa khaotseng.
Phetoho ea enthalpy tlas'a maemo a tloaelehileng (ΔH⁰) ke phetoho ea enthalpy ha li-reactants tsohle le lihlahisoa li le maemong a tsona a tloaelehileng, ke hore, khatellong ea 1 atm le mocheso oa hangata 25°C (298 K).
Lipotso tsa Mehlala le Puisano
Potso ea 1: Ho Chesoa ha Methane
Potso: Bala phetoho e tloaelehileng ea enthalpy (ΔH⁰) bakeng sa ho chesoa ha mole e le 'ngoe ea methane (\(CH_4\)) ho latela equation e latelang:
\[ CH_4(g) + 2O_2(g) \ motsu CO_2(g) + 2H_2O(l) \]
Hoa tsebahala:
– ΔH⁰f \(CH_4(g)\) = -74.8 kJ/mol
– ΔH⁰f \(CO_2(g)\) = -393.5 kJ/mol
– ΔH⁰f \(H_2O(l)\) = -285.8 kJ/mol
Puisano:
Phetoho e tloaelehileng ea enthalpy bakeng sa karabelo ea lik'hemik'hale e ka baloa ho sebelisoa molao oa Hess ka foromo e latelang:
\[ \Delta H⁰ = ∑ ΔH⁰f(sehlahiswa) – ∑ ΔH⁰f(reactant) \]
Taba ea pele, tseba enthalpy e tloaelehileng ea sebopeho bakeng sa ntho e 'ngoe le e 'ngoe karabelong:
– \( ΔH⁰f_{CH_4(g)} = -74.8 \) kJ/mol
– \( ΔH⁰f_{CO_2(g)} = -393.5 \) kJ/mol
– \( ΔH⁰f_{H_2O(l)} = -285.8 \) kJ/mol (×2 bakeng sa di-moles tse pedi \(H_2O\))
Ebe, bala kakaretso ea li-enthalpies tsa tlhahiso ea lihlahisoa le li-reactant:
\[ ∑ ΔH⁰f(sehlahiswa) = [-393.5] + [2(-285.8)]
= -393.5 + (-571.6)
= -965.1 \mongolo{ kJ/mol} \]
\[ ∑ ΔH⁰f(reactant) = [-74.8] + [0] \]
(Metsoako eohle ea motheo ka sebopeho sa eona sa mantlha e na le enthalpy e tloaelehileng ea ho thehoa ha 0 kJ/mol)
Ebe, bala phetoho e tloaelehileng ea enthalpy (ΔH⁰):
\[ ΔH⁰ = -965.1 – (-74.8)
= -965.1 + 74.8
= -890.3 \mongolo{ kJ/mol} \]
Ka hona, phetoho e tloaelehileng ea enthalpy bakeng sa ho chesoa ha mole e le 'ngoe ea methane ke -890.3 kJ/mol.
Potso ea 2: Karabelo ea ho Hlaha ha Metsi
Potso: Bala phetoho e tloaelehileng ea enthalpy (ΔH⁰) bakeng sa ho thehoa ha metsi ho tsoa ho haeterojene le oksijene ho latela equation e latelang:
\[ 2H_2(g) + O_2(g) \ motsu 2H_2O(l) \]
Hoa tsebahala:
– ΔH⁰f \(H_2O(l)\) = -285.8 kJ/mol
Puisano:
Re hloka ho fumana phetoho e tloaelehileng ea enthalpy bakeng sa karabelo ho tloha linthong tse qalang ho ea ho lihlahisoa tse lakatsehang. Ho sebelisoa enthalpy e tloaelehileng ea ho thehoa:
\[ ΔH⁰ = ∑ ΔH⁰f(sehlahiswa) – ∑ ΔH⁰f(reactant) \]
Bala enthalpy ea ho thehoa ha lihlahisoa le li-reactants:
\[
\begin{align}
∑ ΔH⁰f(sehlahisoa) & = [2(-285.8)] \\
∑ ΔH⁰f(sehlahiswa) & = -571.6 \text{ kJ/mol}
\end{align}
\]
\[
ΔH⁰f(H_2(g)) = 0 \text{ kJ/mol} \\
ΔH⁰f(O_2(g)) = 0 \text{ kJ/mol} \\
∑ ΔH⁰f(di-reactant) = [2(0)] + [0] = 0 \text{ kJ/mol}
\]
Ebe, bala phetoho e tloaelehileng ea enthalpy (ΔH⁰):
\[ ΔH⁰ = -571.6 \mongolo{ kJ/mol} \]
Ka hona, phetoho e tloaelehileng ea enthalpy bakeng sa ho thehoa ha metsi ke -571.6 kJ/mol.
Potso ea 3: Ho bola ha Naetrojene Dioxide
Potso: Bala phetoho e tloaelehileng ea enthalpy (ΔH⁰) bakeng sa ho bola ha naetrojene dioxide (\(NO_2\)) ho ba naetrojene monoxide gase (\(NO\)) le khase ea oksijene (O₂) ho latela equation e latelang:
\[ 2NO_2(g) \ motsu 2NO(g) + O_2(g) \]
Hoa tsebahala:
– ΔH⁰f \(NO_2(g)\) = 33.2 kJ/mol
– ΔH⁰f \(NO(g)\) = 90.3 kJ/mol
Puisano:
Lipalo tse tšoanang:
\[ ΔH⁰ = ∑ ΔH⁰f(sehlahiswa) – ∑ ΔH⁰f(reactant) \]
Bala enthalpy ea ho thehoa ha sehlahisoa:
\[
\begin{align}
∑ ΔH⁰f(sehlahiswa) & = [2( ΔH⁰f_{NO(g)} )] + [ ΔH⁰f_{O_2(g)}] \\
& = [2(90.3)] + [0] \\
& = 180.6 \mongolo{ kJ/mol}
\end{align}
\]
Bala enthalpy ea ho thehoa ha li-reactant:
\[
\begin{align}
∑ ΔH⁰f(reactant) & = [2( ΔH⁰f_{NO_2(g)} )] \\
& = [2(33.2)] \\
& = 66.4 \mongolo{ kJ/mol}
\end{align}
\]
Bala phetoho e tloaelehileng ea enthalpy (ΔH⁰):
\[ ΔH⁰ = 180.6 – 66.4 = 114.2 \mongolo{ kJ/mol} \]
Ka hona, phetoho e tloaelehileng ea enthalpy bakeng sa ho bola ha naetrojene dioxide ke 114.2 kJ/mol.
Qetello
Ho bala liphetoho tse tloaelehileng tsa enthalpy (ΔH⁰) ke mokhoa oa bohlokoa ho thermochemistry. Ka ho utloisisa mokhoa oa ho sebelisa li-enthalpies tse tloaelehileng tsa ho thehoa le ho sebelisa molao oa Hess, re ka fumana liphetoho tsa matla liketsong tse fapaneng tsa lik'hemik'hale. Ka mehlala ea mathata a kaholimo, babali ba lebelletsoe ho fumana temohisiso le bokhoni ba ho bala liphetoho tsa enthalpy bakeng sa liketso tse fapaneng tsa lik'hemik'hale. Tsebo ena ha ea bohlokoa feela bakeng sa lithuto tsa thuto empa hape le bakeng sa lits'ebetso tse fapaneng tsa indasteri le lipatlisiso tsa mahlale.
