Tambayoyi Misali Game da Canje-canjen Enthalpy a Yanayin Daidaitacce
Pendahuluan
Canjin enthalpy wani muhimmin ra'ayi ne a fannin thermochemistry wanda ke taka muhimmiyar rawa a cikin hanyoyin sinadarai daban-daban. A cikin wannan labarin, za mu tattauna dalla-dalla yadda ake ƙididdigewa da fahimtar canjin enthalpy a ƙarƙashin yanayi na yau da kullun ta hanyar misalai da yawa na matsaloli da tattaunawa mai zurfi. Wannan labarin zai zama da amfani ga ɗalibai, ɗaliban kwaleji, da duk wanda ke karatun ilmin sunadarai don samun fahimtar wannan batu.
Fahimtar Enthalpy da Canje-canjensa
Enthalpy (H) shine cikakken kuzarin tsarin, wanda ya ƙunshi makamashin ciki da kuma kuzarin da ke da alaƙa da matsin lamba da girma. A yanayin halayen sinadarai, sau da yawa muna sha'awar canjin enthalpy (ΔH), wanda ke nuna jimlar canjin kuzari yayin amsawar a matsin lamba akai-akai.
Canjin enthalpy a ƙarƙashin yanayin da aka saba amfani da shi (ΔH⁰) shine canjin enthalpy lokacin da duk masu amsawa da samfuran ke cikin yanayin da aka saba amfani da shi, wato, a matsin lamba na 1 atm da zafin jiki na yawanci 25°C (298 K).
Tambayoyi da Tattaunawa Samfura
Tambaya ta 1: Konewar Methane
Tambaya: A ƙididdige canjin enthalpy na yau da kullun (ΔH⁰) don ƙonewar mole 1 na methane (\(CH_4\)) bisa ga lissafin da ke ƙasa:
\[ CH_4(g) + 2O_2(g) \kai tsaye CO_2(g) + 2H_2O(l) \]
An sani:
– ΔH⁰f \(CH_4(g)\) = -74.8 kJ/mol
– ΔH⁰f \(CO_2(g)\) = -393.5 kJ/mol
– ΔH⁰f \(H_2O(l)\) = -285.8 kJ/mol
Tattaunawa:
Ana iya ƙididdige canjin enthalpy na daidaitaccen amsawar sinadarai ta amfani da dokar Hess ta hanyar dabarar da ke ƙasa:
\[ \Delta H⁰ = ∑ ΔH⁰f(samfuri) – ∑ ΔH⁰f(mai amsawa) \]
Da farko, gano daidaitaccen enthalpy na samuwar kowane abu a cikin amsawar:
– \( ΔH⁰f_{CH_4(g)} = -74.8 \) kJ/mol
– \( ΔH⁰f_{CO_2(g)} = -393.5 \) kJ/mol
– \( ΔH⁰f_{H_2O(l)} = -285.8 \) kJ/mol (×2 ga moles biyu \(H_2O\))
Sannan, ƙididdige jimlar enthalpies na samuwar samfuran da abubuwan da ke haifar da amsawa:
\[ ∑ ΔH⁰f(samfuri) = [-393.5] + [2(-285.8)]
= -393.5 + (-571.6)
= -965.1 \rubutu{ kJ/mol} \]
\[ ∑ ΔH⁰f(mai amsawa) = [-74.8] + [0] \]
(Duk wani abu mai gina jiki a cikin siffa ta asali yana da daidaitaccen enthalpy na 0 kJ/mol)
Sannan, ƙididdige canjin enthalpy na yau da kullun (ΔH⁰):
\[ ΔH⁰ = -965.1 – (-74.8)
= -965.1 + 74.8
= -890.3 \rubutu{ kJ/mol} \]
Saboda haka, canjin enthalpy na yau da kullun don ƙone mole 1 na methane shine -890.3 kJ/mol.
Tambaya ta 2: Martanin Samuwar Ruwa
Tambaya: Lissafa canjin enthalpy na yau da kullun (ΔH⁰) don samar da ruwa daga hydrogen da oxygen bisa ga lissafin da ke ƙasa:
\[ 2H_2(g) + O_2(g) \arrow 2H_2O(l) \]
An sani:
– ΔH⁰f \(H_2O(l)\) = -285.8 kJ/mol
Tattaunawa:
Muna buƙatar nemo canjin enthalpy na yau da kullun don amsawa daga abubuwan farawa zuwa samfuran da ake so. Ta amfani da daidaitaccen enthalpy na samuwar:
\[ ΔH⁰ = ∑ ΔH⁰f(samfuri) – ∑ ΔH⁰f(mai amsawa) \]
Lissafa enthalpy na samuwar samfuran da abubuwan da ke haifar da amsawa:
\[
\begin{daidai}
∑ ΔH⁰f(samfuri) & = [2(-285.8)] \\
∑ ΔH⁰f(samfuri) & = -571.6 \rubutu{ kJ/mol}
\end{daidai}
\]
\[
ΔH⁰f(H_2(g)) = 0 \rubutu{ kJ/mol} \\
ΔH⁰f(O_2(g)) = 0 \rubutu{ kJ/mol} \\
∑ ΔH⁰f(masu amsawa) = [2(0)] + [0] = 0 \rubutu{ kJ/mol}
\]
Sannan, ƙididdige canjin enthalpy na yau da kullun (ΔH⁰):
\[ ΔH⁰ = -571.6 \rubutu{ kJ/mol} \]
Saboda haka, canjin enthalpy na yau da kullun don samar da ruwa shine -571.6 kJ/mol.
Tambaya ta 3: Rushewar Nitrogen Dioxide
Tambaya: A ƙididdige canjin enthalpy na yau da kullun (ΔH⁰) don bazuwar nitrogen dioxide (\(NO_2\)) zuwa iskar nitrogen monoxide (\(NO\)) da iskar oxygen (O₂) bisa ga lissafin da ke ƙasa:
\[ 2NO_2(g) \kibiya dama 2NO(g) + O_2(g) \]
An sani:
– ΔH⁰f \(NO_2(g)\) = 33.2 kJ/mol
– ΔH⁰f \(NO(g)\) = 90.3 kJ/mol
Tattaunawa:
Lissafi makamantan haka:
\[ ΔH⁰ = ∑ ΔH⁰f(samfuri) – ∑ ΔH⁰f(mai amsawa) \]
Lissafa enthalpy na samuwar samfurin:
\[
\begin{daidai}
∑ ΔH⁰f(samfuri) & = [2(ΔH⁰f_{NO(g)} )] + [ ΔH⁰f_{O_2(g)}] \\
& = [2(90.3)] + [0] \\
& = 180.6 \rubutu{ kJ/mol}
\end{daidai}
\]
Lissafa enthalpy na samuwar sinadaran:
\[
\begin{daidai}
∑ ΔH⁰f(mai amsawa) & = [2( ΔH⁰f_{NO_2(g)} )] \\
& = [2(33.2)] \\
& = 66.4 \rubutu{ kJ/mol}
\end{daidai}
\]
Lissafa canjin enthalpy na yau da kullun (ΔH⁰):
\[ ΔH⁰ = 180.6 – 66.4 = 114.2 \text{ kJ/mol} \]
Saboda haka, canjin enthalpy na yau da kullun don bazuwar nitrogen dioxide shine 114.2 kJ/mol.
Kammalawa
Lissafin canje-canjen enthalpy na yau da kullun (ΔH⁰) wata muhimmiyar dabara ce a fannin thermochemistry. Ta hanyar fahimtar yadda ake amfani da daidaitattun enthalpies na samuwar da kuma amfani da dokar Hess, za mu iya tantance canje-canjen makamashi a cikin nau'ikan halayen sinadarai daban-daban. Ta hanyar misalan matsalolin da ke sama, ana sa ran masu karatu su sami fahimta da ikon ƙididdige canje-canjen enthalpy don nau'ikan halayen sinadarai daban-daban. Wannan ilimin yana da mahimmanci ba kawai ga karatun ilimi ba har ma ga aikace-aikacen masana'antu daban-daban da binciken kimiyya.
