Su'aalo Tusaale ah oo Ka Hadlaya Isbeddellada Enthalpy ee Xaaladaha Caadiga ah
Pendahuluan
Isbeddelka Enthalpy waa fikrad aasaasi ah oo ku saabsan thermochemistry kaas oo door muhiim ah ka ciyaara hababka kiimikada ee kala duwan. Maqaalkan, waxaan si faahfaahsan uga hadli doonnaa sida loo xisaabiyo loona fahmo isbeddelka enthalpy xaaladaha caadiga ah iyada oo loo marayo dhowr dhibaato oo tusaale ah iyo doodo dhammaystiran. Maqaalkani wuxuu waxtar u yeelan doonaa ardayda, ardayda jaamacadda, iyo qof kasta oo baranaya kiimikada si uu si fiican u fahmo mowduucan.
Fahmidda Enthalpy iyo Isbeddelladeeda
Enthalpy (H) waa tamarta guud ee nidaamka, oo ka kooban tamar gudaha ah iyo tamarta la xiriirta cadaadiska iyo mugga. Marka laga hadlayo falcelinta kiimikada, waxaan inta badan xiisaynaynaa isbeddelka enthalpy (ΔH), kaas oo ka tarjumaya isbeddelka tamarta guud inta lagu jiro falcelinta cadaadiska joogtada ah.
Isbeddelka enthalpy ee xaaladaha caadiga ah (ΔH⁰) waa isbeddelka enthalpy marka dhammaan falgalayaasha iyo alaabtu ay ku jiraan xaaladdooda caadiga ah, taas oo ah, cadaadiska 1 atm iyo heerkul caadi ahaan 25°C (298 K).
Su'aalo iyo Doodo Tusaale ah
Su'aal 1aad: Gubista Methane
Su'aal: Xisaabi isbeddelka enthalpy ee caadiga ah (ΔH⁰) ee gubashada 1 mole oo methane ah (\(CH_4\)) iyadoo lagu saleynayo isla'egta soo socota:
\[ CH_4(g) + 2O_2(g) \heerka saxda ah CO_2(g) + 2H_2O(l) \]
Waa la ogyahay:
– ΔH⁰f \(CH_4(g)\) = -74.8 kJ/mol
– ΔH⁰f \(CO_2(g)\) = -393.5 kJ/mol
– ΔH⁰f \(H_2O(l)\) = -285.8 kJ/mol
Dood:
Isbeddelka caadiga ah ee enthalpy ee falgalka kiimikada waxaa lagu xisaabin karaa iyadoo la adeegsanayo sharciga Hess iyada oo loo marayo qaacidada soo socota:
\[ \Delta H⁰ = ∑ ΔH⁰f(alaabta) – ∑ ΔH⁰f(fal-celin) \]
Marka hore, aqoonso enthalpy-ga caadiga ah ee sameynta walax kasta oo ku jirta falgalka:
– \( ΔH⁰f_{CH_4(g)} = -74.8 \) kJ/mol
– \( ΔH⁰f_{CO_2(g)} = -393.5 \) kJ/mol
– \( ΔH⁰f_{H_2O(l)} = -285.8 \) kJ/mol (×2 laba mool \(H_2O\))
Kadib, xisaabi wadarta enthalpies-ka sameynta alaabada iyo falgalayaasha:
\[ ∑ ΔH⁰f(alaabta) = [-393.5] + [2(-285.8)]
= -393.5 + (-571.6)
= -965.1 \qoraal{ kJ/mol} \]
\[ ∑ ΔH⁰f(fal-celin) = [-74.8] + [0] \]
(Dhammaan iskudhisyada curiyaha ee qaabkooda aasaasiga ah waxay leeyihiin enthalpy caadi ah oo ah 0 kJ/mol)
Kadib, xisaabi isbeddelka enthalpy-ga caadiga ah (ΔH⁰):
\[ ΔH⁰ = -965.1 – (-74.8)
= -965.1 + 74.8
= -890.3 \qoraal{ kJ/mol} \]
Sidaa darteed, isbeddelka caadiga ah ee enthalpy ee gubashada 1 mole oo methane ah waa -890.3 kJ/mol.
Su'aal 2: Falcelinta Samaynta Biyaha
Su'aal: Xisaabi isbeddelka enthalpy-ga caadiga ah (ΔH⁰) ee sameynta biyaha ka yimaada haydarojiin iyo oksijiin iyadoo lagu saleynayo isla'egta soo socota:
\[ 2H_2(g) + O_2(g) \heerka saxda ah 2H_2O(l) \]
Waa la ogyahay:
– ΔH⁰f \(H_2O(l)\) = -285.8 kJ/mol
Dood:
Waxaan u baahanahay inaan helno isbeddelka enthalpy-ga caadiga ah ee falcelinta laga bilaabo walxaha bilowga ah ilaa alaabada la rabo. Iyadoo la adeegsanayo enthalpy-ga caadiga ah ee sameynta:
\[ ΔH⁰ = ∑ ΔH⁰f(alaabta) – ∑ ΔH⁰f(falcelin) \]
Xisaabi enthalpy-ga sameynta alaabada iyo falgalayaasha:
\[
\begin{hagaajin}
∑ ΔH⁰f(shey) & = [2(-285.8)] \\
∑ ΔH⁰f(alaabta) & = -571.6 \qoraal{ kJ/mol}
\end{align}
\]
\[
ΔH⁰f(H_2(g)) = 0 \qoraal{ kJ/mol} \\
ΔH⁰f(O_2(g)) = 0 \qoraal{ kJ/mol} \\
∑ ΔH⁰f(fal-celiyeyaasha) = [2(0)] + [0] = 0 \qoraal{ kJ/mol}
\]
Kadib, xisaabi isbeddelka enthalpy-ga caadiga ah (ΔH⁰):
\[ ΔH⁰ = -571.6 \qoraal{ kJ/mol} \]
Sidaa darteed, isbeddelka caadiga ah ee enthalpy ee sameynta biyaha waa -571.6 kJ/mol.
Su'aal 3: Kala-baxa Naytarojiin Dioxide
Su'aal: Xisaabi isbeddelka enthalpy-ga caadiga ah (ΔH⁰) ee kala-goynta nitrogen dioxide (\(NO_2\)) ee gaaska nitrogen monoksaydh (\(NO\)) iyo gaaska oksijiinta (O₂) iyadoo lagu saleynayo isla'egta soo socota:
\[ 2NO_2(g) \ fallaadha midig 2NO(g) + O_2(g) \]
Waa la ogyahay:
– ΔH⁰f \(NO_2(g)\) = 33.2 kJ/mol
– ΔH⁰f \(NO(g)\) = 90.3 kJ/mol
Dood:
Xisaabin la mid ah:
\[ ΔH⁰ = ∑ ΔH⁰f(alaabta) – ∑ ΔH⁰f(falcelin) \]
Xisaabi enthalpy-ga sameynta badeecada:
\[
\begin{hagaajin}
∑ ΔH⁰f(badeecad) & = [2( ΔH⁰f_{NO(g)} )] + [ ΔH⁰f_{O_2(g)}] \\
& = [2(90.3)] + [0] \\
& = 180.6 \qoraal{ kJ/mol}
\end{align}
\]
Xisaabi enthalpy-ga sameynta falgalayaasha:
\[
\begin{hagaajin}
∑ ΔH⁰f(fal-celin) & = [2( ΔH⁰f_{NO_2(g)} )] \\
& = [2(33.2)] \\
& = 66.4 \qoraal{ kJ/mol}
\end{align}
\]
Xisaabi isbeddelka enthalpy-ga caadiga ah (ΔH⁰):
\[ ΔH⁰ = 180.6 – 66.4 = 114.2 \qoraal{ kJ/mol} \]
Sidaa darteed, isbeddelka caadiga ah ee enthalpy ee kala-baxa nitrogen dioxide waa 114.2 kJ/mol.
Gabagabo
Xisaabinta isbeddellada enthalpy-ga caadiga ah (ΔH⁰) waa farsamo muhiim u ah thermochemistry. Marka la fahmo sida loo isticmaalo enthalpies-ka caadiga ah ee sameynta iyo sida loo dabaqo sharciga Hess, waxaan go'aamin karnaa isbeddellada tamarta ee falgallada kiimikada ee kala duwan. Iyada oo loo marayo dhibaatooyinka tusaalaha ah ee kor ku xusan, akhristayaasha waxaa laga filayaa inay helaan aragti iyo awoodda ay ku xisaabiyaan isbeddellada enthalpy ee falgallada kiimikada ee kala duwan. Aqoontani muhiim ma aha oo kaliya daraasadaha tacliinta laakiin sidoo kale codsiyada warshadaha kala duwan iyo cilmi-baarista sayniska.
