Imibuzo eyisibonelo exoxa ngokulingana kwamakhemikhali emhlabeni wezimboni

Imibuzo Eyisibonelo Ekhuluma Ngokulingana Kwamakhemikhali Ezweni Lezimboni

Ukulingana kwamakhemikhali kuwumqondo obalulekile kumakhemikhali futhi kusebenza kabanzi emikhakheni eyahlukene yezimboni. Ekuphenduleni kwamakhemikhali, ukulingana kwenzeka lapho izinga lokusabela phambili lilingana nezinga lokusabela okuphambene, ukuze amazinga ezinto ezisabelayo nemikhiqizo ahlale engaguquki ngokuhamba kwesikhathi. Izimboni eziningi, njengezemithi, amakhemikhali kaphethroli, kanye nokucubungula ukudla, zithembele kakhulu ekuqondeni nasekulawuleni ukulingana kwamakhemikhali ukuze kuthuthukiswe ukukhiqizwa kanye nokusebenza kahle. Lesi sihloko sizoxoxa ngezibonelo eziningana zezinkinga ezihlobene nokulingana kwamakhemikhali esimweni sezimboni nokuthi zingaxazululwa kanjani.

Isibonelo Umbuzo 1: Imboni ye-Ammonia (Inqubo ye-Haber-Bosch)

Umbuzo:
Inqubo yeHaber-Bosch ikhiqiza i-ammonia (NH2)₃) kusuka ku-nitrogen (N₂) kanye ne-hydrogen (H₂) ngokwempendulo:
\[ \umbhalo{N}_2(g) + 3\umbhalo{H}_2(g) \umbhalo we-rightleftharpoons 2\umbhalo{NH}_3(g) \]

Ku-500 K, i-equilibrium constant (K _c ) yalokhu kusabela ingu-6.0 x 10^-2. Uma siqala ngo-1.00 mol N ₂ kanye no-3.00 mol H ₂ ku-reactor enomthamo ongu-1.00 L, bala ukuhlushwa kwengxenye ngayinye ku-equilibrium.

Ingxoxo:
1. Nquma ushintsho ekugxilweni kwengxenye ngayinye ohlelweni.
\[ \umbhalo{N}_2(g) + 3\umbhalo{H}_2(g) \umbhalo we-rightleftharpoons 2\umbhalo{NH}_3(g) \]
Ake u-x abe ama-moles e-NH₃ okwakhiwa ngesikhathi sokulingana, khona-ke ushintsho ekugxileni lungokulandelayo:
- N₂: -x mol/L
- H₂: -3x mol/L
– NH₃: +2x mol/L

2. Hlela i-equation ye-equilibrium ngokusekelwe ku-equilibrium constant (K)_c):
\[
K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = 6.0 \times 10^{-2}
\]
Ukuhlushwa kokuqala kanye noshintsho ekuhlushweni:
– [N₂] = 1.00 – x
– [H₂] = 3.00 – 3x
– [NH]₃] = 2x

3. Faka la manani ku-equation yokulingana:
\[
6.0 \izikhathi eziyi-10^{-2} = \frac{(2x)^2}{(1.00 – x)(3.00 – 3x)^3}
\]

4. Bala inani lika-x usebenzisa amaphutha noma ezinye izindlela zezinombolo ukuxazulula i-equation.

Ngemva kokubala, sithola u-x = 0.46. Ngakho-ke:
– [N₂] = 1.00 – 0.46 = 0.54 mol/L
– [H₂] = 3.00 – 3(0.46) = 1.62 mol/L
– [NH]₃] = 2(0.46) = 0.92 mol/L

Isibonelo Umbuzo 2: Imboni ye-Sulfuric Acid (Inqubo Yokuxhumana)

Umbuzo:
Enqubweni yokuxhumana, ukuguqulwa kwe-sulfur dioxide (SO2)₂) ibe yi-trioxide yesulfure (SO2)₃) ngokusebenzisa impendulo:
\[ 2\umbhalo{SO}_2(g) + \umbhalo{O}_2(g) \umbhalo we-rightleftharpoons 2\umbhalo{SO}_3(g) \]

I-equilibrium constant (K _c ) yalokhu kusabela ku-600 K ingu-350. Uma i-reactor iqukethe u-0.50 mol SO ₂ , u-0.25 mol O ₂ , kanye no-0.10 mol SO ₃ , bala amazinga ezingxenye ku-equilibrium kuvolumu engu-2.00 L.

Ingxoxo:
1. Thola ukuhlushwa kokuqala:
– [SO₂]_awu = 0.50 mol / 2.00 L = 0.25 M
– [O₂]_awu = 0.25 mol / 2.00 L = 0.125 M
– [SO₃]_awu = 0.10 mol / 2.00 L = 0.05 M

2. Ake kube u-x ushintsho ekugxileni kwe-SO.₃ okwakhiwa ngesikhathi sokulingana:
– [SO₂]: 0.25 – x
– [O₂]: 0.125 – \(\frac{x}{2}\)
– [SO₃]: 0.05 + x

3. Xhuma i-equation yokulingana:
\[
350 = \frac{(0.05 + x)^2}{(0.25 – x)^2 \cdot (0.125 – \frac{x}{2})}
\]

4. Ngokuxazulula lesi sibalo (usebenzisa indlela yezinombolo noma usebenzisa isibali esihleliwe), kutholakala ukuthi u-x = 0.165. Bese:
– [SO₂] = 0.25 – 0.165 = 0.085 M
– [O₂] = 0.125 – \(\frac{0.165}{2}\) = 0.0425 M
– [SO₃] = 0.05 + 0.165 = 0.215 M

Isibonelo Umbuzo 3: Ukukhiqizwa kwe-Ethylbenzene

Umbuzo:
Ekukhiqizweni kwe-ethylbenzene, i-styrene ikhiqizwa ngokukhishwa kwe-ethylbenzene (C)₆H₅CH₂CH₃):
\[ \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3(g) \rightleftharpoons \text{C}_6\text{H}_5\text{CH=CH}_2(g) + \text{H}_2(g) \]

Uma i-equilibrium constant (K _c ) yalokhu kusabela ku-700 K ingu-2.5, futhi ekuqaleni kukhona i-1.0 mol ye-ethylbenzene kuvolumu engu-1.0 L, bala ukuhlushwa ku-equilibrium.

Ingxoxo:
1. Thola ukuhlushwa kokuqala:
– [C]₆H₅CH₂CH₃] = 1.0 M
– [C]₆H₅CH=CH₂] = 0 M (ngoba ayikakaboli)
– [H₂] = 0 M

2. Ake kube u-x ushintsho ekugxilweni kuka-C₆H₅CH=CH₂ okwakhiwa ngesikhathi sokulingana:
– [C]₆H₅CH₂CH₃]: 1.0 – x
– [C]₆H₅CH=CH₂]: x
– [H₂]: x

3. Xhuma i-equation yokulingana:
\[
2.5 = \frac{x \cdot x}{1.0 – x} = \frac{x^2}{1.0 – x}
\]

4. Ngokuxazulula lesi sibalo se-quadratic, kutholakala ukuthi u-x = 0.62. Bese:
– [C]₆H₅CH₂CH₃] = 1.0 – 0.62 = 0.38 M
– [C]₆H₅CH=CH₂] = 0.62 M
– [H₂] = 0.62 M

Kulezi zibonelo ezintathu, sibone ukuthi umqondo wokulingana kwamakhemikhali usetshenziswa kanjani ezimweni ezahlukene zezimboni. Ukulingana kwamakhemikhali kuyisimiso esiyisisekelo nesibalulekile ezinqubweni zezimboni, njengoba ukulawulwa okunembile kokulingana kwamakhemikhali kungathuthukisa ukusebenza kahle kokukhiqiza kanye nekhwalithi yomkhiqizo. Ukuqonda okuphelele kokulingana kwamakhemikhali kwenza unjiniyela noma uchwepheshe wezimboni akwazi ukuklama nokusebenzisa izinqubo ngendlela efanele.