Imibuzo Yezibonelo Ekhuluma Ngokulingana Kwamakhemikhali
Ukulingana kwamakhemikhali kuwumqondo obalulekile kumakhemikhali ochaza isimo lapho amazinga okusabela okuya phambili nokuya emuva ekusabela kwamakhemikhali elingana. Ngaphansi kwalesi simo, ukugxila kwama-reactants nemikhiqizo kuhlala kungaguquki. Lesi sihloko sizohlinzeka ngezibonelo eziningana zezinkinga kanye nezixazululo zazo ukusiza ukuqonda umqondo wokulingana kwamakhemikhali.
Imiqondo Eyisisekelo Yokulingana Kwamakhemikhali
Ukulingana kwamakhemikhali kwenzeka lapho ukusabela kwamakhemikhali kungaqhubeka ngendlela eguqukayo noma kuzo zombili izinhlangothi. Ukusabela okuguqukayo kungafanekiswa kanje:
\[ \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD} \]
Di mana:
– A no-B bangama-reactant,
– C no-D kuyimikhiqizo,
– a, b, c, kanye no-d yizinhlanganisela ze-stoichiometric zento ngayinye.
Uma kufinyelelwa ukulingana, izinga lokusabela phambili (ukukhiqiza imikhiqizo) lilingana nezinga lokusabela okuphambene (ukukhiqiza ama-reactant). Kuleli qophelo, nakuba ukusabela kuqhubeka ngokuguquguqukayo, amazinga azo zonke izinto ahlala engashintshi.
Ukulingana Okuhlala Njalo (K)
I-constant ye-equilibrium \(K_c\) yokusabela okungenhla ingachazwa kanje:
\[ K_c = \frac{{[\text{C}]^c [\text{D}]^d}}{{[\text{A}]^a [\text{B}]^b}} \]
Lapho i-[X] ingukuhlushwa kwe-molar kwento X. Uma usebenzisa ukucindezela okungaphelele ku-gas equilibrium, i-equilibrium constant ivezwa njenge-\(K_p\).
Imibuzo Yesibonelo Sokulingana Kwamakhemikhali
Umbuzo 1: Ukusabela Ngedatha Yokuhlushwa
Inani le-mole eli-1 le-N₂ kanye nama-mole amathathu e-H₂ afakwa esitsheni se-1-litre ekushiseni okuthile. Ukusabela kuqhubeka kanje:
\[ \umbhalo{N}_2(g) + 3\umbhalo{H}_2(g) \umbhalo we-rightleftharpoons 2\umbhalo{NH}_3(g) \]
Ngemva kokufinyelelwa kwe-equilibrium, kutholakala i-0,8 mol ye-N₂. Bala i-equilibrium constant \(K_c\).
Ingxoxo:
1. Nquma ushintsho ekugxileni:
Ekuqaleni, inani lama-moles yileli:
– \([\text{N}_2]_{initial} = 1 \, \text{mol/L}\]
– \([\text{H}_2]_{initial} = 3 \, \text{mol/L}\]
– \([\text{NH}_3]_{initial} = 0 \, \text{mol/L}\]
Ekulinganisweni, inani lama-moles:
– \([\umbhalo{N}_2] = 0,8 \, \umbhalo{mol/L}\]
Ushintsho ku-N₂ = 1 – 0,8 = 0,2 mol/L
2. I-Stoichiometry yezinguquko:
\[
\umbhalo{N}_2(g) + 3\umbhalo{H}_2(g) \umbhalo we-rightleftharpoons 2\umbhalo{NH}_3(g)
\]
Ngakho-ke, izinguquko ze-H₂ kanye ne-NH₃:
– \([\text{H}_2] = 3 \cdot 0,2 = 0,6 \, \text{mol/L}\]
– \([\text{NH}_3] = 2 \cdot 0,2 = 0,4 \, \text{mol/L}\]
Ukulingana kwe-Mole:
– \([\text{H}_2] = 3 – 0,6 = 2,4 \, \text{mol/L}\]
– \([\text{NH}_3] = 0 + 0,4 = 0,4 \, \text{mol/L}\]
3. Bala \(K_c\):
\[
K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}
\]
Faka amanani esikhundleni sawo:
– \([\text{NH}_3] = 0,4 \, \text{mol/L}\]
– \([\umbhalo{N}_2] = 0,8 \, \umbhalo{mol/L}\]
– \([\text{H}_2] = 2,4 \, \text{mol/L}\]
\[
K_c = \frac{(0,4)^2}{(0,8)(2,4)^3}
\]
\[
= \frac{0,16}{0,8 \cdot 13,824}
\]
\[
= \frac{0,16}{11,0592}
cishe 0,0145
\]
Umbuzo 2: Umphumela Wezinguquko Ekugxileni
Inani le-N₂O₄(g) libola libe yi-2NO₂(g) esitsheni esivaliwe. Ekushiseni okuthile, i-equilibrium constant \(K_c\) ingu-0,36. Uma i-concentration yokuqala ye-N₂O₄(g) ingu-1,0 M futhi kungekho NO₂(g) ekuqaleni, bala i-concentration ye-NO₂(g) ku-equilibrium.
Ingxoxo:
1. Ukusebenzisa i-ICE Table:
\[
\begin{align }
\text{Reaction:} & \ \ \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \\
\umbhalo{Owokuqala:} kanye \ \ [\umbhalo{N}_2\umbhalo{O}_4]_{0} = 1.0 \, \umbhalo{M}, \ [\umbhalo{NO}_2]_{0} = 0 \\
\text{Shintsha:} & \ \ [\text{N}_2\text{O}_4]_{eq} = 1.0 – x, \ [\text{NO}_2]_{eq} = 2x \\
\end{align }
\]
2. Ukuxhuma ne-\(K_c\):
\[
K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}
= \frac{(2x)^2}{1.0 – x}
= \frac{4x^2}{1 – x}
\]
3. Thola u-x:
\[
K_c = 0,36
\]
Ngakho-ke, ukufaka esikhundleni:
\[
0,36 = \frac{4x^2}{1 – x}
\]
Ukuphindaphinda okuphambene:
\[
0,36(1 – x) = 4x^2
\]
\[
0,36 – 0,36x = 4x^2
\]
Hambisa konke eceleni:
\[
4x^2 + 0,36x – 0,36 = 0
\]
4. Ukuxazulula Izibalo Zesikwele:
Sebenzisa ifomula ye-quadratic:
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
Lapho u-a = 4, u-b = 0,36 kanye no-c = -0,36:
\[
x = \frac{-0,36 \pm \sqrt{(0,36)^2 – 4(4)(-0,36)}}{2(4)}
\]
\[
x = \frac{-0,36 \pm \sqrt{0,1296 + 5,76}}{8}
\]
\[
x = \frac{-0,36 \pm \sqrt{5,8896}}{8}
\]
Njengoba ukuhlushwa kungeke kube kubi, sikhetha impande enhle:
\[
x \cishe 0,36
\]
5. Ukuhlushwa kwe-NO₂:
\[
[\umbhalo{NO}_2]_{eq} = 2x = 2 \cdot 0,36 = 0,72 \, \umbhalo{M}
\]
Isiphetho
Ukuqonda ukulingana kwamakhemikhali kubalulekile ekubikezeleni ukuthi uhlelo lwamakhemikhali luzosabela kanjani ngaphansi kwezimo ezithile. Ngokuzijwayeza nokuqonda okuphelele, singaxazulula izinkinga ezihilela lo mqondo, sichaze isimo sokugcina sohlelo, futhi siqonde ukuguquguquka kokusabela kwamakhemikhali ohlelweni oluvaliwe. Ukuqaphela nokusebenzisa i-equilibrium constant \(K_c\) kuzokwenza kube lula ukubikezela amazinga azo zonke izinhlobo ohlelweni ku-equilibrium.
