Isibonelo Semibuzo Exoxa Ngezinga Lokusabela
Izinga lokusabela liwumqondo oyisisekelo kumakhemikhali odlala indima ebalulekile ezinqubweni ezahlukene, kokubili kwezezimboni nakwansuku zonke. Kulesi sihloko, sizochaza umqondo wezinga lokusabela ngokujulile, sinikeze izibonelo kanye nezingxoxo ezinemininingwane ukuqinisekisa ukuthi abafundi bayakuqonda kahle.
Ukuqonda Izinga Lokusabela
Izinga lokusabela lichazwa njengoshintsho ekugxilweni kwe-reactant noma umkhiqizo ngesikhathi ngasinye. Ku-equation elula, izinga lokusabela lingabhalwa kanje:
\[ \text{Reaction Rate} = \frac{\Delta \text{[Concentration]}}{\Delta t} \]
Ukuhlushwa kuvame ukulinganiswa ngama-moles ngelitha (M) kanti isikhathi sivame ukuba ngamasekhondi (s). Ngakho-ke, amayunithi esilinganiso sokusabela avame ukuba yi-M/s.
Izici Ezithinta Izinga Lokusabela
Okulandelayo ezinye zezici ezithonya izinga lokusabela:
1. Ukuhlushwa kwama-Reactant: Ukwandisa ukuhlushwa kwama-reactant kuvame ukwandisa izinga lokusabela.
2. Izinga Lokushisa: Ukushisa okwandayo kuvame ukusheshisa izinga lokusabela.
3. Indawo Yobuso: Uma indawo yobuso itholakala kakhulu, izinga lokusabela lishesha kakhulu.
4. I-Catalyst: Ama-Catalyst asheshisa izinga lokusabela ngaphandle kokushintsha okuhlala njalo.
5. Ukucindezela: Ekuphenduleni okubandakanya amagesi, ukucindezela okwandayo kuvame ukwandisa izinga lokusabela.
Imibuzo Eyisibonelo Nengxoxo
Isibonelo Umbuzo 1
Ukusabela phakathi kwe-sodium thiosulfate (Na2S2O3) kanye ne-hydrochloric acid (HCl) kungokulandelayo:
\[ \text{Na}_2\text{S}_2\text{O}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{S} + \text{SO}_2 + \text{H}_2\text{O} \]
Kokunye ukuhlola, ukuhlushwa kwe-sodium thiosulfate kuyashintsha kusuka ku-0,10 M kuya ku-0,05 M ngemizuzwana engama-30. Bala isilinganiso sokusabela!
Ingxoxo
Isilinganiso sokusabela singabalwa kusetshenziswa ifomula:
\[ \text{Reaction Rate} = -\frac{\Delta \text{[Na}_2\text{S}_2\text{O}_3\text{]}}{\Delta t} \]
Faka amanani anikeziwe kufomula:
\[ \Delta \text{[Na}_2\text{S}_2\text{O}_3\text{]} = 0,05 \text{M} – 0,10 \text{M} = -0,05 \text{M} \]
\[ \Delta t = 30 \text{s} \]
Ngakho-ke,
\[ \text{Reaction Rate} = -\left(\frac{-0,05 \text{ M}}{30 \text{ s}}\right) = \frac{0,05 \text{ M}}{30 \text{ s}} = 0,00167 \text{ M/s} \]
Ngakho-ke, isilinganiso sokusabela singu-0,00167 M/s.
Isibonelo Umbuzo 2
Ekuphenduleni, izinga lokusabela linikezwa yi-equation yezinga:
\[ \umbhalo{Izinga} = k [A]^m [B]^n \]
Kusukela ekuhlolweni, kutholakale idatha elandelayo:
| Ukuhlolwa | [A] (M) | [B] (M) | Izinga Lokusabela (M/s) |
|————–|————|———————-|
| 1 | 0.10 | 0.10 | 2.0 × 10^-3 |
| 2 | 0.20 | 0.10 | 8.0 × 10^-3 |
| 3 | 0.10 | 0.20 | 2.0 × 10^-3 |
Nquma ama-oda okusabela m no-n bese ubala inani lesilinganiso esingaguquki, k.
Ingxoxo
Ukunquma i-Reaction Order \( m \) kanye \( n \):
1. Kusukela Kuzivivinyo 1 no-2:
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k [A]_2^m [B]_2^n}{k [A]_1^m [B]_1^n} \]
\[ \frac{8.0 \times 10^{-3}}{2.0 \times 10^{-3}} = \frac{(0.20)^m (0.10)^n}{(0.10)^m (0.10)^n} \]
\[ 4 = (2)^m \]
Ngakho-ke, \( m = 2 \).
2. Kusukela Kuzivivinyo 1 no-3:
\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k [A]_3^m [B]_3^n}{k [A]_1^m [B]_1^n} \]
\[ \frac{2.0 \times 10^{-3}}{2.0 \times 10^{-3}} = \frac{(0.10)^m (0.20)^n}{(0.10)^m (0.10)^n} \]
\[ 1 = (2)^n \]
Ngakho-ke, \( n = 0 \).
Ngakho-ke, i-reaction order maqondana no-A ingu-2 kanti maqondana no-B ingu-0.
Ukubala Inani Elihlala Njalo Lesilinganiso \( k \):
Ukusebenzisa idatha evela ekuhlolweni 1:
\[ \umbhalo{Izinga} = k [A]^m [B]^n \]
\[ 2.0 \izikhathi ezingu-10^{-3} = k (0.10)^2 (0.10)^0 \]
\[ 2.0 \izikhathi ezingu-10^{-3} = k (0.01) \]
\[ k = \frac{2.0 \times 10^{-3}}{0.01} \]
\[k = 0.20 \]
Ngakho-ke, izinga elingaguquki \( k \) lingu-0.20 M^{-1} s^{-1}.
Isibonelo Umbuzo 3
Ukusabela kwamakhemikhali kulandela indlela elandelayo:
\[ \text{Reaction 1: } \text{A} \rightarrow \text{B} \quad (k_1 = 1.0 \, \text{s}^{-1}) \]
\[ \text{Reaction 2: } \text{B} \rightarrow \text{C} \quad (k_2 = 0.1 \, \text{s}^{-1}) \]
Uma ekuqaleni ukuhlushwa kuka-A kungu-1 M kanti u-B ungu-0, thola ukuhlushwa kuka-A no-B ngemva kwemizuzwana emi-5.
Ingxoxo
Sisebenzisa umthetho wezinga lokusabela, sinalokhu okulandelayo:
Ukusabela 1: A kuya ku-B
\[ [A] = [A]_0 e^{-k_1 t} \]
\[ [A] = 1 \umbhalo{ M} \izikhathi e^{-1.0 \umbhalo{ s}^{-1} \izikhathi 5 \umbhalo{ s}} \]
\[ [A] = e^{-5} \umbhalo{ M} \]
Ukusabela 2: B kuya ku-C
\[ \frac{d[B]}{dt} = k_1 [A] – k_2 [B] \]
\[ \frac{d[B]}{dt} = 1.0 \text{s}^{-1} \times [A] – 0.1 \text{s}^{-1} \times [B] \]
Ukusebenzisa isixazululo sokuhlaziya noma sezinombolo salesi sibalo sokwehluka (ngokuvamile indlela ye-Euler noma ye-Runge-Kutta):
\[ [B] \cishe 0.316 \umbhalo{ M} \]
Ngakho-ke, ngemva kwemizuzwana emi-5, ukuhlushwa kuka-A cishe kungama-\( e^{-5} \text{ M} \) kanti ukuhlushwa kuka-B cishe kungama-0.316 M.
Isiphetho
Izinga lokusabela liyisihloko esibalulekile kumakhemikhali, esibonisa izinga lapho ukujiya kwama-reactants kushintsha khona kube yimikhiqizo. Ezinkingeni zesibonelo ezingenhla, sixoxe ngendlela yokubala izinga lokusabela elimaphakathi, sinqume ukulandelana kokusabela, futhi sibale izinga lokusabela elingaguquki. Ukuqonda le mibono kusenza sikwazi ukuyisebenzisa ezimweni ezahlukahlukene ezisebenzayo, kokubili elabhorethri kanye nasezinqubweni zezimboni.
