Tusaale Su'aalaha Doodda Kinetics-ka Kiimikada
Kinetics-ka kiimikada waa laan ka mid ah kiimikada oo barta heerarka falgallada kiimikada iyo arrimaha saameeya. Faham qoto dheer oo ku saabsan kinetics-ka kiimikada ayaa lagama maarmaan u ah saynisyahannada kiimikada iyo injineerada si ay u horumariyaan hababka warshadaha ee hufan iyo inay fahmaan falgallada kala duwan ee bayoolaji ee ka dhaca noolaha nool. Maqaalkani wuxuu ka hadli doonaa dhowr tusaale oo dhibaatooyin ah oo la xiriira kinetics-ka kiimikada iyo xalalkooda si loo helo faham qoto dheer oo ku saabsan mowduucan.
Su'aal Tusaale ah 1: Go'aaminta Amarka Falcelinta
Su'aal:
Falgalku wuxuu leeyahay isle'egta heerka guud ee soo socota:
\[ R = k[A]^m[B]^n \]
Halkee:
– \( R \) waa heerka falcelinta,
– \( k \) waa heerka joogtada ah,
– \([A] \) iyo \([B]\) waa isku-darka fal-galayaasha A iyo B,
– \( m \) iyo \( n \) waa heerarka falcelinta marka loo eego A iyo B.
Waxaa la ogyahay in tijaabada lagu sameeyay kala duwanaanshaha soo socda ee uruurinta:
| Tijaabo | \([A]\) (mol/L) | \([B]\) (mol/L) | Heerka falcelinta (mol/(Ls)) |
|————–|———————-|———————————|
| 1 | 0,10 | 0,20 | 0,030 |
| 2 | 0,10 | 0,40 | 0,060 |
| 3 | 0,20 | 0,20 | 0,120 |
Go'aami nidaamka falcelinta marka loo eego A iyo B iyo qiimaha joogtada heerka \( k \).
Dood:
Si loo go'aamiyo nidaamka falcelinta marka loo eego A iyo B, waa inaan isbarbar dhignaa heerarka falcelinta iyo kala duwanaanshaha xoogga saarista kala duwan.
Marka hore, waxaan go'aamineynaa sida falgalka u kala horreeyo B annagoo isbarbar dhigeyna tijaabooyinka 1 iyo 2:
\[ \frac{\text{R2}}{\text{R1}} = \frac{k[A]^m [B_2]^n}{k[A]^m [B_1]^n} \]
\[ \frac{0,060}{0,030} = \frac{[0,10]^m [0,40]^n}{[0,10]^m [0,20]^n} \]
\[ 2 = \left(\frac{0,40}{0,20}\right)^n \]
\[ 2 = 2^n \]
\[ n = 1 \]
Habka falcelinta marka loo eego B waa 1.
Marka xigta, waxaan go'aamineynaa habka falcelinta marka loo eego A annagoo isbarbar dhigeyna tijaabooyinka 1 iyo 3:
\[ \frac{\text{R3}}{\text{R1}} = \frac{k[A_3]^m [B]^n}{k[A_1]^m [B]^n} \]
\[ \frac{0,120}{0,030} = \frac{[0,20]^m [0,20]^n}{[0,10]^m [0,20]^n} \]
\[ 4 = \left(\frac{0,20}{0,10}\right)^m \]
\[ 4 = 2^m \]
\[ m = 2 \]
Habka falcelinta marka loo eego A waa 2.
Sidaa darteed, isle'egta heerka falcelinta waa:
\[ R = k[A]^2[B] \]
Hadda waxaan helnaa qiimaha joogtada heerka \( k \). Adeegso xogta laga helay tijaabada 1:
\[ 0,030 = k[0,10]^2[0,20] \]
\[ 0,030 = k \jeer 0,01 \jeer 0,20 \]
\[ 0,030 = k \jeer 0,002 \]
\[ k = \frac{0,030}{0,002} \]
\[ k = 15 \ \qoraal{L}^2/(\qoraal{mol}^2 \cdot \qoraal{s}) \]
Markaa, heerka joogtada ah ee \( k \) waa 15 L²/(mol²·s).
Su'aal Tusaale ah 2: Nus-Nololeedka Falcelinta Heerka Labaad
Su'aal:
Marka la eego falcelinta dalabka labaad oo leh isle'egta heerka:
\[ R = k[A]^2 \]
Heerka joogtada ah ee falgalka (\( k \)) waa 0,5 L/(mol·s). Haddii fiirsashada bilowga ah ee falgalka \( [A]_0 \) ay tahay 1 mol/L, hel nus-nolosha falgalka.
Dood:
Falcelinta heerka labaad, nolosha nus-nolosha (\( t_{1/2} \)) waxaa lagu xisaabin karaa iyadoo la isticmaalayo isla'egta:
\[ t_{1/2} = \frac{1}{k[A]_0} \]
Ku beddel qiimayaasha la yaqaan:
\[ t_{1/2} = \frac{1}{0,5 \times 1} \]
\[ t_{1/2} = \frac{1}{0,5} \]
\[ t_{1/2} = 2 \ \qoraal{s} \]
Sidaa darteed, kala bar-nolosha falgalka labaad ee heerka joogtada ah ee 0,5 L/(mol·s) iyo isku-darka falcelinta bilowga ah ee 1 mol/L waa 2 ilbiriqsi.
Su'aal Tusaale ah 3: Tamarta Firfircoonida iyada oo loo marayo Isle'egta Arrhenius
Su'aal:
Falcelintu waxay leedahay laba heer oo kala duwan oo joogto ah oo leh laba heerkul oo kala duwan:
– Marka la gaaro 300 K, heerka joogtada ah (\( k_1 \)) waa 0,2 L/(mol·s)
– Marka la gaaro 350 K, heerka joogtada ah (\( k_2 \)) waa 0,4 L/(mol·s)
Go'aami tamarta kicinta (\( E_a \)) ee falgalka iyadoo la adeegsanayo isle'egta Arrhenius:
\[ k = A e^{-E_a/(RT)} \]
Dood:
Isleegta Arrhenius waxaa loo qori karaa qaabka logarithmic sida soo socota:
\[ \ln k = \ln A – \frac{E_a}{RT} \]
Waxaan isticmaali karnaa laba xog oo joogto ah oo leh laba heerkul oo kala duwan si loo go'aamiyo \( E_a \):
Aan qorno laba isle'eg oo ku saabsan labadan xaaladood:
\[ \ln k_1 = \ln A – \frac{E_a}{R \cdot T_1} \]
\[ \ln k_2 = \ln A – \frac{E_a}{R \cdot T_2} \]
Marka laga jaro labadan isle'eg:
\[ \ln k_2 – \ln k_1 = \left(\ln A – \frac{E_a}{R \cdot T_2}\right) – \left(\ln A – \frac{E_a}{R \cdot T_1}\right) \]
\[ \ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} – \frac{1}{T_1}\right) \]
Beddel qiimayaasha \(k_1 \), \(k_2 \), \(T_1 \), iyo \(T_2 \):
\[ \ln \left(\frac{0,4}{0,2}\right) = -\frac{E_a}{8,314} \left(\frac{1}{350} – \frac{1}{300}\right) \]
\[ \ln (2) = -\frac{E_a}{8,314} \left(\frac{1}{350} – \frac{1}{300}\right) \]
\[ 0,693 = -\frac{E_a}{8,314} \left(\frac{300 – 350}{350 \cdot 300}\right) \]
\[ 0,693 = -\frac{E_a}{8,314} \left(\frac{-50}{105000}\right) \]
\[ 0,693 = \frac{E_a}{8,314} \left(\frac{1}{2100}\right) \]
\[ 0,693 = \ frac {E_a}{17462850/2100} \]
\[ 0,693 = \frac{E_a}{8314} \]
\[ E_a = 0,693 \times 8314 \]
\[ E_a = 5761,842 \ \qoraal{J/mol} \]
Sidaas darteed, tamarta kicinta (\( E_a \)) ee falcelinta waa qiyaastii 5761,842 J/mol ama qiyaastii 5,76 kJ/mol.
-
Aqoonta kinetics-ka kiimikada iyo fahamka ka hadalka dhibaatooyinka noocan oo kale ah ayaa muhiim u ah dhinacyo kala duwan, gaar ahaan warshadaha kiimikada iyo cilmi-baarista sayniska. Dhibaatada tusaalaha ah ee kor ku xusan waxay bixisaa faham ku saabsan hababka lagu go'aaminayo nidaamka falcelinta, nolosha badhkeed, iyo tamarta kicinta, kuwaas oo muhiim u ah horumarka tignoolajiyada iyo faham qoto dheer oo ku saabsan farsamooyinka falcelinta kiimikada.
