Su'aalo Tusaale ah oo Ka Hadlaya Sinnaanta Kiimikada ee Dunida Warshadaha
Isku dheelitirka kiimikada waa fikrad muhiim ah oo ku saabsan kiimikada waxaana si weyn loogu dabaqi karaa qaybaha kala duwan ee warshadaha. Falcelinta kiimikada, dheelitirku wuxuu dhacaa marka heerka falcelinta hore uu la mid noqdo heerka falcelinta dib-u-celinta, si heerarka fal-celinta iyo alaabada ay u sii ahaadaan kuwo joogto ah waqti ka dib. Warshado badan, sida daawooyinka, kiimikooyinka batroolka, iyo habaynta cuntada, waxay si weyn ugu tiirsan yihiin fahamka iyo xakamaynta dheelitirka kiimikada si loo wanaajiyo wax soo saarka iyo hufnaanta. Maqaalkani wuxuu ka hadli doonaa dhowr tusaale oo dhibaatooyin ah oo la xiriira dheelitirka kiimikada ee macnaha warshadaha iyo sida loo xalliyo.
Su'aal Tusaale 1aad: Warshadaha Ammooniya (Hab-socodka Haber-Bosch)
Su'aal:
Habka Haber-Bosch wuxuu soo saaraa ammonia (NH)3) laga bilaabo naytaroojiin (N)2) iyo haydarojiin (H)2) sida ku cad falcelinta:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
Marka la gaaro 500 K, joogtada dheelitirka (Kc ) ee falgalkan waa 6.0 x 10^-2. Haddii aan ku bilowno 1.00 mol N 2 iyo 3.00 mol H 2 falgalka oo leh mugga 1.00 L, xisaabi xoogga qayb kasta marka la barbar dhigo.
Dood:
1. Go'aami isbeddelka ku yimaada xoojinta qayb kasta oo nidaamka ku jirta.
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
Ha x noqoto burooyinka NH3 kaas oo lagu sameeyo dheelitirka, markaa isbeddelka ku-fiirsashada waa sidan soo socota:
- N2: -x mol/L
- H2: -3x mol/L
– NH3: +2x mol/L
2. Hagaaji isla'egta dheelitirka iyadoo lagu saleynayo joogtada dheelitirka (K)c):
\[
K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = 6.0 \jeer 10^{-2}
\]
Fiirsashada bilowga ah iyo isbeddelka fiirsashada:
– [N]2] = 1.00 – x
– [H]2] = 3.00 – 3x
– [NH]3] = 2x
3. Qiimayaashan ku xidh isle'egta dheelitirka:
\[
6.0 \jeer 10^{-2} = \frac{(2x)^2}{(1.00 – x)(3.00 – 3x)^3}
\]
4. Xisaabi qiimaha x adoo isticmaalaya tijaabo iyo qalad ama habab kale oo tirooyin ah si aad u xalliso isla'egta.
Kadib xisaabinta, waxaan helnaa x = 0.46. Markaa:
– [N]2] = 1.00 - 0.46 = 0.54 mol/L
– [H]2] = 3.00 - 3 (0.46) = 1.62 mol/L
– [NH]3] = 2(0.46) = 0.92 mol/L
Su'aal Tusaale ah 2: Warshadaha Aashitada Sulfuric (Hab-raaca Xiriirka)
Su'aal:
Habka xiriirka, beddelka sulfur dioxide (SO2)2) galay sulfur trioxide (SO2)3) iyada oo loo marayo falcelinta:
\[ 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \]
Joogtada dheelitirka (Kc ) ee falgalkan 600 K waa 350. Haddii falgalka uu ka kooban yahay 0.50 mol SO2 , 0.25 mol O2 , iyo 0.10 mol SO3 , xisaabi isku-darka qaybaha marka la isku dheelitirayo mugga 2.00 L.
Dood:
1. Go'aami xoogga bilowga ah:
– [SO]2]awal = 0.50 mol / 2.00 L = 0.25 M
– [O]2]awal = 0.25 mol / 2.00 L = 0.125 M
– [SO]3]awal = 0.10 mol / 2.00 L = 0.05 M
2. Ha ahaato isbeddelka ku yimid xoogga SO.3 kaas oo lagu sameeyo dheelitirka:
– [SO]2]: 0.25 – x
– [O]2]: 0.125 – \(\frac{x}{2}\)
– [SO]3]: 0.05 + x
3. Ku xidh isle'egta dheelitirka:
\[
350 = \frac{(0.05 + x)^2}{(0.25 – x)^2 \cdot (0.125 – \frac{x}{2})}
\]
4. Marka la xalliyo isla'egtan (iyadoo la adeegsanayo hab tirooyin ah ama la isticmaalayo xisaabiye barnaamij lagu samayn karo), waxaa la ogaaday in x = 0.165. Kadib:
– [SO]2] = 0.25 – 0.165 = 0.085 M
– [O]2] = 0.125 – \(\frac{0.165}{2}\) = 0.0425 M
– [SO]3] = 0.05 + 0.165 = 0.215 M
Su'aal Tusaale ah 3: Soosaarka Ethylbenzene
Su'aal:
Soo saarista ethylbenzene, styrene waxaa soo saara fuuq-baxa ethylbenzene (C)6H5CH2CH3):
\[ \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3(g) \rightleftharpoons \text{C}_6\text{H}_5\text{CH=CH}_2(g) + \text{H}_2(g) \]
Haddii isku dheelitirka joogtada ah (Kc ) ee falgalkan 700 K uu yahay 2.5, oo ay marka hore jiraan 1.0 mol oo ethylbenzene ah oo ku jira mugga 1.0 L, xisaabi xoogga isku dheelitirka.
Dood:
1. Go'aami xoogga bilowga ah:
– [C]6H5CH2CH3] = 1.0 M
– [C]6H5CH=CH2] = 0 M (sababtoo ah lama burburin)
– [H]2] = 0 M
2. Ha ahaato isbeddelka ku yimid heerka C6H5CH=CH2 kaas oo lagu sameeyo dheelitirka:
– [C]6H5CH2CH3]: 1.0 – x
– [C]6H5CH=CH2]: x
– [H]2]: x
3. Ku xidh isle'egta dheelitirka:
\[
2.5 = \frac{x \cdot x}{1.0 – x} = \frac{x^2}{1.0 – x}
\]
4. Marka la xalliyo isla'egtan labajibbaaran, waxaa la ogaaday in x = 0.62. Kadib:
– [C]6H5CH2CH3] = 1.0 – 0.62 = 0.38 M
– [C]6H5CH=CH2] = 0.62 M
– [H]2] = 0.62 M
Saddexdan tusaale, waxaan ku aragnay sida fikradda dheelitirka kiimikada loogu dabaqo xaaladaha warshadaha ee kala duwan. Isku dheelitirka kiimikada waa mabda' aasaasi ah oo muhiim u ah hababka warshadaha, maadaama xakamaynta saxda ah ee dheelitirka kiimikada ay horumarin karto hufnaanta wax soo saarka iyo tayada wax soo saarka. Faham buuxa oo ku saabsan dheelitirka kiimikada waxay u suurtagelinaysaa injineer ama xirfadle warshadeed inuu si fiican u naqshadeeyo oo uu u shaqeeyo hababka.
