Tusaalaha Su'aalaha Doodda ee Kululeeyaha
Pendahuluan
Heerkulku waa laan ka mid ah kiimikada oo barta xiriirka ka dhexeeya falgallada kiimikada iyo isbeddellada tamarta, gaar ahaan qaabka kulaylka. Waa mawduuc muhiim ah oo ku jira kiimikada jirka sababtoo ah waxay naga caawineysaa inaan fahanno sida tamarta loo wareejiyo loona beddelo inta lagu jiro falgallada kiimikada. Maqaalkan, waxaan ka hadli doonnaa dhowr dhibaato oo tusaale ah oo ku saabsan thermochemistry si aan u caddeyno fikradahan.
Su'aal Tusaale 1aad: Isbeddelka Enthalpy ee Gubashada
Su'aal:
enthalpy-ga gubashada (ΔHc) ee ethanol (C2H5OH) waa -1367 kJ/mol. Go'aami xaddiga tamarta la sii daayo marka 2 moles oo ethanol ah si buuxda loo gubo!
Dood:
Isbeddelka enthalpy ee gubashada (ΔHc) waa tamarta la sii daayo marka hal mole oo walax ah si buuxda loogu gubo oksijiinta. ethanol ahaan, ΔHc = -1367 kJ/mol.
Haddii 2 moles oo ethanol ah la gubo, xaddiga tamarta la sii daayo waa:
\[ \text{Tamarta sii deynta} = 2 \, \text{mol} \times (-1367 \, \text{kJ/mol}) \]
\[ \text{Tamarta Bixinta} = -2734 \, \text{kJ} \]
Marka, marka 2 moles oo ethanol ah si buuxda loo gubo, tamarta la sii daayo waa -2734 kJ.
Su'aal Tusaale 2: Xisaabinta Enthalpy Falcelinta laga soo bilaabo Xogta Enthalpy ee Qaabaynta
Su'aal:
Isticmaal enthalpy-ga caadiga ah ee soo socda ee xogta sameynta si aad u xisaabiso enthalpy-ga falcelinta ee sameynta uumiga biyaha:
– \(\Delta H_{\text{f}}^{\circ}(\text{H}_2(g)) = 0 \, \text{kJ/mol}\)
– \(\Delta H_{\text{f}}^{\circ}(\text{O}_2(g)) = 0 \, \text{kJ/mol}\)
– \(\Delta H_{\text{f}}^{\circ}(\text{H}_2\text{O}(g)) = -241.8 \, \text{kJ/mol}\)
Falcelinta:
\[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g) \]
Dood:
Enthalpy-ga falgalka (\(\Delta H_{\text{rxn}}\)) waxaa lagu xisaabin karaa iyadoo la isticmaalayo enthalpy-ga caadiga ah ee sameynta (\(\Delta H_{\text{f}}^{\circ}\)) iyadoo la adeegsanayo qaacidada:
\[ \Delta H_{\text{rxn}} = \Sigma \Delta H_{\text{f}}^{\circ} \text{product} – \Sigma \Delta H_{\text{f}}^{\circ} \text{reactants} \]
Falcelintan:
\[ \Delta H_{\text{rxn}} = \Delta H_{\text{f}}^{\circ}(\text{H}_2\text{O}(g)) – \left[ \Delta H_{\text{f}}^{\circ}(\text{H}_2(g)) + \frac{1}{2} \Delta H_{\text{f}}^{\circ}(\text{O}_2(g)) \right] \]
Geli enthalpy-ga qiimaha sameynta:
\[ \Delta H_{\text{rxn}} = -241.8 \, \text{kJ/mol} – \left[ 0 + \frac{1}{2} \times 0 \right] \]
\[ \Delta H_{\text{rxn}} = -241.8 \, \text{kJ/mol} \]
Markaa, enthalpy-ga falgalka ee sameynta uumiga biyaha waa -241.8 kJ/mol.
Su'aal Tusaale ah 3: Sharciga Hess
Su'aal:
Go'aami isbeddelka enthalpy (\(\Delta H\)) ee falcelinta soo socota:
\[ \text{C(s)} + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO(g)} \]
Isticmaal xogta soo socota:
1. \(\text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \, \Delta H = -393.5 \, \text{kJ}\)
2. \(\text{CO(g)} + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \, \Delta H = -283.0 \, \text{kJ}\)
Dood:
Adeegso Sharciga Hess kaas oo sheegaya in haddii falgal kiimiko ah lagu sheegi karo wadarta falgallo kale oo dhowr ah, markaas isbeddelka enthalpy ee falgalka guud waa wadarta isbeddellada enthalpy ee falgalladaas.
Tallaabada 1: Qor falcelinta u baahan in la rogo ama la beddelo si ay ula jaanqaado falcelinta bartilmaameedka ah:
– Falcelinta (1): \(\text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \, \Delta H = -393.5 \, \text{kJ}\)
– Falcelinta (2): \(\text{CO(g)} + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \, \Delta H = -283.0 \, \text{kJ}\)
Tallaabada 2: Dib u celi falcelinta (2) oo beddel calaamadda \(\Delta H\):
\[ \text{CO}_2(g) \rightarrow \text{CO(g)} + \frac{1}{2}\text{O}_2(g) \, \Delta H = 283.0 \, \text{kJ} \]
Tallaabada 3: Ku dar falcelinta (1) iyo falcelinta rogan (2):
\[ \text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \, \Delta H = -393.5 \, \text{kJ} \]
\[ \text{CO}_2(g) \rightarrow \text{CO(g)} + \frac{1}{2}\text{O}_2(g) \, \Delta H = 283.0 \, \text{kJ} \]
Tallaabada 4: Ku dar labada falcelin:
\[ \text{C(s)} + \text{O}_2(g) + \text{CO}_2(g) \rightarrow \text{CO}_2(g) + \text{CO(g)} + \frac{1}{2}\text{O}_2(g) \]
Fududayntu waxay keentaa:
\[ \text{C(s)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO(g)} \]
Isbeddelka guud ee enthalpy (\(\Delta H\)):
\[ \Delta H = -393.5 \, \text{kJ} + 283.0 \, \text{kJ} = -110.5 \, \text{kJ} \]
Markaa, isbeddelka enthalpy ee falcelinta waa -110.5 kJ.
Su'aal Tusaale 4: Kalorimeter
Su'aal:
Tijaabada kaloorimetriga, 50 g oo biyo ah (c = 4.18 J/g°C) ayaa lagu kululeeyaa 25°C ilaa 75°C. Xisaabi inta kulaylka (q) ee loo baahan yahay.
Dood:
Cadadka kulaylka (q) ee loo baahan yahay si loo kululeeyo biyaha waxaa lagu xisaabin karaa qaacidada:
\[ q = m \cdot c \cdot \Delta T \]
– m = cufka biyaha = 50 g
– c = awoodda kulaylka gaarka ah = 4.18 J/g°C
– \(\Delta T = \) isbeddelka heerkulka \( = 75°C – 25°C = 50°C \)
\[ q = 50 \, \text{g} \times 4.18 \, \text{J/g°C} \times 50°C \]
\[ q = 10450 \, \qoraal{J} \]
Markaa, kulaylka loo baahan yahay si loo kululeeyo 50 g oo biyo ah laga bilaabo 25°C ilaa 75°C waa 10450 J.
Xiritaanka
Doodda ku saabsan dhibaatooyinka tusaalaha ah ee kor ku xusan waxay bixinaysaa dulmar faahfaahsan oo ku saabsan sida fikradaha heerkulbeegga loogu dabaqi karo xallinta dhibaatooyin kala duwan oo la xiriira falgallada kiimikada iyo isbeddellada tamarta. Kimistariga heerkulbeegga waa mid muhiim u ah dhinacyo kala duwan, oo ay ku jiraan warshadaha, tamarta, iyo cilmi-baarista sayniska, sababtoo ah waxay noo ogolaanaysaa inaan fahanno oo aan maareyno isbeddellada tamarta ee ka dhaca falgallada kiimikada. Iyadoo si fiican loo fahmayo kiimikada heerkulka, waxaan naqshadeyn karnaa habab waxtar badan leh oo aan horumarin karnaa teknoolojiyado cusub oo deegaanka u fiican.
