Su'aalo Tusaale ah oo Ka Hadlaya Isbeddellada Enthalpy iyo Enthalpy
Enthalpy waa fikrad muhiim ah oo ku jirta thermodynamics-ka kiimikada, oo inta badan laga helo mowduucyo kala duwan oo kiimiko ah, laga bilaabo falcelinta kiimikada ilaa isbeddellada marxaladda. Maqaalkan, waxaan dib u eegi doonnaa dhowr dhibaato oo tusaale ah waxaanan ka wada hadli doonnaa isbeddellada enthalpy iyo enthalpy si ay nooga caawiyaan inaan si fiican u fahanno fikradda.
Fahmidda Enthalpy
Enthalpy (H) waa wadarta guud ee tamarta ku jirta nidaamka thermodynamic. Waxay ka kooban tahay ma aha oo kaliya tamarta gudaha ee ku kaydsan walxaha laakiin sidoo kale tamarta loo baahan yahay si loo abuuro meel loogu talagalay walxaha ku jira jawi cadaadis la bixiyay. Enthalpy waxaa lagu cabbiraa joules (J) Nidaamka Caalamiga ah (SI).
Xisaab ahaan, enthalpy waxaa lagu qeexaa sida:
\[ H = U + PV \]
Halkee:
– \( H \) waa enthalpy
– \( U \) waa tamarta gudaha
– \( P \) waa cadaadis
– \(V \) waa mugga
Isbeddelka Enthalpy
Isbeddelka Enthalpy (\( \Delta H \)) wuxuu dhacaa marka falgal kiimiko ama hab-socod jireed dhaco. Isbeddelkan enthalpy waxaa lagu qeexi karaa xaddiga kulaylka ee uu sii daayo ama uu nuugo nidaam cadaadis joogto ah. Xisaab ahaan:
\[ \Delta H = H_{\text{product}} – H_{\text{reactant}} \]
Falgallada kulaylka dibadda waa falgallada kulaylka sii daaya deegaanka, falgalladan, \( \Delta H \) waa taban. Dhanka kale, falgallada kulaylka dibadda waa falgallo nuugaya kulaylka deegaanka, falgalladan, \( \Delta H \) waxay leedahay qiimo togan.
Su'aalo iyo Doodo Tusaale ah
Su'aal Tusaale 1aad: Isbeddelka Enthalpy ee Gubashada
Su'aal:
Waxaa la ogyahay in gubashada buuxda ee 1 mole oo methane ah (\(CH_4\)) ay soo saarto kaarboon laba ogsaydh (\(CO_2\)) iyo biyo (\(H_2O\)). Enthalpy-ga xogta sameynta waa sidan soo socota:
– \( \Delta H_{{f, H_2O (l)}} = -285.8 \qoraal{kJ/mol} \)
– \( \Delta H_{{f, CO_2 (g)}} = -393.5 \qoraal{kJ/mol} \)
– \( \Delta H_{{f, CH_4 (g)}} = -74.8 \qoraal{kJ/mol} \)
Xisaabi isbeddelka enthalpy (\( \Delta H \)) ee falcelinta gubashada.
Dood:
Falcelinta gubashada ee methane waa:
\[ CH_4 (g) + 2 O_2 (g) \heerka saxda ah CO_2 (g) + 2 H_2O (l) \]
Isbeddelka enthalpy ee falgalka (\( \Delta H \)) waxaa lagu xisaabin karaa iyadoo la isticmaalayo enthalpy ee sameynta:
\[ \Delta H = \Sigma \Delta H_f \text{products} – \Sigma \Delta H_f \text{reactants} \]
Badeecada:
\[ \Delta H_f (CO_2 (g)) = -393.5 \qoraal{kJ/mol} \]
\[ \Delta H_f (H_2O (l)) = -285.8 \qoraal{kJ/mol} \]
Wadarta guud ee badeecadaha:
\[ (-393.5 \text{kJ/mol}) + 2\times(-285.8 \text{kJ/mol}) = -393.5 – 571.6 = -965.1 \text{kJ/mol} \]
Falgalayaasha:
\[ \Delta H_f (CH_4 (g)) = -74.8 \qoraal{kJ/mol} \]
\[ \Delta H_f (O_2 (g)) = 0 \qoraal{kJ/mol} \]
(Oksijiin marka la eego xaaladdeeda caadiga ah waxay leedahay enthalpy oo ah formation eber.)
Wadarta guud ee falgalayaasha:
\[ (-74.8 \text{kJ/mol}) + 2\times(0 \text{kJ/mol}) = -74.8 \text{kJ/mol} \]
Haddaba, isbeddelka ku yimid enthalpy (\( \Delta H \)) waa:
\[ \Delta H = -965.1 \text{kJ/mol} - (-74.8 \text{kJ/mol}) \]
\[ \Delta H = -965.1 + 74.8 \]
\[ \Delta H = -890.3 \qoraal{kJ/mol} \]
Markaa, isbeddelka enthalpy ee gubashada 1 mole oo methane ah waa \(-890.3 \text{kJ/mol}\).
Su'aal Tusaale 2: Isbeddellada Enthalpy ee Habraacyada Jirka
Su'aal:
Xisaabi isbeddelka enthalpy marka 50 garaam oo baraf ah (\(H_2O_{(s)}\)) oo heerkulkiisu yahay 0°C lagu dhalaaliyo biyo (\(H_2O_{(l)}\)) heerkulkiisu yahay 0°C. Waxaa la ogyahay in kulaylka dhalaalka barafka (\( \Delta H_{\text{fus}} \)) uu yahay 6.01 kJ/mol miisaanka molar-ka biyuhuna uu yahay 18 g/mol.
Dood:
Tallaabada ugu horreysa waa in la xisaabiyo tirada burooyinka barafka.
\[ \text{Moles of ice} = \frac{50 \text{g}}{18 \text{g/mol}} \approx 2.78 \text{mol} \]
Marka xigta waxaan xisaabinaynaa isbeddelka enthalpy ee dhalaalinta barafka:
\[ \Delta H = n \cdot \Delta H_{\text{fus}} \]
\[ \Delta H = 2.78 \text{mol} \cdot 6.01 \text{kJ/mol} \]
\[ \Delta H \qiyaastii 16.7 \qoraal{kJ} \]
Markaa, isbeddelka enthalpy marka uu dhalaalayo 50 garaam oo baraf ah heerkulkiisu yahay 0°C waa qiyaastii 16.7 kJ. Tani waa hab endothermic ah sababtoo ah barafku wuxuu nuugaa kulaylka si uu u noqdo biyo.
Tusaale 3: Falcelinta Hess
Su'aal:
Isticmaal sharciga Hess si aad u go'aamiso isbeddelka enthalpy ee falcelinta soo socota:
\[ 2 C(garaafit) + 3 H_2(g) \rightarrow C_2H_6(g) \]
Dhawr falgal ayaa lagu yaqaanaa isbeddellada enthalpy-ga ee kala duwan:
1. \( C(garaafit) + O_2(g) \rightarrow CO_2(g), \Delta H = -393.5 \text{kJ} \)
2. \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \Delta H = -285.8 \text{kJ} \)
3. \( 2 C_2H_6(g) + 7 O_2(g) \rightarrow 4 CO_2(g) + 6 H_2O(l), \Delta H = -3119.6 \text{kJ} \)
Dood:
Si loo xisaabiyo isbeddelka enthalpy (\( \Delta H \)) ee falcelinta, waxaan u baahanahay inaan dib u celino oo aan ku dhufanno qaar ka mid ah falcelinta si aan ula jaanqaadno falcelinta bartilmaameedka.
Talaabooyinka:
1. Beddel falcelinta \(C_2H_6 \rightarrow 2 CO_2 + 3 H_2O\):
\[2C_2H_6(g) + 7O_2(g) \arrow 4 CO_2(g) + 6H_2O(l), \Delta H = -3119.6 \text{kJ}\]
Dib u celi oo u qaybi falcelinta:
\[ 4CO_2(g) + 6H_2O(l) \ta toosan 2C_2H_6(g) +7O_2(g), \Delta H = 3119.6/2 = 1559.8 \qoraalka{kJ} \]
2. Falcelinta \(C \rightarrow CO_2\):
\[ 4C(garaafit) + 4O_2(g) \rightarrow 4CO_2(g), \Delta H = 4 \jeer -393.5 = -1574 \qoraal{kJ} \]
3. Falcelinta \(H_2\rightarrow H_2O\):
\[ 6H_2(g) + 3O_2(g) \rightarrow 6H_2O(l), \Delta H = 6 \ times -285.8 = -1714.8 \text{kJ} \]
Marka la isku daro dhammaanteed, waxaan helnaa:
\[2C(garaafka) + 3H_2(g) \heerka midig C_2H_6 \]
\[ \Delta H = 1559.8 – 1574 – 1714.8 = -1729 \qoraal{kJ}\]
Markaa, isbeddelka enthalpy ee falcelinta 2 moles oo graphite ah iyo 3 moles oo \(H_2\) ah una gudbay \(C_2H_6(g)\) waa -1729 kJ.
Sidaa darteed, fahamka fikradda enthalpy iyo ku-dhaqankeeda noocyada kala duwan ee falcelinta waxay bixisaa aragti ku saabsan sida tamartu ugu lug leedahay isbeddellada kiimikada iyo jirka. Dhibaatooyinka kor ku xusan waa tusaalooyin si joogto ah loogu isticmaalo manhajka si loo xoojiyo fahamka fikradahan.
