Enthalpy iyo kalorimitari: cabbirka kulaylka falgalka

Sawirista xadka u dhexeeya nidaamka iyo deegaanka

Xisaab kasta oo tamar ahi waxay ka bilaabmaa dooridda nidaam, badanaa kiimikooyinka is-falgalaya. Wax kasta oo kale waa deegaanka, kaas oo kalorimiter ahaan ka dhigan inta badan biyaha ama xalka iyo weelka. Tamarta waxay ka gudbi kartaa xadka sida kulayl, q, ama shaqo, w, isbeddelka tamarta gudaha ee nidaamkuna waa ΔU = q + w.

Enthalpy, H = U + PV, waxaa loo qeexay si isbeddelkiisu u le'ekaado kulaylka isweydaarsan cadaadis joogto ah marka shaqada kaliya ay tahay shaqada fidinta: ΔH = q_p. Sababtoo ah kimistariga shaybaarka intiisa badan wuxuu ka dhacaa qiyaastii cadaadis hawada joogto ah, kimistariyaashu waxay shaxeeyaan enthalpy halkii tamarta gudaha.

Heshiisyada calaamadda ee kulaylka iyo enthalpy

Calaamadaha had iyo jeer waxaa laga qoraa aragtida nidaamka. Kulaylka u qulqula nidaamka waa togan; kulaylka ka baxaya waa taban. Falgal kulayl-bixiye ahi wuxuu kulayl siidaaya deegaanka, sidaa darteed ΔH-kiisu waa taban deegaankuna wuu kululaadaa. Falgal kulayl-nuugaha ahi wuxuu nuugaa kulayl, sidaa darteed ΔH-kiisu waa togan deegaankuna wuu qabowdaa.

Kalorimiter, waxaad cabbirtaa deegaanka, ma aha falgalka. Haddii biyuhu kululaadaan, biyuhu waxay heleen kulayl, falgalkuna wuxuu lumiyey si sax ah intaas. Tani waxay bixisaa xiriirka dhexe ee kalorimitarka: q_reaction = −q_surroundings.

Awoodda kulaylka iyo isleegga q = mcΔT

Kulaylka loo baahan yahay si loo beddelo heerkulka walax waa q = mcΔT, halkaas oo m ay tahay culeys, c waa kulayl gaar ah (kulaylka loo baahan yahay si hal garaam loo kululeeyo hal darajo), ΔT = T_final − T_initial. Isbeddel hal darajo Celsius ah wuxuu le'eg yahay isbeddel hal kelvin ah, sidaa darteed ΔT wuxuu leeyahay isla qiime labada cabbir. Kulaylada gaarka ah ee hoos loo isticmaalay waxay ka yimaadeen OpenStax Chemistry 2e Shaxda 5.1, oo la cabbiray agagaarka 25 °C; biyaha dareeraha ah, c = 4.184 J/(g·°C).

Tusaalaha la xaliyey 1. Intee le'eg kulayl ayaa kor u qaadaya block silicon ah oo 310.0 g ah laga bilaabo 24.5 °C ilaa 160.0 °C? Kulaylka gaarka ah ee silicon waa 0.712 J/(g·°C). ΔT = 160.0 − 24.5 = 135.5 °C. q = (310.0 g)(0.712 J/(g·°C))(135.5 °C) = 2.99 × 10⁴ J, ama 29.9 kJ. Kulaylka gaarka ahi wuxuu leeyahay saddex tiro oo muhiim ah, sidaa darteed jawaabtuna sidaas.

Tusaalaha la xaliyey 2. Block lead ah oo 450.0 g ah 88.0 °C ayaa la dhigay 60.0 g biyo ah 20.5 °C koob xumbo ah. Kulaylka gaarka ah ee lead waa 0.130 J/(g·°C), kan biyahana waa qiimaha kor ku bixiyey. Kulaylka lead lumiyey wuxuu le'eg yahay kulaylka biyuhu helay, sidaa darteed q_lead + q_water = 0: (58.5 J/°C)(T_f − 88.0) + (251.0 J/°C)(T_f − 20.5) = 0, halkaas oo isku-dhufte kastaa yahay culeys lagu dhufto kulayl gaar ah. Xallinta, T_f = (58.5 × 88.0 + 251.0 × 20.5)/(58.5 + 251.0) = 33.3 °C. Hubin: biyuhu waxay helaan qiyaastii 251.0 × 12.8 ≈ 3.2 kJ, lead-na wuxuu lumiyaa qiyaastii 58.5 × 54.7 ≈ 3.2 kJ. Heerkulka biyaha aad buu uga yar beddelmaa kan lead sababtoo ah awooddiisa kulaylka waa in ka badan afar jibaar.

Kalorimitarka koob-qaxwaha cadaadis joogto ah

Kalorimiterka koob-qaxwaha waa koob dahaaran oo hawada u furan, sidaa darteed habku wuxuu dhacaa cadaadis joogto ah kulaylka la cabbirayna wuxuu si toos ah u bixiyaa ΔH. Xalalka biyaha ee khafiifka ah, malo-awaalka caadiga ahi waa in xalku leeyahay cufnaanta iyo kulaylka gaarka ah ee biyaha iyo in koobku nuugo kulayl aan la tixgelin karin.

Tusaalaha la xaliyey 3. 60.0 mL nitric acid ah 0.800 mol/L 21.42 °C ayaa lagu qasay 60.0 mL potassium hydroxide ah 0.800 mol/L isla heerkulkaas. Isku-darku wuxuu gaaray 26.75 °C. Hel ΔH mool biyo ah oo samaysmay kasta. Culeyska xalka: 120.0 mL × 1.00 g/mL = 120. g. ΔT = 26.75 − 21.42 = 5.33 °C. q_solution = (120. g)(4.184 J/(g·°C))(5.33 °C) = 2.68 × 10³ J. Sidaa darteed q_reaction = −2.68 × 10³ J.

Moolasha biyaha samaysmay: xal kastaa wuxuu bixiyaa 0.0600 L × 0.800 mol/L = 0.0480 mol, falgalka H⁺(aq) + OH⁻(aq) → H₂O(l) waa 1:1, sidaa darteed 0.0480 mol biyo ah ayaa samaysma. ΔH = −2.68 kJ / 0.0480 mol = −55.8 kJ/mol. Hubin ahaan, enthalpy-yada samaysanka caadiga ah ee laga helay OpenStax Lifaaqa G waxay bixiyaan −285.83 − (−230.0) = −55.8 kJ/mol isla falgalka ayoonka saafiga ah, iyadoo ΔH°f ee H⁺(aq) loo qaatay eber heshiis ahaan.

Kalorimitarka bomka iyo farqiga u dhexeeya ΔU iyo ΔH

Gubashada waxaa lagu cabbiraa kalorimiter bom ah, weel bir ah oo xidhan oo ku jira qubeys biyo ah. Sababtoo ah mugga waa go'an yahay, shaqo fidin ah ma dhacdo kulaylka la cabbirayna waa ΔU, ma aha ΔH. Qalabka oo dhan waxaa loo habeeyaa sida cutub, taasoo bixisa joogtada kalorimitarka C_cal ee kJ/°C, sidaa darteed q = C_cal × ΔT iyada oo aan culeys iyo kulayl gaar ah oo gooni ah la isticmaalin.

Tusaalaha la xaliyey 4. Gubidda 0.5000 g benzene dareere ah, C₆H₆, kalorimiter leh C_cal = 10.25 kJ/°C waxay kor u qaadaysaa heerkulka 2.038 °C. Adigoo isticmaalaya culeysyada atamka caadiga ah ee IUPAC oo loo dhoweeyey laba meelood oo tobanle (C 12.01, H 1.01), M = 78.12 g/mol, sidaa darteed n = 0.5000 g / 78.12 g/mol = 0.006400 mol. q_cal = (10.25 kJ/°C)(2.038 °C) = 20.89 kJ, sidaa darteed ΔU = −20.89 kJ / 0.006400 mol = −3264 kJ/mol, afar tiro oo muhiim ah sababtoo ah qiime kasta oo la cabbiray wuxuu leeyahay afar.

Si loogu beddelo enthalpy, isticmaal ΔH = ΔU + Δn_gas RT. C₆H₆(l) + 15/2 O₂(g) → 6 CO₂(g) + 3 H₂O(l), moolasha gaaska waxay ka dhacaan 7.5 ilaa 6, sidaa darteed Δn_gas = −1.5 sixiddana waa (−1.5)(8.314 J/(mol·K))(298.15 K) = −3.72 kJ/mol. ΔH = −3264 − 3.72 = −3268 kJ/mol, taasoo waafaqsan −3267.7 kJ/mol ee laga helay enthalpy-yada samaysanka Lifaaqa G. Sixiddu way kortaa marka falgal sameeyo ama cuno mool badan oo gaas ah.

Akhrinta iyo cabbirinta isleegyada kulaylka kiimikaad

Isleegga kulaylka kiimikaad wuxuu ΔH ku xidhaa falgalka sida saxda ah loo qoray. Laba-laabid isku-dhuftayaasha ΔH waa laba-laabmaa; rogid falgalka calaamaduna way rogmataa. Xaaladaha jidheed way muhiim yihiin: samaynta biyo dareere ah waxay sii daysaa kulayl ka badan samaynta uumi biyo ah, sababtoo ah isku-dhuqiddu lafteedu waa kulayl-bixiye.

Si loo helo kulaylka qadar walax ah oo la bixiyey, u beddel mool oo isticmaal ΔH mool falgal kasta sida arrin beddel. Gubidda 12.5 g benzene, oo leh ΔH = −3267.7 kJ mool benzene kasta, waxay sii daysaa (12.5 g / 78.12 g/mol) × 3267.7 kJ/mol = 523 kJ. U qor sida q = −523 kJ falgalka, ama 523 kJ oo kulayl ah oo la sii daayay.

Khaladaadka kalorimitarka ee rogaya calaamadaha ama lumiya kulaylka

Xisaabta kalorimitarku way gaaban tahay, sidaa darteed khaladaadka intooda badan waa kuwo fikrad ah.

Hubin ugu dambeysa oo la isku halleyn karo waa mid jidheed: haddii biyuhu kululaadeen, falgalku waa inuu leeyahay ΔH taban.

• Calaamad laga marqaaday: heerkulka deegaanku wuu kacay, sidaa darteed q_reaction waa taban. Sheegidda +55.8 kJ/mol dhexdhexaadin kululeysay biyaha waa khalad calaamad.
• Culeys khaldan: kalorimitarka xalka, isticmaal culeyska xalka oo dhan, ma aha walaxda la milay ama mid ka mid ah labada mug ee la isku daray oo keliya.
• Garaam iyo mool: ku qaybi moolasha nooca la cayimay si aad u hesho kJ/mol, oo sheeg nooc kee.
• Isku-qaldidda joule iyo kilojoule: q ee ka yimid mcΔT waa joule; u beddel kahor intaadan ku qaybin jawaab kJ/mol ah.
• U aqoonsiga xogta bomka sida ΔH: bom xidhan wuxuu bixiyaa ΔU; ku dabaq sixidda Δn_gas RT marka ΔH loo baahan yahay.