Tambayoyi Misali Game da Canje-canjen Enthalpy da Enthalpy
Enthalpy muhimmin ra'ayi ne a cikin thermodynamics na sinadarai, wanda ake yawan samu a cikin batutuwa daban-daban na sinadarai, daga halayen sinadarai zuwa canje-canjen matakai. A cikin wannan labarin, za mu sake duba misalai da yawa na matsaloli kuma mu tattauna canje-canjen enthalpy da enthalpy don taimaka mana mu fahimci manufar sosai.
Fahimtar Enthalpy
Enthalpy (H) shine jimlar adadin kuzarin da ke cikin tsarin thermodynamic. Ya ƙunshi ba kawai makamashin ciki da aka adana a cikin barbashi ba, har ma da makamashin da ake buƙata don ƙirƙirar sarari ga barbashi a cikin wani yanayi na matsin lamba. Ana auna Enthalpy a cikin joules (J) a cikin Tsarin Duniya (SI).
A fannin lissafi, an bayyana enthalpy kamar haka:
\[H = U + PV \]
Ina:
– \(H\) shine enthalpy
– \( U \) shine makamashin ciki
– \( P \) shine matsin lamba
– \(V \) shine girman
Canjin Enthalpy
Canjin enthalpy (\( \Delta H \)) yana faruwa ne lokacin da wani abu mai guba ko tsarin jiki ya faru. Wannan canjin enthalpy za a iya bayyana shi a matsayin adadin zafi da aka saki ko aka sha ta hanyar tsarin a matsin lamba akai-akai. A lissafi:
\[ \Delta H = H_{\text{product}} – H_{\text{reactant}} \]
Halayen waje na waje halayen halayen da ke fitar da zafi zuwa muhalli ne, kuma a cikin waɗannan halayen, \( \Delta H \) ba shi da kyau. A halin yanzu, halayen ciki na ciki martani ne da ke shan zafi daga muhalli, kuma a cikin wannan halayen, \( \Delta H \) yana da ƙima mai kyau.
Tambayoyi da Tattaunawa Samfura
Misali Tambaya ta 1: Canji a cikin Enthalpy na Konewa
Tambaya:
An san cewa ƙonewar mole 1 na methane (\(CH_4\)) gaba ɗaya yana samar da carbon dioxide (\(CO_2\)) da ruwa (\(H_2O\)). Enthalpy na bayanan samuwar sune kamar haka:
– \( \Delta H_{{f, H_2O (l)}} = -285.8 \text{kJ/mol} \)
– \( \Delta H_{{f, CO_2 (g)}} = -393.5 \text{kJ/mol} \)
– \( \Delta H_{{f, CH_4 (g)}} = -74.8 \text{kJ/mol} \)
Lissafa canjin enthalpy (\( \Delta H \)) na amsawar konewa.
Tattaunawa:
Haɗarin konewa na methane shine:
\[ CH_4 (g) + 2 O_2 (g) \kai tsaye CO_2 (g) + 2 H_2O (l) \]
Ana iya ƙididdige canjin enthalpy na amsawar (\( \Delta H \)) ta amfani da enthalpy na samuwar:
\[ \Delta H = \Sigma \Delta H_f \text{products} – \Sigma \Delta H_f \text{reactants} \]
Samfuri:
\[ \Delta H_f (CO_2 (g)) = -393.5 \text{kJ/mol} \]
\[ \Delta H_f (H_2O (l)) = -285.8 \text{kJ/mol} \]
Jimlar enthalpy na samfuran:
\[ (-393.5 \text{kJ/mol}) + sau 2(-285.8 \text{kJ/mol}) = -393.5 – 571.6 = -965.1 \text{kJ/mol} \]
Masu amsawa:
\[ \Delta H_f (CH_4 (g)) = -74.8 \text{kJ/mol} \]
\[ \Delta H_f (O_2 (g)) = 0 \text{kJ/mol} \]
(Oxygen a yanayinsa na yau da kullun yana da enthalpy na samuwar sifili.)
Jimlar enthalpy na masu amsawa:
\[ (-74.8 \text{kJ/mol}) + sau 2(0 \text{kJ/mol}) = -74.8 \text{kJ/mol} \]
Don haka, canjin enthalpy (\( \Delta H \)) shine:
\[ \Delta H = -965.1 \rubutu{kJ/mol} - (-74.8 \rubutu{kJ/mol}) \]
\[ \Delta H = -965.1 + 74.8 \]
\[ \Delta H = -890.3 \text{kJ/mol} \]
Don haka, canjin enthalpy don ƙone mole 1 na methane shine \(-890.3 \text{kJ/mol}\).
Misali Tambaya ta 2: Canje-canje a cikin Tsarin Jiki
Tambaya:
Lissafa canjin enthalpy lokacin da aka narke gram 50 na kankara (\(H_2O_{(s)}\)) a 0°C zuwa ruwa (\(H_2O_{(l)}\)) a 0°C. An san cewa zafin narkewar kankara (\( \Delta H_{\text{fus}} \)) shine 6.01 kJ/mol kuma nauyin molar na ruwa shine 18 g/mol.
Tattaunawa:
Mataki na farko shine a ƙididdige adadin moles na kankara.
\[ \text{Moles of ice} = \frac{50 \text{g}}{18 \text{g/mol}} \approx 2.78 \text{mol} \]
Na gaba za mu ƙididdige canjin enthalpy don narke kankara:
\[ \Delta H = n \cdot \Delta H_{\text{fus}} \]
\[ \Delta H = 2.78 \text{mol} \cdot 6.01 \text{kJ/mol} \]
\[ \Delta H \kimanin 16.7 \text{kJ} \]
Don haka, canjin enthalpy lokacin da ake narkar da gram 50 na kankara a 0°C shine kimanin 16.7 kJ. Wannan tsari ne na endothermic saboda kankara tana shan zafi don ta koma ruwa.
Misali na 3: Amsar Hess
Tambaya:
Yi amfani da dokar Hess don tantance canjin enthalpy don amsawar da ke tafe:
\[ 2 C(graphite) + 3 H_2(g) \rightarrow C_2H_6(g) \]
An san halayen da dama tare da canje-canjen enthalpy:
1. \( C(graphite) + O_2(g) \rightarrow CO_2(g), \Delta H = -393.5 \text{kJ} \)
2. \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \Delta H = -285.8 \text{kJ} \)
3. \( 2 C_2H_6(g) + 7 O_2(g) \rightarrow 4 CO_2(g) + 6 H_2O(l), \Delta H = -3119.6 \text{kJ} \)
Tattaunawa:
Domin ƙididdige canjin enthalpy (\( \Delta H \)) na amsawar, muna buƙatar juyawa da ninka wasu daga cikin martanin don daidaita martanin da aka yi niyya.
Matakai:
1. Canja martanin \(C_2H_6 \rightarrow 2 CO_2 + 3 H_2O\):
\[2C_2H_6(g) + 7O_2(g) \arrow 4 CO_2(g) + 6H_2O(l), \Delta H = -3119.6 \rubutu{kJ}\]
Juya da raba halayen:
\[ 4CO_2 (g) + 6H_2O (l) \madaidaicin 2C_2H_6(g) + 7O_2(g), \Delta H = 3119.6/2 = 1559.8 \rubutu{kJ} \]
2. Reaction \(C \rightarrow CO_2\):
\[ 4C(graphite) + 4O_2(g) \rightarrow 4CO_2(g), \Delta H = 4 \sau -393.5 = -1574 \text{kJ} \]
3. Reaction \(H_2\rightarrow H_2O\):
\[ 6H_2(g) + 3O_2(g) \arrow 6H_2O(l), \Delta H = 6 \ lokuta -285.8 = -1714.8 \ rubutu {kJ} \]
Idan muka haɗa shi duka, za mu samu:
\[2C(graphite) + 3H_2(g) \kirawar dama C_2H_6\]
\[ \Delta H = 1559.8 – 1574 – 1714.8 = -1729 \text{kJ}\]
Don haka, canjin enthalpy na amsawar moles 2 na graphite da moles 3 na \(H_2\) zuwa \(C_2H_6(g)\) shine -1729 kJ.
Saboda haka, fahimtar manufar enthalpy da kuma amfani da shi ga nau'ikan halayen daban-daban yana ba da haske game da yadda makamashi ke shiga cikin canje-canjen sinadarai da na zahiri. Matsalolin da ke sama wasu misalai ne da ake yawan amfani da su a cikin manhajar karatu don ƙarfafa fahimtar waɗannan ra'ayoyi.
