Imibuzo Yesibonelo Ekhuluma Ngezinguquko Ze-Enthalpy Ne-Enthalpy
I-Enthalpy ingumqondo oyinhloko ku-thermodynamics yamakhemikhali, evame ukutholakala ezihlokweni ezahlukene ze-chemistry, kusukela ekuphenduleni kwamakhemikhali kuya ekushintsheni kwesigaba. Kulesi sihloko, sizobuyekeza izinkinga eziningana zezibonelo futhi sixoxe ngezinguquko ze-enthalpy kanye ne-enthalpy ukuze sisize siqonde kangcono lo mqondo.
Ukuqonda i-Enthalpy
I-Enthalpy (H) inani eliphelele lamandla ohlelweni lwe-thermodynamic. Ayihlanganisi nje kuphela amandla angaphakathi agcinwe ezinhlayiyeni kodwa futhi namandla adingekayo ukudala isikhala sezinhlayiyeni endaweni ethile yokucindezela. I-Enthalpy ilinganiswa ngama-joules (J) ku-International System (SI).
Ngokwezibalo, i-enthalpy ichazwa ngokuthi:
\[ H = U + PV \]
Kuphi:
– \( H \) yi-enthalpy
– \( U \) amandla angaphakathi
– \( P \) ingcindezi
– \( V \) yivolumu
Ushintsho lwe-Enthalpy
Ushintsho lwe-enthalpy (\( \Delta H \)) lwenzeka lapho kwenzeka ukusabela kwamakhemikhali noma inqubo yomzimba. Lolu shintsho lwe-enthalpy lungachazwa njengenani lokushisa okukhishwa noma okumuncwa uhlelo ngokucindezela okungaguquki. Ngokwezibalo:
\[ \Delta H = H_{\text{product}} – H_{\text{reactant}} \]
Ukusabela kwe-exothermic yizimpendulo ezikhipha ukushisa endaweni ezungezile, futhi kulezi zimpendulo, \( \Delta H \) ayilungile. Okwamanje, ukusabela kwe-endothermic kuyimpendulo emunca ukushisa endaweni ezungezile, futhi kulokhu kusabela, \( \Delta H \) inenani elihle.
Imibuzo Eyisibonelo Nengxoxo
Isibonelo Umbuzo 1: Ushintsho ku-Enthalpy Yokusha
Umbuzo:
Kuyaziwa ukuthi ukusha okuphelele kwe-mole eyi-1 ye-methane (\(CH_4\)) kukhiqiza i-carbon dioxide (\(CO_2\)) kanye namanzi (\(H_2O\)). I-enthalpy yedatha yokwakheka imi kanje:
– \( \Delta H_{{f, H_2O (l)}} = -285.8 \text{kJ/mol} \)
– \( \Delta H_{{f, CO_2 (g)}} = -393.5 \text{kJ/mol} \)
– \( \Delta H_{{f, CH_4 (g)}} = -74.8 \text{kJ/mol} \)
Bala ushintsho lwe-enthalpy (\( \Delta H \)) lokusabela kokushisa.
Ingxoxo:
Ukusabela kokushiswa kwe-methane yilokhu:
\[ CH_4 (g) + 2 O_2 (g) \umcibisholo wangakwesokudla CO_2 (g) + 2 H_2O (l) \]
Ushintsho lwe-enthalpy lokusabela (\( \Delta H \)) lungabalwa kusetshenziswa i-enthalpy yokwakheka:
\[ \Delta H = \Sigma \Delta H_f \text{products} – \Sigma \Delta H_f \text{reactants} \]
Umkhiqizo:
\[ \Delta H_f (CO_2 (g)) = -393.5 \text{kJ/mol} \]
\[ \Delta H_f (H_2O (l)) = -285.8 \text{kJ/mol} \]
Ingqikithi ye-enthalpy yemikhiqizo:
\[ (-393.5 \umbhalo{kJ/mol}) + izikhathi ezi-2(-285.8 \umbhalo{kJ/mol}) = -393.5 – 571.6 = -965.1 \umbhalo{kJ/mol} \]
Izinto ezisabelayo:
\[ \Delta H_f (CH_4 (g)) = -74.8 \text{kJ/mol} \]
\[ \Delta H_f (O_2 (g)) = 0 \text{kJ/mol} \]
(I-oksijini esimweni sayo esijwayelekile inokwakheka kwe-enthalpy okungu-zero.)
I-enthalpy ephelele yama-reactants:
\[ (-74.8 \umbhalo{kJ/mol}) + izikhathi ezi-2(0 \umbhalo{kJ/mol}) = -74.8 \umbhalo{kJ/mol} \]
Ngakho-ke, ushintsho ku-enthalpy (\( \Delta H \)) luyi:
\[ \Delta H = -965.1 \umbhalo{kJ/mol} – (-74.8 \umbhalo{kJ/mol}) \]
\[ \Delta H = -965.1 + 74.8 \]
\[ \Delta H = -890.3 \umbhalo{kJ/mol} \]
Ngakho-ke, ushintsho lwe-enthalpy lokushiswa kwe-mole eyi-1 ye-methane luyi-\(-890.3 \text{kJ/mol}\).
Isibonelo Umbuzo 2: Izinguquko ze-Enthalpy Ezinqubweni Zomzimba
Umbuzo:
Bala ushintsho lwe-enthalpy lapho amagremu angu-50 eqhwa (\(H_2O_{(s)}\)) ku-0°C encibilikiswa emanzini (\(H_2O_{(l)}\)) ku-0°C. Kuyaziwa ukuthi ukushisa kokuncibilika kweqhwa (\( \Delta H_{\text{fus}} \)) kungu-6.01 kJ/mol kanti isisindo samanzi esinama-molar singama-18 g/mol.
Ingxoxo:
Isinyathelo sokuqala ukubala inani lama-moles eqhwa.
\[ \text{Moles of ice} = \frac{50 \text{g}}{18 \text{g/mol}} \approx 2.78 \text{mol} \]
Okulandelayo sibala ushintsho lwe-enthalpy lokuncibilikisa iqhwa:
\[ \Delta H = n \cdot \Delta H_{\text{fus}} \]
\[ \Delta H = 2.78 \text{mol} \cdot 6.01 \text{kJ/mol} \]
\[ \Delta H \cishe 16.7 \text{kJ} \]
Ngakho-ke, ushintsho lwe-enthalpy lapho kuncibilikiswa amagremu angu-50 eqhwa ku-0°C lungaba ngu-16.7 kJ. Lena inqubo ye-endothermic ngoba iqhwa limunca ukushisa ukuze liphenduke amanzi.
Isibonelo 3: Ukusabela kukaHess
Umbuzo:
Sebenzisa umthetho kaHess ukuze unqume ushintsho lwe-enthalpy kulokho okulandelayo:
\[ 2 C(i-graphite) + 3 H_2(g) \umcibisholo ongakwesokudla C_2H_6(g) \]
Kuyaziwa ukusabela okuningana ngezinguquko zazo ze-enthalpy:
1. \( C(i-graphite) + O_2(g) \umcibisholo ongakwesokudla CO_2(g), \i-Delta H = -393.5 \umbhalo{kJ} \)
2. \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \Delta H = -285.8 \text{kJ} \)
3. \( 2 C_2H_6(g) + 7 O_2(g) \umcibisholo ongakwesokudla 4 CO_2(g) + 6 H_2O(l), \Delta H = -3119.6 \umbhalo{kJ} \)
Ingxoxo:
Ukuze sibale ushintsho lwe-enthalpy (\( \Delta H \)) lokusabela, sidinga ukuhlehlisa nokuphindaphinda eminye yemiphumela ukuze ifane nokusabela okuqondiwe.
I-Langkah-langkah:
1. Shintsha ukusabela \(C_2H_6 \rightarrow 2 CO_2 + 3 H_2O\):
\[2C_2H_6(g) + 7O_2(g) \umcibisholo wangakwesokudla 4 CO_2(g) + 6H_2O(l), \Delta H = -3119.6 \umbhalo{kJ}\]
Ukusabela okuphambene nokuhlukanisa:
\[ 4CO_2(g) + 6H_2O(l) \umcibisholo wangakwesokudla 2C_2H_6(g) + 7O_2(g), \Delta H = 3119.6/2 = 1559.8 \umbhalo{kJ} \]
2. Ukusabela \(C \umcibisholo ongakwesokudla CO_2\):
\[ 4C(i-graphite) + 4O_2(g) \umcibisholo ongakwesokudla 4CO_2(g), \i-Delta H = 4 \izikhathi -393.5 = -1574 \umbhalo{kJ} \]
3. Ukusabela \(H_2\umcibisholo ongakwesokudla H_2O\):
\[ 6H_2(g) + 3O_2(g) \umcibisholo wangakwesokudla 6H_2O(l), \Delta H = 6 \izikhathi -285.8 = -1714.8 \umbhalo{kJ} \]
Uma sihlanganisa konke, sithola:
\[2C(graphite) + 3H_2(g) \umcibisholo ongakwesokudla C_2H_6\]
\[ \Delta H = 1559.8 – 1574 – 1714.8 = -1729 \text{kJ}\]
Ngakho-ke, ushintsho lwe-enthalpy lokusabela kwama-moles amabili e-graphite nama-moles amathathu e-\(H_2\) kuya ku-\(C_2H_6(g)\) luyi--1729 kJ.
Ngakho-ke, ukuqonda umqondo we-enthalpy kanye nokusetshenziswa kwawo ezinhlotsheni ezahlukene zokusabela kunikeza ukuqonda kokuthi amandla ahileleke kanjani ezinguqukweni zamakhemikhali nezomzimba. Izinkinga ezingenhla ziyizibonelo ezivame ukusetshenziswa kuhlelo lokufunda ukuqinisa ukuqonda kwale mibono.
