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8 stepsSawa, Ignatius. Tuangalie swali hili la kemia.
Part a
Mlinganyo uliosawazishwa wa mmenyuko wa uchomaji wa ethane (C₂H₆) ni
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)
Part b· 5 steps
Ili kuhesabu mabadiliko ya joto (ΔH) kwa ajili ya mmenyuko huu, tutatumia sheria ya Hess na maadili ya enthalpy ya uchomaji yaliyotolewa
ΔH = Σ(ΔH°f bidhaa) - Σ(ΔH°f reactants)
Tambua bidhaa na reactants.
Bidhaa: 4 mol CO₂(g) na 6 mol H₂O(g)
Reactants: 2 mol C₂H₆(g) na 7 mol O₂(g)Tumia maadili ya enthalpy ya uchomaji (ΔH°f).
ΔH°f [CO₂(g)] = -393.5 kJ/mol
ΔH°f [H₂O(g)] = -241.8 kJ/mol
ΔH°f [C₂H₆(g)] = -84.7 kJ/mol
ΔH°f [O₂(g)] = 0 kJ/mol (kwa sababu ni elementi katika hali yake ya kawaida)Hesabu jumla ya enthalpy ya bidhaa.
Jumla ΔH°f bidhaa = [4 mol × (-393.5 kJ/mol)] + [6 mol × (-241.8 kJ/mol)]
Jumla ΔH°f bidhaa = [-1574 kJ] + [-1450.8 kJ]
Jumla ΔH°f bidhaa = -3024.8 kJHesabu jumla ya enthalpy ya reactants.
Jumla ΔH°f reactants = [2 mol × (-84.7 kJ/mol)] + [7 mol × (0 kJ/mol)]
Jumla ΔH°f reactants = [-169.4 kJ] + [0 kJ]
Jumla ΔH°f reactants = -169.4 kJHesabu mabadiliko ya jumla ya enthalpy (ΔH).
ΔH = (-3024.8 kJ) - (-169.4 kJ)
ΔH = -3024.8 kJ + 169.4 kJ
ΔH = -2855.4 kJKwa hiyo, mabadiliko ya joto kwa ajili ya mmenyuko huu ni -2855.4 kJ.
Part c· 3 steps
Ili kuhesabu kiwango cha joto kinachotolewa wakati wa kuchoma gramu 10 za ethane, tutafuata hatua hizi
Badilisha gramu za ethane kuwa moles.
Masi ya molar ya C₂H₆ = (2 × 12.01 g/mol) + (6 × 1.01 g/mol) = 24.02 + 6.06 = 30.08 g/mol
Moles za C₂H₆ = (10 g) / (30.08 g/mol) ≈ 0.332 molTumia uwiano wa stoichiometric kutoka kwa mlinganyo uliosawazishwa.
Kutoka kwa mlinganyo: 2 mol C₂H₆ hutoa -2855.4 kJ ya joto.
Hivyo, 1 mol C₂H₆ hutoa (-2855.4 kJ) / 2 = -1427.7 kJ.Hesabu joto linalotolewa kwa moles 0.332 za ethane.
Joto linalotolewa = 0.332 mol × (-1427.7 kJ/mol)
Joto linalotolewa ≈ -473.8 kJKiwango cha joto kinachotolewa wakati wa kuchoma gramu 10 za ethane ni takriban 473.8 kJ. (Tunatoa ishara ya minus kwa sababu tunauliza "kiwango cha joto kinachotolewa", ambalo ni thamani chanya).
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