Calculate the enthalpy of reaction of this coupling, given that:
C6H12O6 ∆Hcombustion = -2805 kJ/mol
CO2(g) ∆HF = -393.5 kJ/mol
H2O(l) ∆HF = -285.8 kJ/mol
C12H22O11 ∆HF = -2247.1 kJ/mol
Maltose is formed according to the equation 2C6H12O6(s) → C12H22O11(s) + H2O(l) Calculate the enthalpy of reaction, given that:?
2020-05-11 3:22 pm
回答 (1)
2020-05-11 4:52 pm
Write the following four thermochemical equation according to the given ΔH values.
2C₆H₁₂O₆(s) + 12O₂(g) → 12CO₂(g) + 12H₂O(ℓ) ΔH₁ = -2805 kJ
12CO₂(g) → 12C(s) + 12O₂(g) ΔH₂ = -12(-393.5) = +4722 kJ
11H₂O(ℓ) → 11H₂(g) + (11/2)O₂(g) ΔH₃ = -11(-285.8) = +3143.8 kJ
12C(s) + 11H₂(g) + (11/2)O₂(g) → C₁₂H₂₂O₁₁(s) ΔH₄ = -2247.1 kJ
Add the above four thermochemical equation, and cancel (35/2)O₂(g), 11H₂O(ℓ),
11H₂(g), 12CO₂(g) and 12C(s) on each side. The thermochemical equation is:
2C₆H₁₂O₆(s) → C₁₂H₂₂O₁₁(s) + H₂O(ℓ) ΔHrxn
ΔHrxn
= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄
= (-2805 + 4722 + 3143.8 - 2247.1) kJ
= +2813.7 kJ
2C₆H₁₂O₆(s) + 12O₂(g) → 12CO₂(g) + 12H₂O(ℓ) ΔH₁ = -2805 kJ
12CO₂(g) → 12C(s) + 12O₂(g) ΔH₂ = -12(-393.5) = +4722 kJ
11H₂O(ℓ) → 11H₂(g) + (11/2)O₂(g) ΔH₃ = -11(-285.8) = +3143.8 kJ
12C(s) + 11H₂(g) + (11/2)O₂(g) → C₁₂H₂₂O₁₁(s) ΔH₄ = -2247.1 kJ
Add the above four thermochemical equation, and cancel (35/2)O₂(g), 11H₂O(ℓ),
11H₂(g), 12CO₂(g) and 12C(s) on each side. The thermochemical equation is:
2C₆H₁₂O₆(s) → C₁₂H₂₂O₁₁(s) + H₂O(ℓ) ΔHrxn
ΔHrxn
= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄
= (-2805 + 4722 + 3143.8 - 2247.1) kJ
= +2813.7 kJ
收錄日期: 2021-04-18 18:29:58
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