Respuesta :
Answer:
Standard Heat of Reaction 1 = -136.2 kJ/mol
Standard Heat of Reaction 2 = -41.166 kJ/mol
Standard Heat of Reaction 3 = -136.07 kJ/mol
Standard Heat of Reaction 4 = 279.448kJ/mol
Explanation:
C₂H₄ (g) + H₂ (g) → C₂H₆ (g)
CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)
3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)
Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)
The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.
C₂H₄ (g), 52.5 kJ/mol
H₂ (g), 0 kJ/mol
C₂H₆ (g), -83.7 kJ/mol
CO (g), -110.525 kJ/mol
H₂O (g), -241.818 kJ/mol
H₂ (g), 0 kJ/mol
CO₂ (g), -393.509 kJ/mol
NO₂ (g), 33.2 kJ/mol
H₂O (l), -285.8 kJ/mol
HNO₃ (aq), -206.28 kJ/mol
NO (g), 90.29 kJ/mol
Cr₂O₃ (s), -1128.4 kJ/mol
CO (g), -110.525 kJ/mol
Cr (s), 0 kJ/mol
CO₂ (g), -393.509 kJ/mol
Note that
ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)
C₂H₄ (g) + H₂ (g) → C₂H₆ (g)
ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)
ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol
ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol
ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol
CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)
ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)
ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol
ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol
ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol
3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)
ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)
ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol
ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol
ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol
Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)
ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)
ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol
ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol
ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol
Hope this Helps!!!
