WebCalculate ΔG∘ (in kJ) for the following reaction at 1 atm and 25 °C: C2H6 (g) + O2 (g) → CO2 (g) + H2O (l) (unbalanced) ΔHf∘ C2H6 (g) = -84.7 kJ/mol; S∘ C2H6 (g) = 229.5 J/K ∙ mol; ΔHf∘ CO2 (g) = -393.5 kJ/mol; S∘ CO2 (g) = 213.6 J/K ∙ mol; ΔHf∘ H2O (l) = -285.8 kJ/mol; S∘H2O (l) = 69.9 J/K ∙ mol; S∘O2 (g) = 205.0 J/K ∙ mol Expert Answer 1st step WebJan 30, 2024 · C (s) + O2 (g) → CO2 (g); ΔHf = -393.5 kJ/mol S (s) + O2 (g) → SO2 (g); ΔHf = -296.8 kJ/mol C (s) + 2 S (s) → CS2 (l); ΔHf = 87.9 kJ/mol Advertisement aditya881653 Explanation: ENTHALPY OF REACTION [1ΔHf (CO2 (g)) + 2ΔHf (SO2 (g))] - [1ΔHf (CS2 (ℓ)) + 3ΔHf (O2 (g))] [1 (-393.51) + 2 (-296.83)] - [1 (89.7) + 3 (0)] = -1076.87 kJ
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WebJun 17, 2024 · We rearranging and add equations (1, 2, and 3) in such a way as to end up with the needed equation: equation (1) be as it is: (1) C (graphite) + O₂ (g) → CO₂ (g), ΔHf₁° = -393.5 kJ/mol . equation (2) should be multiplied by (2) and also the value of ΔHf₂°: (2) 2H2 (g) + O₂ (g) → 2H₂O (l), ΔHf₂° = 2x (-285.8 kJ/mol ). WebAug 3, 2024 · C(s) + O 2(g) → CO 2(g) ΔH = − 393.5kJ According to the first corollary, the first reaction has an energy change of two times −393.5 kJ, or −787.0 kJ: 2C(s) + 2O 2(g) → 2CO 2(g) ΔH = − 787.0kJ The second reaction in the combination is related to the combustion of CO (g): 2CO(g) + O 2(g) → 2CO 2(g) ΔH = − 566.0kJ matt bowling track
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WebAnswer: DU= +13 KJ Explanation: Pa brainliest po plsss :< 13. For the reaction 2NaHCO3 → Na2CO3 + CO2 + H2O ∆H = 129.37 kJ/mole Find the standard heat of formation of NaHCO3 given the following ∆H of formation: Na2CO3 = -1,130.94 kJ/mol; CO2 = -393.5 kJ/mol; H2O = -241.8 kJ/mol. Answer: Chemistry 5 points 5.0 2 Answer shineri1234 • … WebAug 6, 2024 · Calculate the standard enthalpy of combustion of propane Given the following data (1) C (s) + O2 (g) ==> CO2 (g) ΔHθ = -393 kJ/mol-1 (2) H2 (g) + 1/2O2 (g) ==> H2O (l) ΔHθ = -286 kJ/mol-1 (3) 3C (s) + 4H2 (g) ==> C3H8 (g) ΔHθ = -104 kJ/mol-1 Calculate the standard enthalpy of combustion of propane Follow • 2 Add comment … WebConsider the following thermochemical equations. PCl5 (s)→PCl3 (g)+Cl2 (g)2P (s)+3Cl2 (g)→2PCl3 (g)ΔH∘rxnΔH∘rxn=87.9kJmol=−574kJmol. Using this data, determine the … herborium group stock