What is the enthalpy change (in kJ/mol) for the combustion of propane (C3H8) using Hess's Law? Use the following information: C3H8 + 5O2 → 3CO2 + 4H2O ∆H = −2043 kJ/mol 2CO + O2 → 2CO2 ∆H = −566 kJ/mol H2 + 1/2O2 → H2O ∆H = −285.8 kJ/mol

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What is the enthalpy change (in kJ/mol) for the combustion of propane (C3H8) using Hess's Law? Use the following information: C3H8 + 5O2 → 3CO2 + 4H2O ∆H = −2043 kJ/mol 2CO + O2 → 2CO2 ∆H = −566 kJ/mol H2 + 1/2O2 → H2O ∆H = −285.8 kJ/mol

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RodrigoHopwo · Answer 1 · 2025-02-20T01:46:22+0000

The enthalpy change for the combustion of propane C3H8 is already given in the information provided:C3H8 + 5O2 3CO2 + 4H2O H = 2043 kJ/molSo, the enthalpy change for the combustion of propane is -2043 kJ/mol.

What is the enthalpy change (in kJ/mol) for the combustion of propane (C3H8) using Hess's Law? Use the following information: C3H8 + 5O2 → 3CO2 + 4H2O ∆H = −2043 kJ/mol 2CO + O2 → 2CO2 ∆H = −566 kJ/mol H2 + 1/2O2 → H2O ∆H = −285.8 kJ/mol

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