Question

Calculate the enthalpy of oxidation of methane gas in kJ/mol if the reaction produces carbon dioxide gas and water vapor, and the enthalpy changes for the following reactions are known: - Enthalpy change of combustion of methane = -890.3 kJ/mol - Enthalpy change of formation of carbon dioxide = -393.5 kJ/mol - Enthalpy change of formation of water vapor = -241.8 kJ/mol

Answer 1

To calculate the enthalpy of oxidation of methane gas, we need to use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction. In this case, we are given the enthalpy change of combustion of methane, the enthalpy change of formation of carbon dioxide, and the enthalpy change of formation of water vapor.The balanced equation for the oxidation of methane is:CH g  + 2O g   CO g  + 2HO g First, we need to find the enthalpy change for the formation of methane  Hf[CH] . We can do this by rearranging the enthalpy change of combustion of methane equation:Hc[CH] = Hf[CO] + 2Hf[HO] - Hf[CH]Now, we can plug in the given values:-890.3 kJ/mol =  -393.5 kJ/mol  + 2 -241.8 kJ/mol  - Hf[CH]Solve for Hf[CH]:Hf[CH] =  -393.5 kJ/mol  + 2 -241.8 kJ/mol  + 890.3 kJ/molHf[CH] = -393.5 kJ/mol - 483.6 kJ/mol + 890.3 kJ/molHf[CH] = 13.2 kJ/molNow that we have the enthalpy change for the formation of methane, we can calculate the enthalpy of oxidation of methane:H[oxidation] = Hf[CO] + 2Hf[HO] - Hf[CH]Plug in the values:H[oxidation] =  -393.5 kJ/mol  + 2 -241.8 kJ/mol  - 13.2 kJ/molH[oxidation] = -393.5 kJ/mol - 483.6 kJ/mol - 13.2 kJ/molH[oxidation] = -890.3 kJ/molThe enthalpy of oxidation of methane gas is -890.3 kJ/mol.