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What is the standard enthalpy change for the combustion of methane (CH4) gas to form carbon dioxide (CO2) gas and water (H2O) at 298 K? Use quantum chemistry calculations to determine the necessary thermodynamic properties of CH4, CO2, and H2O to solve the problem.

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As an AI language model, I am unable to perform quantum chemistry calculations directly. However, I can provide you with the standard enthalpy change for the combustion of methane using known values from experimental data.The balanced chemical equation for the combustion of methane is:CH g  + 2O g   CO g  + 2HO g Using standard enthalpies of formation  Hf  at 298 K, we can calculate the standard enthalpy change  H  for this reaction. The standard enthalpies of formation for the compounds involved are:Hf CH  = -74.81 kJ/molHf CO  = -393.51 kJ/molHf HO  = -241.82 kJ/mol  for gaseous water The standard enthalpy change for the reaction can be calculated using the following formula:H =  [Hf products ] -  [Hf reactants ]H = [1   -393.51  + 2   -241.82 ] - [1   -74.81  + 2  0]H =  -393.51 - 483.64  -  -74.81 H = -802.34 kJ/molSo, the standard enthalpy change for the combustion of methane at 298 K is approximately -802.34 kJ/mol.

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