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What is the change in enthalpy (in kilojoules) for the combustion reaction of methane (CH4) at standard conditions (298 K, 1 atm)?

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To calculate the change in enthalpy  H  for the combustion of methane  CH4  at standard conditions, we need to consider the balanced chemical equation for the combustion reaction:CH4 g  + 2 O2 g   CO2 g  + 2 H2O l Next, we need to know the standard enthalpies of formation  Hf  for each compound involved in the reaction. The standard enthalpies of formation are as follows:Hf  CH4  = -74.8 kJ/molHf  O2  = 0 kJ/mol  since O2 is in its standard state Hf  CO2  = -393.5 kJ/molHf  H2O  = -285.8 kJ/molNow, we can calculate the change in enthalpy  H  for the reaction using the following equation:H =  [Hf  products ] -  [Hf  reactants ]H = [1 mol   -393.5 kJ/mol  + 2 mol   -285.8 kJ/mol ] - [1 mol   -74.8 kJ/mol  + 2 mol   0 kJ/mol ]H =  -393.5 - 2  285.8  -  -74.8 H = -965.1 kJ/molSo, the change in enthalpy  H  for the combustion reaction of methane  CH4  at standard conditions  298 K, 1 atm  is -965.1 kJ/mol.

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