Question

A student needs to determine the change in enthalpy (ΔH) for the reaction between methane (CH4) and oxygen (O2) to form carbon dioxide (CO2) and water (H2O) at a temperature of 25°C and a pressure of 1 atm. The initial concentrations of CH4 and O2 are both 0.1 M, and the final concentrations of CO2 and H2O are both 0.05 M. What is the change in enthalpy for this reaction at the given conditions?

Answer 1

To determine the change in enthalpy  H  for the reaction between methane  CH4  and oxygen  O2  to form carbon dioxide  CO2  and water  H2O , we need to use the balanced chemical equation and the standard enthalpies of formation for each compound involved in the reaction.The balanced chemical equation for the combustion of methane is:CH4 g  + 2 O2 g   CO2 g  + 2 H2O g Next, we need the standard enthalpies of formation  Hf  for each compound at 25C  298 K . These values can be found in a standard thermodynamic table:Hf  CH4  = -74.8 kJ/molHf  O2  = 0 kJ/mol  since O2 is in its standard state Hf  CO2  = -393.5 kJ/molHf  H2O  = -241.8 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 CO2   -393.5 kJ/mol  + 2 mol H2O   -241.8 kJ/mol ] - [1 mol CH4   -74.8 kJ/mol  + 2 mol O2   0 kJ/mol ]H =  -393.5 - 2  241.8 + 74.8  kJH =  -393.5 - 483.6 + 74.8  kJH = -802.3 kJTherefore, the change in enthalpy  H  for the reaction between methane and oxygen to form carbon dioxide and water at the given conditions is -802.3 kJ.