Question

Calculate the standard enthalpy change for the formation of 2 moles of water vapor (H2O(g)) if the combustion of 1 mole of methane gas (CH4(g)) results in the formation of 2 moles of water vapor and 1 mole of carbon dioxide gas (CO2(g)). Given that the standard enthalpy change of combustion of methane is -890.4 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the formation of 2 moles of water vapor  H2O g  , we can use the given information about the combustion of methane  CH4 g  .The balanced chemical equation for the combustion of methane is:CH4 g  + 2O2 g   CO2 g  + 2H2O g The standard enthalpy change of combustion of methane is given as -890.4 kJ/mol. This value corresponds to the formation of 1 mole of CO2 g  and 2 moles of H2O g  from 1 mole of CH4 g .Since we are only interested in the enthalpy change for the formation of 2 moles of water vapor, we can ignore the formation of CO2 g  in this case.The enthalpy change for the formation of 2 moles of H2O g  is the same as the enthalpy change for the combustion of 1 mole of CH4 g , which is -890.4 kJ/mol.Therefore, the standard enthalpy change for the formation of 2 moles of water vapor  H2O g   is -890.4 kJ/mol.