Question

Calculate the enthalpy change for the reaction of burning 1 mole of methane using Hess's Law, given the following information:1) Enthalpy of formation for methane = -74.8 kJ/mol2) Enthalpy of combustion for methane = -890.3 kJ/mol3) Enthalpy of formation for water = -285.8 kJ/mol4) Enthalpy of formation for carbon dioxide = -393.5 kJ/mol

Answer 1

To calculate the enthalpy change for the reaction of burning 1 mole of methane using Hess's Law, we first need to write the balanced chemical equation for the combustion of methane:CH g  + 2 O g   CO g  + 2 HO l Now, we can use Hess's Law to find the enthalpy change for this reaction. Hess's Law states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for the formation of the products minus the sum of the enthalpy changes for the formation of the reactants.H reaction  =  H products  -  H reactants In this case, the enthalpy change for the reaction is:H reaction  = [1  H CO  + 2  H HO ] - [1  H CH  + 2  H O ]Since the enthalpy of formation for O is zero  it is in its standard state , the equation becomes:H reaction  = [1  H CO  + 2  H HO ] - [1  H CH ]Now, we can plug in the given values for the enthalpy of formation:H reaction  = [1   -393.5 kJ/mol  + 2   -285.8 kJ/mol ] - [1   -74.8 kJ/mol ]H reaction  =  -393.5 - 2  285.8 + 74.8  kJ/molH reaction  =  -393.5 - 571.6 + 74.8  kJ/molH reaction  = -890.3 kJ/molTherefore, the enthalpy change for the reaction of burning 1 mole of methane is -890.3 kJ/mol.