Question

Calculate the standard enthalpy change for the formation of methane (CH4) given the following equations:C(s) + O2(g) ⟶ CO2(g); ΔH = -393.5 kJ/mol2H2(g) + O2(g) ⟶ 2H2O(l); ΔH = -571.6 kJ/mol

Answer 1

To calculate the standard enthalpy change for the formation of methane  CH4 , we need to manipulate the given equations to form the desired equation: C s  + 2H2 g   CH4 g .First, reverse the first equation to get CO2 g  to C s  + O2 g :CO2 g   C s  + O2 g ; H = +393.5 kJ/molNext, divide the second equation by 2 to get 1 mol of H2O l  on the product side:H2 g  + 1/2 O2 g   H2O l ; H = -571.6 kJ/mol  2 = -285.8 kJ/molNow, we need to convert H2O l  to H2O g  using the enthalpy of vaporization of water, which is 40.7 kJ/mol:H2O l   H2O g ; H = +40.7 kJ/molAdd the last two equations to get H2 g  + 1/2 O2 g   H2O g :H2 g  + 1/2 O2 g   H2O g ; H = -285.8 kJ/mol + 40.7 kJ/mol = -245.1 kJ/molNow, we need two moles of H2 g  on the reactant side, so multiply the last equation by 2:2H2 g  + O2 g   2H2O g ; H = -245.1 kJ/mol  2 = -490.2 kJ/molFinally, add the first and last equations to get the desired equation:CO2 g  + 2H2 g  + O2 g   C s  + 2H2O g  + 2H2 g  + O2 g ; H = +393.5 kJ/mol - 490.2 kJ/molC s  + 2H2 g   CH4 g ; H = -96.7 kJ/molThe standard enthalpy change for the formation of methane  CH4  is -96.7 kJ/mol.