Question

Calculate the change in enthalpy for the following chemical reaction: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) Given that the enthalpies of formation for CH4(g), CO2(g), and H2O(l) are -74.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol respectively.

Answer 1

To calculate the change in enthalpy  H  for the given chemical reaction, we can use the formula:H =  H products  -  H reactants where H products  is the sum of the enthalpies of formation of the products, and H reactants  is the sum of the enthalpies of formation of the reactants.For the given reaction:CH4 g  + 2O2 g  -> CO2 g  + 2H2O l The enthalpies of formation are:Hf CH4  = -74.8 kJ/molHf CO2  = -393.5 kJ/molHf H2O  = -285.8 kJ/molThe enthalpy of formation for O2 g  is 0 kJ/mol since it is in its standard state.Now, we can calculate the change in enthalpy:H = [1   -393.5  + 2   -285.8 ] - [1   -74.8  + 2  0]H =  -393.5 - 571.6  -  -74.8 H = -965.1 + 74.8H = -890.3 kJ/molSo, the change in enthalpy for the given chemical reaction is -890.3 kJ/mol.