Question

Calculate the standard enthalpy change for the reaction: CH3OH (l) + 3/2 O2 (g) -> CO2 (g) + 2H2O (l) given the standard enthalpies of formation for CH3OH (l), CO2 (g) and H2O (l) as -238.6 kJ/mol, -393.5 kJ/mol and -285.8 kJ/mol respectively.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the formula:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation.For the given reaction:CH3OH  l  + 3/2 O2  g  -> CO2  g  + 2H2O  l The standard enthalpies of formation are:Hf CH3OH  = -238.6 kJ/molHf CO2  = -393.5 kJ/molHf H2O  = -285.8 kJ/molThe standard enthalpy change for the reaction is:H = [1   -393.5  + 2   -285.8 ] - [1   -238.6  + 3/2  0]Note that the standard enthalpy of formation for O2  g  is 0 kJ/mol since it is in its standard state.H = [-393.5 - 571.6] - [-238.6]H = [-965.1] - [-238.6]H = -726.5 kJ/molThe standard enthalpy change for the reaction is -726.5 kJ/mol.