Question

Calculate the standard enthalpy change of formation for ethanoic acid, given the following standard enthalpy of combustion data:- Ethanol: -1367 kJ/mol- Carbon dioxide: -393.5 kJ/mol- Water: -285.8 kJ/molThe chemical equation for the formation of ethanoic acid is:2C2H5OH + 2O2 → 2CH3COOH + 2H2O

Answer 1

To calculate the standard enthalpy change of formation for ethanoic acid, we need to use Hess's Law. We can use the given standard enthalpy of combustion data to create a series of reactions that will ultimately give us the desired reaction.First, let's write the combustion reactions for ethanol, ethanoic acid, and water:1  C2H5OH + 3O2  2CO2 + 3H2O  H = -1367 kJ/mol2  CH3COOH + 2O2  2CO2 + 2H2O  H = x  unknown 3  H2 + 1/2O2  H2O  H = -285.8 kJ/molNow, we need to manipulate these reactions to obtain the desired reaction:2C2H5OH + 2O2  2CH3COOH + 2H2OFirst, multiply reaction 3 by 2 to get 2 moles of water:3'  2H2 + O2  2H2O  H = -571.6 kJ/molNow, reverse reaction 1 and multiply it by 2:1'  4CO2 + 6H2O  4C2H5OH + 6O2  H = 2734 kJ/molFinally, add reactions 1', 2, and 3' together:4CO2 + 6H2O  4C2H5OH + 6O2  H = 2734 kJ/molCH3COOH + 2O2  2CO2 + 2H2O  H = x  unknown 2H2 + O2  2H2O  H = -571.6 kJ/mol---------------------------------------------2C2H5OH + 2O2  2CH3COOH + 2H2O  H = x + 2162.4 kJ/molNow, we can solve for the unknown enthalpy change of formation for ethanoic acid  x :x + 2162.4 kJ/mol = 0x = -2162.4 kJ/molHowever, this is the enthalpy change for the formation of 2 moles of ethanoic acid. To find the enthalpy change for the formation of 1 mole of ethanoic acid, we need to divide this value by 2:Hf  CH3COOH  = -2162.4 kJ/mol / 2 = -1081.2 kJ/molSo, the standard enthalpy change of formation for ethanoic acid is -1081.2 kJ/mol.