Question

Determine the standard enthalpy of formation for acetic acid (CH3COOH) using Hess's law and the following thermochemical equations:1) C(s) + O2(g) → CO2(g) ΔH = -394 kJ/mol2) 2H2(g) + O2(g) → 2H2O(l) ΔH = -572 kJ/mol3) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -1234 kJ/mol4) CH3COOH(l) + O2(g) → CO2(g) + H2O(l) ΔH = ?

Answer 1

To find the standard enthalpy of formation for acetic acid  CH3COOH , we need to manipulate the given thermochemical equations to match the formation of acetic acid from its elements in their standard states. The formation reaction for acetic acid is:C s  + 2H2 g  + O2 g   CH3COOH l First, we need to find the enthalpy change for the formation of one mole of H2O l  from its elements. We can do this by dividing equation 2 by 2:1/2 *  2H2 g  + O2 g   2H2O l   => H2 g  + 1/2O2 g   H2O l  H = -286 kJ/molNext, we need to find the enthalpy change for the formation of one mole of C2H5OH l  from its elements. We can do this by reversing equation 3 and dividing it by 2:1/2 *  2CO2 g  + 3H2O l   C2H5OH l  + 3O2 g   => CO2 g  + 3/2H2O l   C2H5OH l  + 3/2O2 g  H = 617 kJ/molNow, we can sum the enthalpy changes for the formation of CO2 g , H2O l , and C2H5OH l  to find the enthalpy change for the formation of CH3COOH l :C s  + O2 g   CO2 g  H = -394 kJ/molH2 g  + 1/2O2 g   H2O l  H = -286 kJ/molCO2 g  + 3/2H2O l   C2H5OH l  + 3/2O2 g  H = 617 kJ/molAdding these equations together, we get:C s  + 2H2 g  + O2 g   CH3COOH l  + C2H5OH l Now, we need to find the enthalpy change for the reaction:CH3COOH l  + C2H5OH l   CH3COOH l  + O2 g To do this, we can subtract the enthalpy change for the formation of C2H5OH l  from the enthalpy change for the formation of CH3COOH l :H CH3COOH  = H CO2  + H H2O  - H C2H5OH H CH3COOH  =  -394 kJ/mol  +  -286 kJ/mol  -  617 kJ/mol H CH3COOH  = -680 kJ/molTherefore, the standard enthalpy of formation for acetic acid  CH3COOH  is -680 kJ/mol.