Question

Calculate the standard enthalpy change for the reaction: 2C2H5OH(l) + 9O2(g) -> 6CO2(g) + 8H2O(l) given that the standard enthalpy of formation for C2H5OH(l) is -278 kJ/mol, for CO2(g) is -394 kJ/mol and for H2O(l) is -286 kJ/mol.

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 reactants:2 moles of C2H5OH l : 2   -278 kJ/mol  = -556 kJ9 moles of O2 g : 9  0 kJ/mol  since O2 is in its standard state  = 0 kJFor the products:6 moles of CO2 g : 6   -394 kJ/mol  = -2364 kJ8 moles of H2O l : 8   -286 kJ/mol  = -2288 kJNow, we can plug these values into the formula:H =  -2364 kJ +  -2288 kJ   -  -556 kJ + 0 kJ H =  -4652 kJ  -  -556 kJ H = -4096 kJThe standard enthalpy change for the reaction is -4096 kJ.