Question

Find the standard enthalpy change for the reaction: 2C2H5OH(l) + 9O2(g) --> 6CO2(g) + 8H2O(l) given the following information: Standard enthalpy of formation (ΔHfº) of C2H5OH(l) = -277 kJ/mol Standard enthalpy of formation (ΔHfº) of CO2(g) = -393.5 kJ/mol Standard enthalpy of formation (ΔHfº) of H2O(l) = -285.8 kJ/mol Assume that the reaction occurs at standard conditions.

Answer 1

To find the standard enthalpy change for the reaction, we can use the following equation:H_reaction =  Hf_products -  Hf_reactantsFor the products, we have 6 moles of CO2 g  and 8 moles of H2O l : Hf_products =  6 mol  -393.5 kJ/mol  +  8 mol  -285.8 kJ/mol  = -2361 kJ - 2286.4 kJ = -4647.4 kJFor the reactants, we have 2 moles of C2H5OH l  and 9 moles of O2 g . The standard enthalpy of formation for O2 g  is 0 kJ/mol since it is in its elemental form: Hf_reactants =  2 mol  -277 kJ/mol  +  9 mol  0 kJ/mol  = -554 kJNow, we can find the standard enthalpy change for the reaction:H_reaction =  Hf_products -  Hf_reactants = -4647.4 kJ -  -554 kJ  = -4093.4 kJThe standard enthalpy change for the reaction is -4093.4 kJ.