Calculate the standard enthalpy change for the reaction between 2.00 moles of liquid ethanol (C2H5OH) and 3.50 moles of gaseous oxygen (O2) to form carbon dioxide (CO2) gas and liquid water (H2O) at 25°C and standard pressure. The balanced chemical equation for the reaction is C2H5OH + 3O2 → 2CO2 + 3H2O. (The molar enthalpies of formation for C2H5OH, CO2, and H2O are -277.6 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively.)

Question

Calculate the standard enthalpy change for the reaction between 2.00 moles of liquid ethanol (C2H5OH) and 3.50 moles of gaseous oxygen (O2) to form carbon dioxide (CO2) gas and liquid water (H2O) at 25°C and standard pressure. The balanced chemical equation for the reaction is C2H5OH + 3O2 → 2CO2 + 3H2O. (The molar enthalpies of formation for C2H5OH, CO2, and H2O are -277.6 kJ/mol, -393.5 kJ/m

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.First, we need to determine the moles of each product and reactant involved in the balanced chemical equation:C2H5OH: 2.00 molesO2: 3.50 molesCO2: 2 moles  since there are 2 moles of CO2 produced for every mole of C2H5OH H2O: 3 moles  since there are 3 moles of H2O produced for every mole of C2H5OH Now, we can plug these values into the formula:H = [2 moles CO2   -393.5 kJ/mol CO2  + 3 moles H2O   -285.8 kJ/mol H2O ] - [2.00 moles C2H5OH   -277.6 kJ/mol C2H5OH  + 3.50 moles O2   0 kJ/mol O2 ]H = [-787 kJ +  -857.4 kJ ] - [-555.2 kJ + 0 kJ]H =  -1644.4 kJ  -  -555.2 kJ H = -1089.2 kJThe standard enthalpy change for the reaction between 2.00 moles of liquid ethanol and 3.50 moles of gaseous oxygen to form carbon dioxide gas and liquid water at 25C and standard pressure is -1089.2 kJ.