c2h6o + 3o2 → 2co2 + 3h2o
Use the bond enthalpies to calculate the enthalpy change for this reaction. Is the reaction endothermic or exothermic?
SHOW WORK!!
To calculate the enthalpy change for this reaction using bond enthalpies, you need to determine the sum of the bond energies of the reactants and the products, and then find the difference between the two.
First, we'll list the bonds broken and formed in this reaction:
Reactants: C2H6O (ethanol) and 3O2 (oxygen gas)
Products: 2CO2 (carbon dioxide) and 3H2O (water)
Next, we'll find the bond energies for each bond involved in the reaction. Here are the average bond enthalpy values:
C-C = 347 kJ/mol
C-O = 358 kJ/mol
O-O = 498 kJ/mol
O-H = 463 kJ/mol
C=O = 805 kJ/mol
Now, let's calculate the total bond energy for the reactants by summing the bond energies of all the bonds in ethanol (C2H6O) and oxygen gas (O2):
Reactant bond energies:
C-C (2) = 2 * 347 kJ/mol = 694 kJ/mol
C-O (1) = 1 * 358 kJ/mol = 358 kJ/mol
O-O (3) = 3 * 498 kJ/mol = 1494 kJ/mol
O-H (6) = 6 * 463 kJ/mol = 2778 kJ/mol
Total reactant bond energy = 694 kJ/mol + 358 kJ/mol + 1494 kJ/mol + 2778 kJ/mol = 5324 kJ/mol
Next, we calculate the total bond energy for the products by summing the bond energies of all the bonds in carbon dioxide (CO2) and water (H2O):
Product bond energies:
C=O (4) = 4 * 805 kJ/mol = 3220 kJ/mol
O-H (6) = 6 * 463 kJ/mol = 2778 kJ/mol
Total product bond energy = 3220 kJ/mol + 2778 kJ/mol = 5998 kJ/mol
Finally, we find the enthalpy change (ΔH) by taking the difference between the total bond energy of the reactants and the total bond energy of the products:
ΔH = Total product bond energy - Total reactant bond energy
= 5998 kJ/mol - 5324 kJ/mol
= 674 kJ/mol
Since the enthalpy change (ΔH) is positive (674 kJ/mol), the reaction is endothermic, meaning it requires an input of energy to proceed.