Calculate (triangle)H for the following gas-phase reaction: 2 H-C (3dashes) C-H + 5 O=O --> 4 O=C=O = 2 H-O-H
-Using bond energies
To calculate the enthalpy change (∆H) using bond energies, you need to know the bond energies of the bonds involved in the reaction. Bond energy is the energy required to break a bond, and it is usually given in units of kilojoules per mole (kJ/mol).
Here are the bond energies for the bonds involved in the reaction:
- H-C: 415 kJ/mol
- C-H: 413 kJ/mol
- O=O: 498 kJ/mol
- O=C: 805 kJ/mol
- C=O: 745 kJ/mol
- H-O: 464 kJ/mol
Now, we can calculate the enthalpy change (∆H) using the following formula:
∆H = (Sum of bond energies in reactants) - (Sum of bond energies in products)
First, let's calculate the sum of bond energies in the reactants:
2(H-C) + 5(O=O) = 2(415 kJ/mol) + 5(498 kJ/mol) = 830 kJ/mol + 2490 kJ/mol = 3320 kJ/mol
Next, calculate the sum of bond energies in the products:
4(O=C=O) + 2(H-O-H) = 4(745 kJ/mol) + 2(464 kJ/mol) = 2980 kJ/mol + 928 kJ/mol = 3908 kJ/mol
Finally, calculate the ∆H:
∆H = (Sum of bond energies in reactants) - (Sum of bond energies in products)
∆H = 3320 kJ/mol - 3908 kJ/mol = -588 kJ/mol
Therefore, the ∆H for the given gas-phase reaction using bond energies is approximately -588 kJ/mol.