Calculate ∆H˚ for the following reaction using the given bond dissociation energies.
CH4 (g) + 2O2 (g) --> CO2 (g) + 2H2O (g)
Bond.......∆H˚ (kJ/mol)
O-O........142
H-O.........459
C-H.........411
C=O........799
O=O........498
C-O.........358
∆H˚ = ________kJ/mol
This reaction is endothermic or exothermic?
Please help! I'm so confused!
See your previous post.
To calculate ∆H˚ for the given reaction using bond dissociation energies, you need to calculate the sum of the bond energies of the bonds broken minus the sum of the bond energies of the bonds formed.
In the reaction: CH4 (g) + 2O2 (g) --> CO2 (g) + 2H2O (g)
Bonds broken: 4 C-H bonds and 2 O=O bonds
Bonds formed: 2 C=O bonds and 4 O-H bonds
Now, you need to multiply the number of bonds in each category by the corresponding bond energy.
For bonds broken:
4 C-H bonds * 411 kJ/mol = 1644 kJ/mol
2 O=O bonds * 498 kJ/mol = 996 kJ/mol
For bonds formed:
2 C=O bonds * 799 kJ/mol = 1598 kJ/mol
4 O-H bonds * 459 kJ/mol = 1836 kJ/mol
Now, add up the bond energies for the bonds broken and subtract the sum of the bond energies for the bonds formed:
Total energy for bonds broken: 1644 kJ/mol + 996 kJ/mol = 2640 kJ/mol
Total energy for bonds formed: 1598 kJ/mol + 1836 kJ/mol = 3434 kJ/mol
∆H˚ = Total energy for bonds broken - Total energy for bonds formed
∆H˚ = 2640 kJ/mol - 3434 kJ/mol
∆H˚ = -794 kJ/mol
The value for ∆H˚ is -794 kJ/mol.
Since ∆H˚ is negative, it means the reaction is exothermic.
To calculate ∆H˚ for the reaction, we need to analyze the bonds broken and formed during the reaction.
In the reactants:
- 4 C-H bonds are broken (4 * 411 kJ/mol)
- 2 O=O bonds are broken (2 * 498 kJ/mol)
In the products:
- 2 C=O bonds are formed (2 * 799 kJ/mol)
- 4 O-H bonds are formed (4 * 459 kJ/mol)
Now, let's calculate the total energy change by summing up the bond energies:
∆H˚ = (4 * 411) + (2 * 498) - (2 * 799) - (4 * 459)
∆H˚ = 1644 + 996 - 1598 - 1836
∆H˚ = -794 kJ/mol
Since the value of ∆H˚ is negative (-794 kJ/mol), the reaction is exothermic.