To determine the enthalpy change ([tex]\( \Delta H \)[/tex]) for the given reaction:
[tex]\[ C + O_2 \rightarrow CO_2 \][/tex]
we need to consider the bond energies provided.
- Bond energy for [tex]\( C = O \)[/tex] is [tex]\( 799 \, kJ/mol \)[/tex].
- Bond energy for [tex]\( O = O \)[/tex] is [tex]\( 494 \, kJ/mol \)[/tex].
The change in enthalpy for a reaction can be calculated using the bond energies of the bonds broken and formed. The formula used is:
[tex]\[ \Delta H = \Sigma (\text{Bond energies of bonds broken}) - \Sigma (\text{Bond energies of bonds formed}) \][/tex]
1. Identify the bonds broken:
In the reactants, we have one [tex]\( O_2 \)[/tex] molecule, which forms an [tex]\( O = O \)[/tex] double bond.
So, it involves breaking one [tex]\( O = O \)[/tex] bond.
- Bond energy of [tex]\( O = O \)[/tex]: [tex]\( 494 \, kJ/mol \)[/tex]
2. Identify the bonds formed:
In the products, we have one [tex]\( CO_2 \)[/tex] molecule, which has two [tex]\( C = O \)[/tex] double bonds.
So, for forming one [tex]\( CO_2 \)[/tex], it involves forming two [tex]\( C = O \)[/tex] bonds.
- Bond energy of [tex]\( C = O \)[/tex]: [tex]\( 799 \, kJ/mol \)[/tex]
3. Calculate the total bond energies:
- Bonds broken: [tex]\( 1 \times 494 = 494 \, kJ \)[/tex]
- Bonds formed: [tex]\( 2 \times 799 = 1598 \, kJ \)[/tex]
4. Calculate the enthalpy change ([tex]\( \Delta H \)[/tex]):
Substituting these values into our formula:
[tex]\[ \Delta H = 494 \, kJ - 1598 \, kJ \][/tex]
This results in:
[tex]\[ \Delta H = -1104 \, kJ \][/tex]
Therefore, the enthalpy change for the reaction is [tex]\(-1104 \, kJ\)[/tex].
