To calculate the enthalpy change (ΔH) for the given chemical reaction:
[tex]\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \][/tex]
we will use the bond energies provided:
- Bond energy for [tex]\( \text{C}=\text{O} \)[/tex] is 799 kJ/mol.
- Bond energy for [tex]\( \text{O}=\text{O} \)[/tex] is 494 kJ/mol.
### Steps for Calculating the Enthalpy Change:
1. Identify the bonds broken and formed in the reaction.
In this reaction, we are breaking one [tex]\( \text{O}=\text{O} \)[/tex] bond in [tex]\( \text{O}_2 \)[/tex] and forming two [tex]\( \text{C}=\text{O} \)[/tex] bonds in [tex]\( \text{CO}_2 \)[/tex].
2. Determine the total energy required to break the bonds.
- Energy required to break one [tex]\( \text{O}=\text{O} \)[/tex] bond = 494 kJ/mol.
3. Determine the total energy released in forming the new bonds.
- Energy released in forming two [tex]\( \text{C}=\text{O} \)[/tex] bonds = 2 × 799 kJ/mol = 1598 kJ/mol.
4. Calculate the enthalpy change (ΔH).
The enthalpy change ΔH is given by the difference between the energy of the bonds broken and the energy of the bonds formed:
[tex]\( \Delta H = (\text{Energy of bonds broken}) - (\text{Energy of bonds formed}) \)[/tex]
Substituting the values:
[tex]\( \Delta H = 494 \, \text{kJ/mol} - 1598 \, \text{kJ/mol} \)[/tex]
[tex]\( \Delta H = -1104 \, \text{kJ/mol} \)[/tex]
### Final Answer:
The change in enthalpy for the given reaction is [tex]\(-1104\)[/tex] kJ.
