Answer :
To calculate the enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction [tex]\(C + O_2 \rightarrow CO_2\)[/tex], we need to consider the bond energies involved in breaking and forming the bonds.
First, identify the bonds broken and formed:
Bonds broken:
- 1 [tex]\(O=O\)[/tex] bond in [tex]\(O_2\)[/tex]
Bonds formed:
- 2 [tex]\(C=O\)[/tex] bonds in [tex]\(CO_2\)[/tex]
Now use the bond energies provided:
- Bond energy of [tex]\(C=O\)[/tex]: 799 kJ/mol
- Bond energy of [tex]\(O=O\)[/tex]: 494 kJ/mol
Calculate the total energy required to break the bonds:
[tex]\[ \text{Energy for broken bonds} = 1 \times 494 \text{ kJ/mol} = 494 \text{ kJ/mol} \][/tex]
Calculate the total energy released when new bonds are formed:
[tex]\[ \text{Energy for formed bonds} = 2 \times 799 \text{ kJ/mol} = 1598 \text{ kJ/mol} \][/tex]
The enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction is the energy of the bonds broken minus the energy of the bonds formed:
[tex]\[ \Delta H = \text{Energy for broken bonds} - \text{Energy for formed bonds} \][/tex]
Substitute the calculated values:
[tex]\[ \Delta H = 494 \text{ kJ/mol} - 1598 \text{ kJ/mol} = -1104 \text{ kJ/mol} \][/tex]
The change in enthalpy for the given reaction is [tex]\(-1104\)[/tex] kJ.
First, identify the bonds broken and formed:
Bonds broken:
- 1 [tex]\(O=O\)[/tex] bond in [tex]\(O_2\)[/tex]
Bonds formed:
- 2 [tex]\(C=O\)[/tex] bonds in [tex]\(CO_2\)[/tex]
Now use the bond energies provided:
- Bond energy of [tex]\(C=O\)[/tex]: 799 kJ/mol
- Bond energy of [tex]\(O=O\)[/tex]: 494 kJ/mol
Calculate the total energy required to break the bonds:
[tex]\[ \text{Energy for broken bonds} = 1 \times 494 \text{ kJ/mol} = 494 \text{ kJ/mol} \][/tex]
Calculate the total energy released when new bonds are formed:
[tex]\[ \text{Energy for formed bonds} = 2 \times 799 \text{ kJ/mol} = 1598 \text{ kJ/mol} \][/tex]
The enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction is the energy of the bonds broken minus the energy of the bonds formed:
[tex]\[ \Delta H = \text{Energy for broken bonds} - \text{Energy for formed bonds} \][/tex]
Substitute the calculated values:
[tex]\[ \Delta H = 494 \text{ kJ/mol} - 1598 \text{ kJ/mol} = -1104 \text{ kJ/mol} \][/tex]
The change in enthalpy for the given reaction is [tex]\(-1104\)[/tex] kJ.