To calculate the enthalpy change (\(\Delta H_{rxn}\)) for the combustion of ethanol, you can use the bond energy values provided. The reaction is:
\[CH_3CH_2OH(l) + 3 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(g)\]
The enthalpy change (\(\Delta H_{rxn}\)) can be calculated using the following equation:
\[\Delta H_{rxn} = \sum \text{(bond energies of bonds broken)} - \sum \text{(bond energies of bonds formed)}\]
First, identify the bonds broken and formed in the reaction:
**Bonds Broken:**
- 1 C-H bond in CH3CH2OH
- 3 O=O bonds in O2
**Bonds Formed:**
- 2 C=O bonds in CO2
- 3 O-H bonds in H2O
Now, substitute the bond energy values into the equation:
\[\Delta H_{rxn} = (1 \times 411) + (3 \times 498) - (2 \times 805) - (3 \times 464)\]
Calculate the values:
\[\Delta H_{rxn} = 411 + 1494 - 1610 - 1392\]
\[\Delta H_{rxn} = -97 \, \text{kJ}\]
Therefore, the enthalpy change (\(\Delta H_{rxn}\)) for the combustion of ethanol is -97 kJ. The negative sign indicates that the reaction is exothermic.