Explanation:
The reaction is as follows.
[tex]CH_{3}CH_{2}OH(g) + 3O_{2}(g) \rightarrow 2CO_{2} + 3H_{2}O(g)[/tex]
Standard values of bond energies are as follows.
C-C = 347 kJ/mol
C-H = 414 kJ/mol
C-O = 360 kJ/mol
O-H = 464 kJ/mol
O=O = 498 kJ/mol
C=O = 799 kJ/mol
Hence, calculate the change in enthalpy of the reaction as follows.
[tex]\Delta H_{rxn} = \sum \text{bond energy of reactants} - \sum \text{bond energy of products}[/tex]
= [5(C-H) + 1(C-C) + 1(C-O) + 1(O-H) + 3(O=O)] - [4(C=O) + 6(O-H)]
= [tex][5 \times 414 + 1 \times 347 + 1 \times 360 + 1 \times 464 + 3 \times 498] kJ - [4 \times 799 + 6 \times 464]kJ[/tex]
= (4735 - 5980) kJ
= -1245 kJ
Thus, we can conclude that the enthalpy of given reaction is -1245 kJ.
The enthalpy for the combustion of ethanol, calculated using average bond energies, is -1245 kJ/mol.
Let's consider the following combustion equation.
CH₃CH₂OH(g) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g)
Given the average bond energies, we can calculate the enthalpy of the reaction (ΔHrxn) using the following equation.
[tex]\Delta H_{rxn} = \Sigma broken\ bonds - \Sigma formed\ bonds[/tex]
The broken bonds, with their bond energies, are:
CH₃CH₂OH
O₂
The formed bonds, with their bond energies, are:
CO₂
H₂O
The enthalpy of the reaction is:
[tex]\Delta H_{rxn} = 5\Delta H(C-H) + 1\Delta H(C-C) + 1\Delta H(C-O) + 1\Delta H(O-H)+3\Delta H(O=O)-4\Delta H(C=O)-6\Delta H(O-H)\\\\\Delta H_{rxn} = 5(414kJ/mol) + 1(347kJ/mol) + 1(360 kJ/mol) + 1(464kJ/mol)+3(498kJ/mol)-4(799kJ/mol)-6(464kJ/mol)= -1245kJ/mol[/tex]
The enthalpy for the combustion of ethanol, calculated using average bond energies, is -1245 kJ/mol.
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