Answer :
To calculate the enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) for the combustion of ethanol, we need to use bond energies supplied for each type of bond:
[tex]\[ \begin{array}{|c|c|} \hline \text{Bond} & \text{Bond energy (kJ/mol)} \\ \hline \text{H-H} & 436 \\ \hline \text{C-H} & 414 \\ \hline \text{C-C} & 347 \\ \hline \text{C-O} & 360 \\ \hline \text{O-H} & 464 \\ \hline \text{O=O} & 498 \\ \hline \text{C=O} & 736 \\ \hline \text{C=O in CO}_{2} & 799 \\ \hline \end{array} \][/tex]
The balanced chemical equation for the combustion of ethanol ([tex]\(\text{CH}_3\text{CH}_2\text{OH}\)[/tex]) is:
[tex]\[ \text{CH}_3\text{CH}_2\text{OH}(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \][/tex]
To determine [tex]\(\Delta H_{\text{rxn}}\)[/tex], we need to calculate the sum of the bond energies of the reactants and the products separately, and then find the difference.
### Step 1: List all bonds in reactants
1. Bonds in one molecule of ethanol [tex]\((\text{CH}_3\text{CH}_2\text{OH})\)[/tex]:
- C-C: 1
- C-H: 5 (3 from [tex]\(\text{CH}_3\)[/tex] and 2 from [tex]\(\text{CH}_2\)[/tex])
- C-O: 1
- O-H: 1
2. Bonds in oxygen ([tex]\(3 \text{O}_2\)[/tex]):
- O=O: 3
So, the total bonds in the reactants are:
[tex]\[ \begin{align*} 1 \, (\text{C-C}) & \\ 5 \, (\text{C-H}) & \\ 1 \, (\text{C-O}) & \\ 1 \, (\text{O-H}) & \\ 3 \, (\text{O}=O) & \\ \end{align*} \][/tex]
### Step 2: Calculate the total bond energy for reactants
[tex]\[ \begin{align*} \text{Total bond energy of reactants} &= 1 \times 347 \, (\text{C-C}) + 5 \times 414 \, (\text{C-H}) + 1 \times 360 \, (\text{C-O}) + 1 \times 464 \, (\text{O-H}) + 3 \times 498 \, (\text{O}=O) \\ &= 347 + 2070 + 360 + 464 + 1494 \\ &= 4735 \, \text{kJ/mol} \end{align*} \][/tex]
### Step 3: List all bonds in products
1. Bonds in two molecules of carbon dioxide [tex]\((2 \text{CO}_2)\)[/tex]:
- C=O: 4
2. Bonds in three molecules of water [tex]\((3 \text{H}_2\text{O})\)[/tex]:
- O-H: 6
So, the total bonds in the products are:
[tex]\[ \begin{align*} 4 \, (\text{C=O in CO}_{2}) & \\ 6 \, (\text{O-H}) & \\ \end{align*} \][/tex]
### Step 4: Calculate the total bond energy for products
[tex]\[ \begin{align*} \text{Total bond energy of products} &= 4 \times 799 \, (\text{C=O in CO}_{2}) + 6 \times 464 \, (\text{O-H}) \\ &= 3196 + 2784 \\ &= 5980 \, \text{kJ/mol} \end{align*} \][/tex]
### Step 5: Calculate [tex]\(\Delta H_{\text{rxn}}\)[/tex]
[tex]\[ \begin{align*} \Delta H_{\text{rxn}} &= \text{Total bond energy of reactants} - \text{Total bond energy of products} \\ &= 4735 \, \text{kJ/mol} - 5980 \, \text{kJ/mol} \\ &= -1245 \, \text{kJ/mol} \end{align*} \][/tex]
### Conclusion
The enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) for the combustion of ethanol is approximately [tex]\(-1245 \, \text{kJ/mol}\)[/tex].
[tex]\[ \begin{array}{|c|c|} \hline \text{Bond} & \text{Bond energy (kJ/mol)} \\ \hline \text{H-H} & 436 \\ \hline \text{C-H} & 414 \\ \hline \text{C-C} & 347 \\ \hline \text{C-O} & 360 \\ \hline \text{O-H} & 464 \\ \hline \text{O=O} & 498 \\ \hline \text{C=O} & 736 \\ \hline \text{C=O in CO}_{2} & 799 \\ \hline \end{array} \][/tex]
The balanced chemical equation for the combustion of ethanol ([tex]\(\text{CH}_3\text{CH}_2\text{OH}\)[/tex]) is:
[tex]\[ \text{CH}_3\text{CH}_2\text{OH}(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \][/tex]
To determine [tex]\(\Delta H_{\text{rxn}}\)[/tex], we need to calculate the sum of the bond energies of the reactants and the products separately, and then find the difference.
### Step 1: List all bonds in reactants
1. Bonds in one molecule of ethanol [tex]\((\text{CH}_3\text{CH}_2\text{OH})\)[/tex]:
- C-C: 1
- C-H: 5 (3 from [tex]\(\text{CH}_3\)[/tex] and 2 from [tex]\(\text{CH}_2\)[/tex])
- C-O: 1
- O-H: 1
2. Bonds in oxygen ([tex]\(3 \text{O}_2\)[/tex]):
- O=O: 3
So, the total bonds in the reactants are:
[tex]\[ \begin{align*} 1 \, (\text{C-C}) & \\ 5 \, (\text{C-H}) & \\ 1 \, (\text{C-O}) & \\ 1 \, (\text{O-H}) & \\ 3 \, (\text{O}=O) & \\ \end{align*} \][/tex]
### Step 2: Calculate the total bond energy for reactants
[tex]\[ \begin{align*} \text{Total bond energy of reactants} &= 1 \times 347 \, (\text{C-C}) + 5 \times 414 \, (\text{C-H}) + 1 \times 360 \, (\text{C-O}) + 1 \times 464 \, (\text{O-H}) + 3 \times 498 \, (\text{O}=O) \\ &= 347 + 2070 + 360 + 464 + 1494 \\ &= 4735 \, \text{kJ/mol} \end{align*} \][/tex]
### Step 3: List all bonds in products
1. Bonds in two molecules of carbon dioxide [tex]\((2 \text{CO}_2)\)[/tex]:
- C=O: 4
2. Bonds in three molecules of water [tex]\((3 \text{H}_2\text{O})\)[/tex]:
- O-H: 6
So, the total bonds in the products are:
[tex]\[ \begin{align*} 4 \, (\text{C=O in CO}_{2}) & \\ 6 \, (\text{O-H}) & \\ \end{align*} \][/tex]
### Step 4: Calculate the total bond energy for products
[tex]\[ \begin{align*} \text{Total bond energy of products} &= 4 \times 799 \, (\text{C=O in CO}_{2}) + 6 \times 464 \, (\text{O-H}) \\ &= 3196 + 2784 \\ &= 5980 \, \text{kJ/mol} \end{align*} \][/tex]
### Step 5: Calculate [tex]\(\Delta H_{\text{rxn}}\)[/tex]
[tex]\[ \begin{align*} \Delta H_{\text{rxn}} &= \text{Total bond energy of reactants} - \text{Total bond energy of products} \\ &= 4735 \, \text{kJ/mol} - 5980 \, \text{kJ/mol} \\ &= -1245 \, \text{kJ/mol} \end{align*} \][/tex]
### Conclusion
The enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) for the combustion of ethanol is approximately [tex]\(-1245 \, \text{kJ/mol}\)[/tex].
