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Sure, let's go through the detailed step-by-step solution to find the enthalpy of combustion of 1 mole of [tex]\( C_6H_6 \)[/tex] (benzene) when it completely reacts with oxygen.
Given the balanced chemical equation for the combustion of benzene:
[tex]\[ 2 C_6H_6(g) + 15 O_2(g) \rightarrow 12 CO_2(g) + 6 H_2O(g) \][/tex]
We are also given the standard enthalpies of formation ([tex]\(\Delta H_f^\circ\)[/tex]) of the compounds involved:
- [tex]\(C_6H_6(g)\)[/tex]: [tex]\( \Delta H_f^\circ = 82.90 \, \text{kJ/mol} \)[/tex]
- [tex]\(CO_2(g)\)[/tex]: [tex]\( \Delta H_f^\circ = -393.50 \, \text{kJ/mol} \)[/tex]
- [tex]\(H_2O(g)\)[/tex]: [tex]\( \Delta H_f^\circ = -241.82 \, \text{kJ/mol} \)[/tex]
The enthalpy of the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) is calculated using the enthalpies of formation of the products and reactants.
First, calculate the total enthalpy of formation of the reactants and products:
1. Reactants:
- For [tex]\( C_6H_6(g) \)[/tex]:
[tex]\[ \text{Total} = 2 \times 82.90 \, \text{kJ/mol} \][/tex]
2. Products:
- For [tex]\( CO_2(g) \)[/tex]:
[tex]\[ \text{Total} = 12 \times (-393.50) \, \text{kJ/mol} \][/tex]
- For [tex]\( H_2O(g) \)[/tex]:
[tex]\[ \text{Total} = 6 \times (-241.82) \, \text{kJ/mol} \][/tex]
Now sum up these values to calculate the total enthalpy for the reactants and products:
- Total enthalpy of reactants:
[tex]\[ 2 \times 82.90 = 165.80 \, \text{kJ} \][/tex]
- Total enthalpy of products:
[tex]\[ (12 \times -393.50) + (6 \times -241.82) = -4722.00 - 1450.92 = -6172.92 \, \text{kJ} \][/tex]
The enthalpy change of the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) can be calculated using:
[tex]\[ \Delta H_{\text{reaction}} = \text{Total enthalpy of products} - \text{Total enthalpy of reactants} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -6172.92 \, \text{kJ} - 165.80 \, \text{kJ} = -6338.72 \, \text{kJ} \][/tex]
This is the enthalpy change for the combustion of 2 moles of benzene. Therefore, to find the enthalpy change for 1 mole of benzene, we divide this result by 2:
[tex]\[ \Delta H_{\text{combustion}} = \frac{\Delta H_{\text{reaction}}}{2} = \frac{-6338.72 \, \text{kJ}}{2} = -3169.36 \, \text{kJ/mol} \][/tex]
Thus, the enthalpy of combustion for 1 mole of [tex]\( C_6H_6 \)[/tex] is:
[tex]\[ -3169.36 \, \text{kJ/mol} \][/tex]
Therefore, the correct answer from the given options is:
[tex]\[ \boxed{-3169 \, \text{kJ/mol}} \][/tex]
This indicates that when 1 mole of benzene completely reacts with oxygen, the enthalpy of combustion is approximately [tex]\( -3169 \, \text{kJ/mol} \)[/tex].
Given the balanced chemical equation for the combustion of benzene:
[tex]\[ 2 C_6H_6(g) + 15 O_2(g) \rightarrow 12 CO_2(g) + 6 H_2O(g) \][/tex]
We are also given the standard enthalpies of formation ([tex]\(\Delta H_f^\circ\)[/tex]) of the compounds involved:
- [tex]\(C_6H_6(g)\)[/tex]: [tex]\( \Delta H_f^\circ = 82.90 \, \text{kJ/mol} \)[/tex]
- [tex]\(CO_2(g)\)[/tex]: [tex]\( \Delta H_f^\circ = -393.50 \, \text{kJ/mol} \)[/tex]
- [tex]\(H_2O(g)\)[/tex]: [tex]\( \Delta H_f^\circ = -241.82 \, \text{kJ/mol} \)[/tex]
The enthalpy of the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) is calculated using the enthalpies of formation of the products and reactants.
First, calculate the total enthalpy of formation of the reactants and products:
1. Reactants:
- For [tex]\( C_6H_6(g) \)[/tex]:
[tex]\[ \text{Total} = 2 \times 82.90 \, \text{kJ/mol} \][/tex]
2. Products:
- For [tex]\( CO_2(g) \)[/tex]:
[tex]\[ \text{Total} = 12 \times (-393.50) \, \text{kJ/mol} \][/tex]
- For [tex]\( H_2O(g) \)[/tex]:
[tex]\[ \text{Total} = 6 \times (-241.82) \, \text{kJ/mol} \][/tex]
Now sum up these values to calculate the total enthalpy for the reactants and products:
- Total enthalpy of reactants:
[tex]\[ 2 \times 82.90 = 165.80 \, \text{kJ} \][/tex]
- Total enthalpy of products:
[tex]\[ (12 \times -393.50) + (6 \times -241.82) = -4722.00 - 1450.92 = -6172.92 \, \text{kJ} \][/tex]
The enthalpy change of the reaction ([tex]\(\Delta H_{\text{reaction}}\)[/tex]) can be calculated using:
[tex]\[ \Delta H_{\text{reaction}} = \text{Total enthalpy of products} - \text{Total enthalpy of reactants} \][/tex]
[tex]\[ \Delta H_{\text{reaction}} = -6172.92 \, \text{kJ} - 165.80 \, \text{kJ} = -6338.72 \, \text{kJ} \][/tex]
This is the enthalpy change for the combustion of 2 moles of benzene. Therefore, to find the enthalpy change for 1 mole of benzene, we divide this result by 2:
[tex]\[ \Delta H_{\text{combustion}} = \frac{\Delta H_{\text{reaction}}}{2} = \frac{-6338.72 \, \text{kJ}}{2} = -3169.36 \, \text{kJ/mol} \][/tex]
Thus, the enthalpy of combustion for 1 mole of [tex]\( C_6H_6 \)[/tex] is:
[tex]\[ -3169.36 \, \text{kJ/mol} \][/tex]
Therefore, the correct answer from the given options is:
[tex]\[ \boxed{-3169 \, \text{kJ/mol}} \][/tex]
This indicates that when 1 mole of benzene completely reacts with oxygen, the enthalpy of combustion is approximately [tex]\( -3169 \, \text{kJ/mol} \)[/tex].
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