-3169 kJ/mol is the enthalpy of combustion when 1 mol [tex]C_6H_6(g)[/tex] completely reacts with oxygen.
Explanation:
Given:
The reaction of combustion of [tex]C_6H_6(g)[/tex] along with enthalpies of formation of the compounds.
To find:
The enthalpy of combustion when 1 mol [tex]C_6H_6(g)[/tex] completely reacts with oxygen.
Solution:
[tex]2C_6H_6(g) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(g)[/tex]
Enthalpy of the reaction:
[tex]\Delta H^o_{rxn}=\sum [\Delta H^o_{f,products}]-\sum [\Delta H^o_{f,reactants}]\\=[12mol\times \Delta H^o_{f.CO_2(g)}+6mol\times \Delta H^o_{f,H_2O(g)}]-[2 mol\times \Delta H^o_{f.C_6H_6(g)}+15\times \Delta H^o_{f,O_2(g)}]\\=[12mol\times (-393.50 kJ/mol)+6mol\times (-241.82 kJ/mol)]-[2mol\times 82.90kJ/mol+15mol\times 0 kJ/mol]\\=-6338.72 kJ[/tex]
[tex]2C_6H_6(g) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(g).\Delta H^o_{rxn}=-6338.72 kJ[/tex]
When 1 mole of [tex]C_6H_6(g)[/tex] reacts with oxygen gas:
[tex]=\frac{\Delta H^o_{rxn}}{\text{2 mol of } C_6H_6}\\=\frac{-6338.72 kJ}{2 mol}\\=-3169.36 kJ/mol\approx -3169 kJ/mol[/tex]
-3169 kJ/mol is the enthalpy of combustion when 1 mol [tex]C_6H_6(g)[/tex] completely reacts with oxygen.
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