Answer: The Enthalpy of combustion of 1 mol of butane is -2657.5 kJ/mol.
Explanation:
[tex]\Delta H_{f,CO_2}=-393.5 kJ/mol[/tex]
[tex]\Delta H_{f,H_2O}=-241.82 kJ/mol[/tex]
[tex]\Delta H_{f,C_4H_{10}}=-125.6 kJ/mol[/tex]
[tex]\Delta H_{f,O_2}=0 kJ/mol[/tex]
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
Enthalpy of Combustion of 2 moles of butane :
[tex]=\sum(\Delta H_f\text{of products})-\sum(\Delta H_f\text{of reactants})[/tex]
[tex]\Delta H_c=(8\Delta H_{f,CO_2}+10\Delta H_{f,H_2O})-(2\Delta H_{f,C_4H_{10}}-13\Delta H_{f,O_2})[/tex]
[tex]=(8 mol\times -393.5 kJ/mol+10 mol\times (-241.82 kJ/mol))-(2 mol\times (-125.6 kJ/mol)+13 mol\times 0 kJ/mol)=-5315 kJ[/tex]
Enthalpy of Combustion of 1 moles of butane :
[tex]\Delta H_c=\frac{5315 kJ}{2mol}=-2657.5 kJ/mol[/tex]
The Enthalpy of combustion of 1 mol of butane is -2657.5 kJ/mol.
Answer:
A. –2,657.5 kJ/mol
Explanation:
This is correct on ed-genuity, hope this helps! :)
