Answer: The enthalpy change for formation of butane is -125 kJ/mol
Explanation:
The balanced chemical reaction is,
[tex]C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}][/tex]
Putting the values we get :
[tex]\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}][/tex]
[tex]-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0][/tex]
[tex]H_f_{C_4H_{10}=-125kJ/mol[/tex]
Thus enthalpy change for formation of butane is -125 kJ/mol
