Answer: The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol
Explanation:
The balanced reaction for combustion of isooctane is:
[tex]C_8H_{18}(l)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{(CO_2(g))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_8H_{18}(g))})+(\frac{25}{2}\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_{C_8H_{18}(l)}=-258.07kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(8\times (-393.51))+(9\times (-285.8))]-[(1\times (-258.07))+(\frac{25}{2}\times (0))]\\\Delta H^o_{rxn}=-5462.2kJ/mol[/tex]
The enthalpy of combustion of iso-octane in excess oxygen at 298 K is -5462.2kJ/mol
