For a particular isomer of C8H18,C8H18, the combustion reaction produces 5099.5 kJ 5099.5 kJ of heat per mole of C8H18(g)C8H18(g) consumed, under standard conditions. C8H18(g)+252O2(g)⟶8CO2(g)+9H2O(g)ΔH∘rxn=−5099.5 kJ/mol C8H18(g)+252O2(g)⟶8CO2(g)+9H2O(g)ΔHrxn°=−5099.5 kJ/mol What is the standard enthalpy of formation of this isomer of C8H18(g)

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]

The equation used to calculate enthalpy change is of a reaction is:

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

The equilibrium reaction of combustion of [tex]C_8H_{18}[/tex] follows:

[tex]C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightleftharpoons 8CO_2(g)+9H_2O(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_8H_{18})}\times \Delta H^o_f_{(C_8H_{18})})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})][/tex]

We are given:

[tex]\Delta H^o_f_{(C_8H_{18}(g))}=?\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\\\Delta H^o_{rxn}=-5099.5kJ/mol[/tex]

Putting values in above equation, we get:

[tex]-5099.5=[(8\times -393.5)+(9\times -241.5)]-[(1\times \Delta H^o_f_{(C_8H_{18})}))+(\frac{25}{2}\times 0)][/tex]

[tex]\Delta H^o_f_{(C_8H_{18})})=-222kJ/mol[/tex]

Therefore, the standard enthalpy of formation of this isomer of [tex]C_8H_{18}[/tex] is -222 kJ/mol