Respuesta :
Answer: The heat of formation of oleic acid is -94.12 kJ/mol
Explanation:
We are given:
Heat of combustion of oleic acid = [tex]-1.11\times 10^4kJ/mol[/tex]
The chemical equation for the combustion of oleic acid follows:
[tex]C_{18}H_{34}O_2(l)+\frac{51}{2}O_2(g)\rightarrow 18CO_2(g)+17H_2O(g);\Delta H^o=-1.11\times 10^4kJ[/tex]
Enthalpy change 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 equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(18\times \Delta H^o_f_{(CO_2(g))})+(17\times \Delta H^o_f_{(H_2O)})]-[(1\times \Delta H^o_f_{(C_{18}H_{34}O_2(l))})+(\frac{51}{2}\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_{rxn}=-1.11\times 10^4kJ[/tex]
Putting values in above equation, we get:
[tex]-1.11\times 10^4=[(18\times (-393.51))+(17\times (-241.82))]-[(1\times \Delta H^o_f_{(C_{18}H_{34}O_2(l))})+(\frac{51}{2}\times 0)]\\\\\Delta H^o_f_{(C_{18}H_{34}O_2(l))}=-94.12kJ/mol[/tex]
Hence, the heat of formation of oleic acid is -94.12 kJ/mol
