Question

The heat of combustion of oleic

Answer 1

Answer: The heat of formation of oleic acid is -94.12 kJ/mol

Explanation:

We are given:

Heat of combustion of oleic acid = $-1.11\times 10^4kJ/mol$

The chemical equation for the combustion of oleic acid follows:

$C_{18}H_{34}O_2(l)+\frac{51}{2}O_2(g)\rightarrow 18CO_2(g)+17H_2O(g);\Delta H^o=-1.11\times 10^4kJ$

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 $\Delta H^o$

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

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

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

$\Delta H^o_{rxn}=[(18\times \Delta H^o_f_{(CO_2(g))})+(17\times \Delta H^o_f_{(H_2O)})]-[(1\times \Delta H^o_f_{(C_{18}H_{34}O_2(l))})+(\frac{51}{2}\times \Delta H^o_f_{(O_2(g))})]$

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_{rxn}=-1.11\times 10^4kJ$

Putting values in above equation, we get:

$-1.11\times 10^4=[(18\times (-393.51))+(17\times (-241.82))]-[(1\times \Delta H^o_f_{(C_{18}H_{34}O_2(l))})+(\frac{51}{2}\times 0)]\\\\\Delta H^o_f_{(C_{18}H_{34}O_2(l))}=-94.12kJ/mol$

Hence, the heat of formation of oleic acid is -94.12 kJ/mol