Gasoline is a mature of hydrocarbons mostly octane When comb

Gasoline is a mature of hydrocarbons (mostly \'octane\') When combined with oxygen and a spark is added, most (but not all) of these reactants are converted rightarrow CO_2 + H_2O What is the favored direction of this reaction, and why do the reactants not entirely convert to CO_2 + H_2O?

Solution

Gasoline : (used in automobiles)

The combustion reaction for Octane is : C₈H₁₈(g) + 25/2 0₂ (g) ------------> 8 C02 (g) + 9 H20 (l)   G = 5641.4 kj/mol it is an irreversible process i.e., combustion reaction is an example of irreversible process

Let us consider above reaction free energy G is large , negative number indicates that the reaction is spontaneous reaction goes down hill in free energy, which that product are more stable.   To determine the free energy for a reaction in opposite direction i.e., going from water and carbon dioxide back to octane and oxygen we just need to filp the sign G so for the reversible reaction

8 C0₂ (g) + 9 H₂0 (l) ---------------> C₈ H₁₈ (g) + 25/2 02 (g) G^o = + 5641.4 KJ/mol

Now, G^o is a large positive number, which means it is very unspontaneous so the reversible is very unlikely to happen thus combution of gasoline is essentially irreversible.