Using the equations 2 c₆h₆ (l) + 15 o₂ (g) → 12 co₂ (g) + 6 h₂o (g)∆h° = -6271 kj/mol c (s) + o₂ (g) → co₂ (g) ∆h° = -393.5 kj/mol 2 h₂ (g) + o₂ (g) → 2 h₂o (g) ∆h° = -483.6 kj/mol determine the enthalpy for the reaction 6 c (s) + 3 h₂ (g) → c₆h₆ (l).

Answer : The enthalpy for the reaction is 49.1 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of will be,

The intermediate balanced chemical reaction will be,

(1)    

(2)    

(3)    

Now we will reverse the reaction 1 and divide by 2, multiply reaction 2 by 6 and reaction 3 by 3/2 then adding all the equations, we get :

(1)    

(2)    

(3)    

The expression for enthalpy of formation of will be,

Therefore, the enthalpy for the reaction is 49.1 kJ/mol