Given that H 2 ( g ) + F 2 ( g ) ⟶ 2 HF ( g ) Δ H ∘ rxn = − 546.6 kJ ⋅ mol − 1 2 H 2 ( g ) + O 2 ( g ) ⟶ 2 H 2 O ( l ) Δ H ∘ rxn = − 571.6 kJ ⋅ mol − 1 calculate the value of Δ H ∘ rxn for 2 F 2 ( g ) + 2 H 2 O ( l ) ⟶ 4 HF ( g ) + O 2 ( g )

Answer : The enthalpy of reaction is, -521.6 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 main chemical reaction is,

The intermediate balanced chemical reaction will be,

(1)    

(2)    

Now we are multiplying reaction 1 by 2 and reversing reaction 2 and then adding all the equations, we get :

(1)    

(2)    

The expression for enthalpy of main reaction is,

Therefore, the enthalpy of reaction is, -521.6 kJ/mol