Given that H2 (g) + F2 (g) ⟶ 2HF (g) ΔH ∘ rxn = − 546.6 kJ 2H2 (g) + O2 (g) ⟶ 2H2O (l) ΔH∘rxn = − 571.6 kJ calculate the value of ΔH∘rxn for 2F2 (g) + 2H2O (l) ⟶ 4HF (g) + O2 (g)

Answer : The value of for the reaction is, -521.6 kJ

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 given main 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 change will be,

Thus, the value of for the reaction is, -521.6 kJ

2b2t

explanation:

2b2t