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

definition

                    entropy is a measure of the randomness or the extent of disorder of a chemical process.

                      enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

                        entropy is measured in joule per kelvin (jk⁻¹)   whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

                            entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

                              in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.