Answer:
The enthalpy change (∆H) for the reaction is -108.7 kJ
Explanation:
Hess's law can be stated as: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps. Then, Hess's Law states that the enthalpy of one reaction can be achieved by algebraically adding the enthalpies of other reactions.
So, to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them.
Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants
2 ClF + O₂ → Cl₂O + F₂O ∆H=167.4kJ
Cl₂O + 3 F₂O → 2 ClF₃ + 2 O₂ ∆H= -341.4kJ
The previous equation must be inverted, and the enthalpy value is also inverted, that is, the sign is changed.
2 F₂ + O₂ →2 F₂O ∆H=-43.4kJ
Reactants and products are added or canceled, taking into account that certain substances sometimes appear as a reagent and others as a product, so they are totally eliminated (there is nothing left of them anywhere in the reaction, if the same amount in reagents and products) or partially (this substance remains, in less quantity, only on one side), obtaining:
2 ClF + 2 F₂ → 2 ClF₃
Then, as all the reactants and products have a stoichiometric coefficient of 2, dividend by that number is obtained:
ClF + F₂ → ClF₃
Adding the enthalpies algebraically, and dividing by 2, because to get the "data" reaction you had to multiply by two, you get:
ΔH= [167.4 kJ - 341.4 kJ - 43.3 kJ]÷2
ΔH= -108.7 kJ
<u><em>The enthalpy change (∆H) for the reaction is -108.7 kJ</em></u>
= 41.43 mL