Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.
Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.
In this case you want to calculate the enthalpy change of:
C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)
which occurs in three stages.
You know the following reactions, with their corresponding enthalpies:
Equation 1: H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol
Equation 2: C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol
Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol
Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.
<h3 /><h3>FIRST STEP</h3>First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).
When an equation is inverted, the sign of ΔH° also changes.
<h3>SECOND STEP</h3>Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).
Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.
<h3>THIRD STEP</h3>Finally, you need 4 moles of HF (g) on the product side. The first equation has 2 moles of HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.
Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.
<h3>SUMMARY</h3>In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g) ∆H° = -158.4 kJ/mol
Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g) ∆H° = 282.6 kJ/mol
Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g) ∆H° = 97.6 kJ/mol
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g) ΔH°= 221.8 kJ/mol
Finally, the enthalpy change for the reaction is 221.8 kJ/mol.
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