Respuesta :
Answer:
[tex]\Delta H[/tex] for the given reaction is -2486.3 kJ
Explanation:
The given equation can be written as a combination of the following equation:
[tex]2H_{2}(g)+2F_{2}(g)\rightarrow 4HF(g)[/tex] ; [tex]\Delta H_{1}= (2\times -537kJ)=-1074 kJ[/tex]
[tex]2C(s)+4F_{2}(g)\rightarrow 2CF_{4}(g)[/tex] ; [tex]\Delta H_{2}=(2\times -680kJ)=-1360kJ[/tex]
[tex]C_{2}H_{4}(g)\rightarrow 2C(s)+2H_{2}(g)[/tex] ; [tex]\Delta H_{3}=-52.3kJ[/tex]
-----------------------------------------------------------------------------------
[tex]C_{2}H_{4}(g)+6F_{2}(g)\rightarrow 2CF_{4}(g)+4HF(g)[/tex]
[tex]\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=(-1074-1360-52.3)kJ=-2486.3kJ[/tex]
Considering the Hess's Law, the enthalpy change for the reaction is -2486.3 kJ.
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) ΔH=−537 kJ
Equation 2: C (s) + 2 F₂ (g) → CF₄ (g) ΔH=−680 kJ
Equation 3: 2 C (s) + 2 H₂ (g) → C₂H₄ (g) ΔH=+52.3 kJ
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.
In this case, first, to get the enthalpy of the desired chemical reaction, you need 1 mole of C₂H₄ (g) on the reactant side and it is present in the third equation, so let's write this as such. Since this equation the compound is on the product side, it is necessary to place it on the reactant side (invert it). And when an equation is reversed, the sign of delta H also changes.
Now, in the reaction to obtain, there is no C(s) present. So to cancel the C you must multiply the second equation by 2, in order to cancel it with the equation previously raised. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiplied by 2, the variation of enthalpy also.
Finally, to cancel the H₂(g), you need to multiply the first equation by 2, so delta H is also multiplied by 2.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g) ΔH=−1074 kJ
Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g) ΔH=−1360 kJ
Equation 3: C₂H₄ (g) → 2 C (s) + 2 H₂ (g) ΔH=-52.3kJ
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= -2486.3 kJ
Finally, the enthalpy change for the reaction is -2486.3 kJ.
Learn more:
brainly.com/question/5976752?referrer=searchResults
brainly.com/question/13707449?referrer=searchResults
brainly.com/question/13707449?referrer=searchResults
brainly.com/question/6263007?referrer=searchResults
brainly.com/question/14641878?referrer=searchResults
brainly.com/question/2912965?referrer=searchResults
