Answer :
Answer: The reaction enthalpy of given reaction is -890.7KJ.
Solution:
Given:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l),\Delta H= ?[/tex]...(1)
[tex]C(s)+2H_2(g)\rightarrow CH_4(g),\Delta H=-74.8KJ[/tex]...(2)
[tex]C(s)+O_2(g)\rightarrow CO_2(g), \Delta H=-393.5KJ[/tex]...(3)
[tex]2H_2(g)+O_2(g)\rightarrow2H_2O(g), \Delta H=-484.0KJ[/tex]...(4)
[tex]H_2O(l)\rightarrow H_2O(g),\Delta H=44.0KJ[/tex]...(5)
According to Hess law: "The enthalpy of the chemical reaction remains the same whether reaction is carried out in single step or in multiple steps."
[tex]\Delta H\text{of equation}(1)=(3)+(4)-(2)-2\times (5)[/tex]
[tex]\Delta H\text{of equation}(1)=(-393.5)+(-484.0)-(-74.8)-2\times (44.0)=-890.7KJ[/tex]
The reaction enthalpy of given reaction:
[tex]CH_4(g)+2O_2(g)\rightarrowCO_2(g)+2H_2O(l),\Delta H=-890.7KJ[/tex]
Considering the Hess's Law, the enthalpy change for the reaction is -890.7 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:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)
You know the following reactions, with their corresponding enthalpies:
Equation 1: C(s) + 2 H₂(g) → CH₄(g) ΔH = –74.8 kJ
Equation 2: C(s) + O₂(g) → CO₂(g) ΔH = –393.5 kJ
Equation 3: 2 H₂(g) + O₂(g) → 2 H₂O(g) ΔH = -484 kJ
Equation 4: H₂O(l) → H₂O(g) ΔH = 44 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 a mole of CH₄(g) on the reactant side. Said compound is present in the first equation, but as a product. Therefore, it is convenient to write said reaction in reverse, in order to place the CH₄(g) compound on the reactant side. When an equation is inverted, the sign of ΔH also changes.
Now O₂(g) must be a reactant and is present in the second and third equations on the reactant side. So in this case it is not necessary to modify these chemical equations.
Finally, to obtain 2 moles of H₂O(l) in the products, they are found in the fourth equation. Then it is necessary to invert this equation and the sign of ΔH also changes. And since 2 moles are necessary, you must multiply by 2 the expression of the fourth equation. As 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.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: CH₄(g) → C(s) + 2 H₂(g) ΔH = 74.8 kJ
Equation 2: C(s) + O₂(g) → CO₂(g) ΔH = –393.5 kJ
Equation 3: 2 H₂(g) + O₂(g) → 2 H₂O(g) ΔH = -484 kJ
Equation 4: 2 H₂O(g) → 2 H₂O(l) ΔH = -88 kJ
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH= -890.7 kJ
Finally, the enthalpy change for the reaction is -890.7 kJ.
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