Respuesta :
Answer:- [tex]\Delta H=-304.1kJ[/tex]
Solution:- The enthalpy changed asked to calculate for the reaction, [tex]NO(g)+O(g)\rightarrow NO_2(g)[/tex]
It is based on Hess's law where we use the given equations to make the desired equation and calculate it's enthalpy change.
For our equation we need one NO(g) on reactant side and it is present in first equation so let's write this as such.
[tex]NO(g)+O_3(g)\rightarrow NO_2(g)+O_2(g)\Delta H=-198.9kJ[/tex]
Now we need O(g) on reactant side which is present in third equation. Third equation has two O on product side so we need to divide it by 2 and also reverse it so that we could get one O on reactant side. When an equation is reversed then sign of delta H also gets changed.
[tex]O(g)\rightarrow \frac{1}{2}O_2(g)\Delta H=-247.5kJ[/tex]
To cancel ozone, we need to reverse second equation.
[tex]1.5O_2(g)\rightarrow O_3(g)\Delta H=142.3kJ[/tex]
Let's add all these three equations now:
[tex]NO(g)+O_3(g)\rightarrow NO_2(g)+O_2(g)\Delta H=-198.9kJ[/tex]
[tex]O(g)\rightarrow \frac{1}{2}O_2(g)\Delta H=-247.5kJ[/tex]
[tex]1.5O_2(g)\rightarrow O_3(g)\Delta H=142.3kJ[/tex]
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[tex]NO(g)+O(g)\rightarrow NO_2(g)\Delta H=-304.1kJ[/tex]
So, the enthalpy change for the given equation asked to calculate is -304.1 kJ.
Considering the Hess's Law, the enthalpy change for the reaction is -304.1 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:
NO + O → NO₂
which occurs in three stages.
You know the following reactions, with their corresponding enthalpies:
Equation 1: NO + O₃ → NO₂ + O₂ ΔH = –198.9 kJ
Equation 2: O₃ → 1.5 O₂ ΔH = –142.3 kJ
Equation 3: O₂ → 2 O ΔH = 495.0 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 obtain the enthalpy of the desired chemical reaction you need one NO on reactant side and it is present in first equation so let's write this as such.
Now, O must be a reactant and is present in the third equation. Since this equation has 2 O on the product side, it is necessary to locate the O on the reactant side (invert it) and divide it by 2 to obtain an O on the reactant side. When an equation is inverted, the sign of delta H also changes. And since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is divided by 2, the variation of enthalpy also.
Finally, to cancel ozone O₃, you need to reverse second equation, the sign of delta H also changes.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: NO + O₃ → NO₂ + O₂ ΔH = –198.9 kJ
Equation 2: 1.5 O₂ → O₃ ΔH = 142.3 kJ
Equation 3: O → 0.5 O₂ ΔH = - 247.5 kJ
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
NO + O → NO₂ ΔH= -304.2 kJ
Finally, the enthalpy change for the reaction is -304.1 kJ.
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