Look at the following data provided below:
1 answer:
Considering the Hess's Law, the enthalpy change for the reaction is -84.4 kJ.
<h3>Hess's Law</h3>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.
<h3>Enthalpy change for the reaction in this case</h3>In this case you want to calculate the enthalpy change of:
2 C (graphite) + 3 H₂(g) → C₂H₆(g)
which occurs in three stages.
You know the following reactions, with their corresponding enthalpies:
Equation 1: C₂H₆(g) + O₂(g) → 2 CO₂(g) + 3 H₂O(l) ; ΔH° = –1560 kJ
Equation 2: H₂(g) + O₂(g) → H₂O(l) ; ΔH° = –285.8 kJ
Equation 3: C(graphite) + O₂(g) → CO₂(g) ; ΔH° = –393.5 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 2 moles of C(graphite) on reactant side and it is present in third equation. In this case it is necessary to multiply it by 2 to obtain the necessary amount. 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.
Now, you need 3 moles of H₂(g) on reactant side and it is present in second equation. In this case it is necessary to multiply it by 3 to obtain the necessary amount and the variation of enthalpy also is multiplied by 3.
Finally, 1 mole of C₂H₆(g) must be a product and is present in the first equation. Since this equation has 1 mole of C₂H₆(g) on the reactant side, it is necessary to locate the C₂H₆(g) on the reactant side (invert it). When an equation is inverted, the sign of delta H also changes.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: 2 CO₂(g) + 3 H₂O(l) → C₂H₆(g) + O₂(g); ΔH° = 1560 kJ
Equation 2: 3 H₂(g) + O₂(g) → 3 H₂O(l) ; ΔH° = –857.4 kJ
Equation 3: 2 C(graphite) + 2 O₂(g) → 2 CO₂(g) ; ΔH° = –787 kJ
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
2 C (graphite) + 3 H₂(g) → C₂H₆(g) ΔH= -84.4 kJ
Finally, the enthalpy change for the reaction is -84.4 kJ.
Learn more about enthalpy for a reaction:
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Answer:
a. Disminuye.
b. Disminuye.
Explanation:
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En este caso, considerando que tenemos una reacción en la cual se podría asumir que tanto la concentración de monóxido de carbono y la de oxígeno contribuye a la velocidad de reacción, de acuerdo a la siguiente cinética elemental:
Al aplicar los siguientes cambios, resultaría:
a. En este caso, bien sabemos que la velocidad de reacción es aumentada al adicionar un catalizador ya que la constante de velocidad aumenta al disminuir la energía de activación; no obstante, al retirar el catalizador, la energía de activación aumentaría y por consiguiente la velocidad de reacción disminuiría.
b. En este caso, dado que la relación entre el oxígeno y la rapidez de reacción es directamente proporcional, es possible para nosotros determinar que al eliminar parte del oxígeno, la concentración disminuye y por ende la velocidad de reacción también.
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Because when anything is exposed to air, it come in contact with bacteria which begin to break down the proteins in the beans. But when it’s in a can, it can’t get exposed to bacteria which means it won’t break down.
Hope this sheds some light ♥︎
Hope this sheds some light ♥︎