Answer:
∆H° = 790 kJ/mol
Explanation:
Hello,
In this case, by using the Hess law, we should handle the given reactions as shown below:
- The reaction:
2 Sr(s) + O₂ (g) → 2 SrO (s)
Should be inverted as:
2 SrO (s) → 2 Sr(s) + O₂ (g)
So the enthalpy of reaction changes to ∆H° = 1184 kJ/mol
- The reaction:
CO₂ (g) → C (s) + O₂ (g)
Should be also inverted as:
C (s) + O₂ (g) → CO₂ (g)
So the enthalpy of reaction changes to ∆H° = -394 kJ/mol
- Then, we add the modified reactions to obtain the desired reaction:
2 SrO (s) + C (s) + O₂ (g) → 2 Sr(s) + O₂ (g) CO₂ (g)
C(s) + 2 SrO(s) → CO₂ (g) + 2 Sr(s)
Therefore, the resulting enthalpy of reaction is:
∆H° = 1184 kJ/mol - 394 kJ/mol
∆H° = 790 kJ/mol
Best regards.
Considering the Hess's Law, the enthalpy change for the reaction is 790 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:
C(s) + 2 SrO(s) → CO₂ (g) + 2 Sr(s)
which occurs in two stages.
You know the following reactions, with their corresponding enthalpies:
Equation 1: 2 Sr (s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol
Equation 2: CO₂ (g) → C (s) + O₂ (g) ∆H° = 394 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.
In this case, first, to obtain the enthalpy of the desired chemical reaction you need 1 mole od C (s) on reactant side and it is present in second equation. But since this equation has 1 mole of C (s) on the product side, it is necessary to locate this on the reactant side (invert it). And when an equation is inverted, the sign of delta H also changes.
Now, 2 moles of SrO (s) must be a reactant and is present in the first equation. Since this equation has 2 moles of Sr0 (s) on the product side, it is necessary to locate the O on the reactant side (invert it) and the sign of delta H also changes.
In summary, you know that two equations with their corresponding enthalpies are:
Equation 1: 2 SrO (s) → 2 Sr (s) + O₂ (g) ∆H° = 1184 kJ/mol
Equation 2: C (s) + O₂ (g) → CO₂ (g) ∆H° = -394 kJ/mol
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
C(s) + 2 SrO(s) → CO₂ (g) + 2 Sr(s) ΔH= 790 kJ/mol
Finally, the enthalpy change for the reaction is 790 kJ/mol.
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