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 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)
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
In this case, first, to obtain the enthalpy of the desired chemical reaction you need 1 mole of 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 the reaction). And when an equation is inverted, the sign of ΔH 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 the reaction) and the sign of ΔH 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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