Using the equations 2 sr(s) + o₂ (g) → 2 sro (s) ∆h° = -1184 kj/mol co₂ (g) → c (s) + o₂ (g) ∆h° = 394 kj/mol determine the molar enthalpy (in kj/mol) for the reaction c(s) + 2 sro(s) → co₂ (g) + 2 sr(s).

Answers

Answer 1

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  

First step

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.  

Second step

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.

Summary

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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