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.