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 enthalpy for the reaction C(s) + 2 SrO(s) → CO₂ (g) + 2 Sr(s).

∆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.

the answer is 1 i believe