To calculate the enthalpy change (Δ_fH°) for the formation of 1.0 mol of strontium carbonate (SrCO₃) from its elements, we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.
### Step-by-Step Solution:
1. **Identify the Target Reaction**:
The target reaction for the formation of strontium carbonate from its elements is:
\[
\text{Sr(s)} + \text{C(graphite)} + \frac{3}{2} \text{O}_2(g) \rightarrow \text{SrCO}_3(s)
\]
2. **List the Given Reactions**:
We have the following reactions with their enthalpy changes:
- Reaction A:
\[
2\text{Sr(s)} + \text{O}_2(g) \rightarrow 2\text{SrO(s)}, \quad \Delta_rH° = -1180 \, \text{kJ/mol}
\]
- Reaction B:
\[
\text{SrCO}_3(s) \rightarrow \text{SrO(s)} + \text{CO}_2(g), \quad \Delta_rH° = 234 \, \text{kJ/mol}
\]
- Reaction C:
\[
2\text{C(s)} + 2\text{O}_2(g) \rightarrow 2\text{CO}_2(g), \quad \Delta_rH° = -788 \, \text{kJ/mol}
\]
3. **Manipulate the Reactions**:
- For Reaction A, we need only 1 mole of Sr, so we will divide the entire reaction by 2:
\[
\text{Sr(s)} + \frac{1}{2}\text{O}_2(g) \rightarrow \text{SrO(s)}, \quad \Delta_rH° = \frac{-1180}{2} = -590 \, \text{kJ/mol}
\]
- For Reaction C, we need only 1 mole of C, so we will also divide this reaction by 2:
\[
\text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g), \quad \Delta_rH° = \frac{-788}{2} = -394 \, \text{kJ/mol}
\]
- For Reaction B, we need to reverse it since we want to form SrCO₃:
\[
\text{SrO(s)} + \text{CO}_2(g) \rightarrow \text{SrCO}_3(s), \quad \Delta_rH° = -234 \, \text{kJ/mol} \, (\text{reversed, so it becomes } -234)
\]
4. **Combine the Reactions**:
Now we can add the modified reactions:
- From Reaction A:
\[
\text{Sr(s)} + \frac{1}{2}\text{O}_2(g) \rightarrow \text{SrO(s)} \quad (-590 \, \text{kJ})
\]
- From Reaction C:
\[
\text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad (-394 \, \text{kJ})
\]
- From Reaction B (reversed):
\[
\text{SrO(s)} + \text{CO}_2(g) \rightarrow \text{SrCO}_3(s) \quad (-234 \, \text{kJ})
\]
Adding these reactions together, we get:
\[
\text{Sr(s)} + \text{C(s)} + \frac{3}{2}\text{O}_2(g) \rightarrow \text{SrCO}_3(s)
\]
5. **Calculate the Total Enthalpy Change**:
Now, we sum the enthalpy changes:
\[
\Delta_fH° = (-590) + (-394) + (-234) = -1218 \, \text{kJ/mol}
\]
### Final Answer:
The enthalpy change for the formation of 1.0 mol of strontium carbonate (SrCO₃) from its elements is:
\[
\Delta_fH° = -1218 \, \text{kJ/mol}
\]
