To calculate the enthalpy change (Δ_rH°) for the formation of 1.0 mol of strontium carbonate (SrCO₃) from its elements, we will use Hess's law and the given reactions. Here’s the step-by-step solution:
### Step 1: Write the Target Reaction
We need to find the enthalpy change for the formation of strontium carbonate from its elements:
\[ \text{Sr(s)} + \frac{3}{2} \text{O}_2(g) + \text{C(graphite)} \rightarrow \text{SrCO}_3(s) \]
### Step 2: Identify the Relevant Reactions
We have the following reactions with their enthalpy changes:
1. \( 2 \text{Sr(s)} + \text{O}_2(g) \rightarrow 2 \text{SrO(s)} \), \( \Delta_rH° = -1180 \, \text{kJ} \)
2. \( \text{SrCO}_3(s) \rightarrow \text{CO}_2(g) + \text{SrO(s)} \), \( \Delta_rH° = 234 \, \text{kJ} \)
3. \( 2 \text{C(s)} + 2 \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) \), \( \Delta_rH° = -788 \, \text{kJ} \)
### Step 3: Manipulate the Reactions
To obtain the desired reaction, we need to manipulate the second and third reactions:
**For Reaction 2:**
- Reverse the reaction to get strontium carbonate on the products side:
\[ \text{CO}_2(g) + \text{SrO(s)} \rightarrow \text{SrCO}_3(s) \]
- Change the sign of Δ_rH°:
\[ \Delta_rH° = -234 \, \text{kJ} \]
**For Reaction 1:**
- Since we need only 1 mole of strontium, we divide the entire reaction by 2:
\[ \text{Sr(s)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{SrO(s)} \]
- Adjust the enthalpy change:
\[ \Delta_rH° = \frac{-1180}{2} = -590 \, \text{kJ} \]
**For Reaction 3:**
- We need to produce 1 mole of CO₂ from 1 mole of carbon, so we also divide this reaction by 2:
\[ \text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \]
- Adjust the enthalpy change:
\[ \Delta_rH° = \frac{-788}{2} = -394 \, \text{kJ} \]
### Step 4: Combine the Reactions
Now we can add the adjusted reactions:
1. \( \text{Sr(s)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{SrO(s)} \), \( \Delta_rH° = -590 \, \text{kJ} \)
2. \( \text{CO}_2(g) + \text{SrO(s)} \rightarrow \text{SrCO}_3(s) \), \( \Delta_rH° = -234 \, \text{kJ} \)
3. \( \text{C(s)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \), \( \Delta_rH° = -394 \, \text{kJ} \)
Adding these reactions gives:
\[ \text{Sr(s)} + \frac{3}{2} \text{O}_2(g) + \text{C(s)} \rightarrow \text{SrCO}_3(s) \]
### Step 5: Calculate the Total Enthalpy Change
Now, we sum the enthalpy changes:
\[
\Delta_rH° = -590 \, \text{kJ} + (-234 \, \text{kJ}) + (-394 \, \text{kJ})
\]
\[
\Delta_rH° = -590 - 234 - 394 = -1218 \, \text{kJ}
\]
### Final Answer
The enthalpy change for the formation of 1.0 mol of strontium carbonate from its elements is:
\[
\Delta_rH° = -1218 \, \text{kJ/mol}
\]
