Given the experimental information below:
`2Sr(s) + O_(2)(g) rarr 2SrO(s), " "Delta_(r )H^(@) = -1180 kJ//mol`
`SrCO_(3)(s) rarr CO_(2)(g) + SrO(s) Delta_(r)H^(@) = 234 kJ//mol`
`2O_(2)(g) + 2C(s) rarr 2CO_(2)(g), " "Delta_(r )H^(@) = -788 kJ//mol`
Calculate the enthalpy change `Delta_(r) H^(@)` for the formation of 1.0 mol of strontium carbonate, the material that gives red color in fireworks, from its elements.
`Sr(s) + 3/2 O_(2)(g) + C("graphite") rarr SrCO_(3)(s)`.

Options

A. `-740 kJ//mol`
B. `+714 kJ//mol`
C. `-1218 kJ//mol`
D. `-2436 kJ//mol`

Explanation

To calculate the enthalpy change \( \Delta_r H^\circ \) for the formation of 1 mole of strontium carbonate \( \text{SrCO}_3(s) \) from its elements, we will use Hess's law. We will manipulate the given reactions to derive the desired reaction.

### Given Reactions:
1. \( 2 \text{Sr}(s) + \text{O}_2(g) \rightarrow 2 \text{SrO}(s) \)
\( \Delta_r H^\circ = -1180 \, \text{kJ/mol} \)

2. \( \text{SrCO}_3(s) \rightarrow \text{CO}_2(g) + \text{SrO}(s) \)
\( \Delta_r H^\circ = 234 \, \text{kJ/mol} \)

3. \( 2 \text{O}_2(g) + 2 \text{C}(s) \rightarrow 2 \text{CO}_2(g) \)
\( \Delta_r H^\circ = -788 \, \text{kJ/mol} \)

### Target Reaction:
\[ \text{Sr}(s) + \frac{3}{2} \text{O}_2(g) + \text{C}(s) \rightarrow \text{SrCO}_3(s) \]

### Step 1: Reverse Reaction 2
To get \( \text{SrCO}_3(s) \) on the left side, we reverse Reaction 2:
\[ \text{CO}_2(g) + \text{SrO}(s) \rightarrow \text{SrCO}_3(s) \]
The enthalpy change for this reaction will be:
\[ \Delta_r H^\circ = -234 \, \text{kJ/mol} \]

### Step 2: Adjust Reaction 1
We need only 1 mole of strontium and 1 mole of strontium oxide. Therefore, we divide Reaction 1 by 2:
\[ \text{Sr}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{SrO}(s) \]
The enthalpy change for this reaction will be:
\[ \Delta_r H^\circ = \frac{-1180}{2} = -590 \, \text{kJ/mol} \]

### Step 3: Adjust Reaction 3
We also need to adjust Reaction 3 to get 1 mole of carbon dioxide. We divide Reaction 3 by 2:
\[ \text{O}_2(g) + \text{C}(s) \rightarrow \text{CO}_2(g) \]
The enthalpy change for this reaction will be:
\[ \Delta_r H^\circ = \frac{-788}{2} = -394 \, \text{kJ/mol} \]

### Step 4: Combine the Adjusted Reactions
Now we combine the adjusted reactions:
1. \( \text{Sr}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{SrO}(s) \)
\( \Delta_r H^\circ = -590 \, \text{kJ/mol} \)

2. \( \text{O}_2(g) + \text{C}(s) \rightarrow \text{CO}_2(g) \)
\( \Delta_r H^\circ = -394 \, \text{kJ/mol} \)

3. \( \text{CO}_2(g) + \text{SrO}(s) \rightarrow \text{SrCO}_3(s) \)
\( \Delta_r H^\circ = -234 \, \text{kJ/mol} \)

### Step 5: Write the Overall Reaction
Adding these reactions together, we get:
\[ \text{Sr}(s) + \frac{3}{2} \text{O}_2(g) + \text{C}(s) \rightarrow \text{SrCO}_3(s) \]

### Step 6: Calculate the Total Enthalpy Change
Now we sum the enthalpy changes:
\[ \Delta_r H^\circ = (-590) + (-394) + (-234) = -1218 \, \text{kJ/mol} \]

### Final Answer
The enthalpy change \( \Delta_r H^\circ \) for the formation of 1 mole of strontium carbonate from its elements is:
\[ \Delta_r H^\circ = -1218 \, \text{kJ/mol} \]