Consider the following reactions.
I. `C("graphite") + O_(2)(g) rarr CO_(2)(g) , DeltaH^@ = -x_(1)cal`
II> `CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g) , Delta H^(@) = -x_(2) cal`
Based on the above data, `DeltaH_(f)^(@)(CO_(2))` is:

Options

A. `-x_(1)cal`
B. `-x_(2) cal`
C. `(x_2 - x_1) cal`
D. `(x_1 - x_2) cal`

Explanation

To find the standard enthalpy of formation of carbon dioxide, \( \Delta H_f^\circ(CO_2) \), we can use the given reactions and their enthalpy changes.

### Step-by-Step Solution:

1. **Write the Given Reactions:**
- Reaction I:
\[ C(\text{graphite}) + O_2(g) \rightarrow CO_2(g), \quad \Delta H = -x_1 \text{ cal} \]
- Reaction II:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g), \quad \Delta H = -x_2 \text{ cal} \]

2. **Understand the Enthalpy of Formation:**
- The standard enthalpy of formation \( \Delta H_f^\circ \) for a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states.

3. **Combine the Reactions:**
- We can manipulate the reactions to find the overall enthalpy change for the formation of \( CO_2 \) from graphite and oxygen.
- To do this, we can add the two reactions:
\[ C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \quad \text{(from Reaction I)} \]
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \text{(from Reaction II)} \]
- However, we need to reverse Reaction II to express it in terms of the formation of \( CO_2 \):
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g), \quad \Delta H = +x_2 \text{ cal} \]

4. **Add the Reactions:**
- Now, we can add Reaction I and the reversed Reaction II:
\[ C(\text{graphite}) + O_2(g) + CO_2(g) \rightarrow CO_2(g) + CO(g) + \frac{1}{2} O_2(g) \]
- This simplifies to:
\[ C(\text{graphite}) + \frac{3}{2} O_2(g) \rightarrow 2CO_2(g) \]

5. **Calculate the Overall Enthalpy Change:**
- The overall enthalpy change for this reaction is:
\[ \Delta H = -x_1 + x_2 \]

6. **Determine the Standard Enthalpy of Formation:**
- Since we are forming 2 moles of \( CO_2 \), the enthalpy of formation for one mole of \( CO_2 \) is:
\[ \Delta H_f^\circ(CO_2) = \frac{-x_1 + x_2}{2} \text{ cal} \]

### Final Answer:
\[ \Delta H_f^\circ(CO_2) = \frac{-x_1 + x_2}{2} \text{ cal} \]