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 (ΔH_f°(CO₂)), we will analyze the given reactions step by step.

### Step 1: Understanding the Reactions
We have two reactions:
1. \( C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \) with \( \Delta H^\circ = -x_1 \, \text{cal} \)
2. \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \) with \( \Delta H^\circ = -x_2 \, \text{cal} \)

### Step 2: Identifying the Formation Reaction
The standard enthalpy of formation (ΔH_f°) of a compound is defined as the enthalpy change when one mole of the compound is formed from its elements in their standard states. For carbon dioxide (CO₂), the formation reaction is:
\[ C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \]

### Step 3: Relating the Given Reactions to the Formation Reaction
From the first reaction, we see that it directly represents the formation of one mole of CO₂ from its elements (carbon in graphite form and oxygen gas). The enthalpy change for this reaction is given as \( -x_1 \, \text{cal} \).

### Step 4: Conclusion
Since the first reaction directly corresponds to the formation of CO₂ from its standard state elements, we can conclude that:
\[ \Delta H_f^\circ(CO_2) = -x_1 \, \text{cal} \]

Thus, the standard enthalpy of formation of CO₂ is \( -x_1 \, \text{cal} \).

### Final Answer
\[ \Delta H_f^\circ(CO_2) = -x_1 \, \text{cal} \]

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