Form the thermochemical reactions.
C(graphite) `+1//2 O_(2)(g) to CO(g), triangleH=-110.5kJ`
`CO(g)+1//2O_(2)(g) to CO_(2)(g), triangleH=-283.2kJ`
The enthalpy ofreaction of C(graphite) `+O_(2)(g) to CO_(2)" is "`

Options

A. `393.3 kJ`
B. `-393.7 kJ`
C. `-172.7 kJ`
D. `+172.7 kJ`

Answer

To find the enthalpy of the reaction \( C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \), we will use the given thermochemical equations and their corresponding enthalpy changes.

### Step-by-Step Solution:

1. **Write the Given Reactions:**
- Reaction 1:
\[ C(\text{graphite}) + \frac{1}{2} O_2(g) \rightarrow CO(g) \quad \Delta H_1 = -110.5 \, \text{kJ} \]
- Reaction 2:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \Delta H_2 = -283.2 \, \text{kJ} \]

2. **Identify the Target Reaction:**
- We need to find the enthalpy change for the reaction:
\[ C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \]

3. **Manipulate the Given Reactions:**
- We can add the two given reactions to derive the target reaction.
- First, we keep Reaction 1 as it is:
\[ C(\text{graphite}) + \frac{1}{2} O_2(g) \rightarrow CO(g) \quad \Delta H_1 = -110.5 \, \text{kJ} \]
- Next, we keep Reaction 2 as it is:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \quad \Delta H_2 = -283.2 \, \text{kJ} \]

4. **Add the Reactions:**
- When we add these two reactions, the \( CO(g) \) on the product side of Reaction 1 cancels with the \( CO(g) \) on the reactant side of Reaction 2:
\[ C(\text{graphite}) + \frac{1}{2} O_2(g) + CO(g) + \frac{1}{2} O_2(g) \rightarrow CO(g) + O_2(g) \rightarrow CO_2(g) \]
- This simplifies to:
\[ C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \]

5. **Calculate the Enthalpy Change:**
- The total enthalpy change for the reaction can be found by adding the enthalpy changes of the two reactions:
\[ \Delta H = \Delta H_1 + \Delta H_2 = -110.5 \, \text{kJ} + (-283.2 \, \text{kJ}) = -393.7 \, \text{kJ} \]

6. **Final Answer:**
- The enthalpy of reaction for \( C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \) is:
\[ \Delta H = -393.7 \, \text{kJ} \]