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 Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### 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. **Add the Reactions**:
To find the desired reaction, we need to add the two reactions together. We will keep the first reaction as it is and reverse the second reaction to eliminate \( CO(g) \).

- Reversed Reaction 2:
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g), \quad \Delta H = +283.2 \, \text{kJ} \]

3. **Combine the Reactions**:
Now, we can add the two reactions:
\[ C(\text{graphite}) + \frac{1}{2} O_2(g) \rightarrow CO(g) \quad (\Delta H_1 = -110.5 \, \text{kJ}) \]
\[ CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g) \quad (\Delta H = +283.2 \, \text{kJ}) \]
Adding these gives:
\[ C(\text{graphite}) + O_2(g) \rightarrow CO_2(g) \]

4. **Calculate the Total Enthalpy Change**:
Now, we can calculate the total enthalpy change for the overall reaction:
\[ \Delta H = \Delta H_1 + \Delta H_2 \]
\[ \Delta H = -110.5 \, \text{kJ} + (-283.2 \, \text{kJ}) = -393.7 \, \text{kJ} \]

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