To calculate the enthalpy change (ΔH) for the reaction \( C(g) + O_2(g) \rightarrow CO_2(g) \), we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps of the reaction.
### Given Reactions:
1. \( H_2O(g) + C(g) + H_2(g) \rightarrow \Delta H_1 = +131 \, \text{kJ} \)
2. \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \, \Delta H_2 = -242 \, \text{kJ} \)
3. \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \, \Delta H_3 = -242 \, \text{kJ} \)
### Step-by-Step Solution:
**Step 1: Write down the target reaction.**
We need to find ΔH for the reaction:
\[ C(g) + O_2(g) \rightarrow CO_2(g) \]
**Step 2: Rearrange the given reactions to derive the target reaction.**
We need to manipulate the given reactions to cancel out the unwanted species and arrive at the target reaction.
- From reaction 3, we can reverse it to get:
\[ H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \]
This will change the sign of ΔH:
\[ \Delta H = +242 \, \text{kJ} \]
- Now, we can add this reversed reaction to reaction 1:
\[ H_2O(g) + C(g) + H_2(g) \rightarrow \Delta H_1 = +131 \, \text{kJ} \]
Combining these gives:
\[ C(g) + H_2O(g) \rightarrow H_2(g) + CO(g) \]
- Now we need to add reaction 2:
\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \, \Delta H_2 = -242 \, \text{kJ} \]
**Step 3: Combine the reactions.**
Now we combine:
1. \( C(g) + H_2O(g) \rightarrow H_2(g) + CO(g) \) (ΔH = +131 kJ)
2. \( CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \) (ΔH = -242 kJ)
When we add these two reactions, \( H_2(g) \) and \( CO(g) \) will cancel out:
\[ C(g) + O_2(g) \rightarrow CO_2(g) \]
**Step 4: Calculate the total ΔH.**
Now we sum the ΔH values:
\[ \Delta H = +131 \, \text{kJ} + 242 \, \text{kJ} - 242 \, \text{kJ} \]
\[ \Delta H = +131 - 242 \]
\[ \Delta H = -393 \, \text{kJ} \]
### Final Answer:
The enthalpy change (ΔH) for the reaction \( C(g) + O_2(g) \rightarrow CO_2(g) \) is:
\[ \Delta H = -393 \, \text{kJ} \]
