From Q62. above `Delta H_(f)^(@)(CO)` is:

Options

A. `-x_1`
B. `-x_2`
C. `x_2 - x_1`
D. `x_1 - x_2`

Explanation

To find the standard enthalpy of formation (ΔH_f^0) of carbon monoxide (CO), 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. **Identify the Reactions:**
We have two reactions provided with their respective ΔH values:
- Reaction 1: \( C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \) with ΔH = ΔH1
- Reaction 2: \( CO_{2(g)} \rightarrow C_{(s)} + \frac{1}{2} O_{2(g)} \) with ΔH = ΔH2

2. **Write the Main Reaction:**
The main reaction we want to derive is:
\[ C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \]
This is the formation of one mole of CO from its elements in their standard states.

3. **Manipulate the Given Reactions:**
To obtain the main reaction, we can reverse Reaction 2 and then add it to Reaction 1:
- Reverse Reaction 2:
\[ C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \quad \text{(ΔH = -ΔH2)} \]
- Add Reaction 1 and the reversed Reaction 2:
\[ (C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}) + (C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}) \]
- The CO2 will cancel out, leading to:
\[ C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \]

4. **Calculate the Enthalpy Change:**
According to Hess's law, the enthalpy change for the main reaction is the sum of the enthalpy changes of the individual reactions:
\[ ΔH_f^0 (CO) = ΔH1 + (-ΔH2) = ΔH1 - ΔH2 \]

5. **Conclusion:**
The standard enthalpy of formation of CO is given by:
\[ ΔH_f^0 (CO) = ΔH1 - ΔH2 \]

### Final Answer:
Thus, the value of ΔH_f^0 (CO) is \( ΔH1 - ΔH2 \).

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