Compute the heat of formation of liquid methyl alcohol in kilojoules per mole, using the following data. Heat of vapourisation of liquid methyl alcohol= 38 kJ/mol. Heat of formation of gaseous atoms from the elements in their standard states, H= 218kJ/mol, C= 715 kJ/mol, O= 249 kJ/mol. Average bond energies, C-H = 415 kJ/mol, C-O = 365 kJ/mol, O-H= 463 kJ/mol

Answer

To compute the heat of formation of liquid methyl alcohol (CH₃OH), we will use the provided data and apply Hess's Law. The heat of formation of a compound can be determined by considering the energy changes associated with the formation of the compound from its elements.

### Step-by-Step Solution:

1. **Write the formation reaction of methyl alcohol:**
\[ C_{(s)} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow CH_3OH_{(l)} \]

2. **Identify the heat of formation (\( \Delta H_f \)) we want to calculate:**
We need to find \( \Delta H_f \) for \( CH_3OH_{(l)} \).

3. **Use the heat of vaporization:**
The heat of vaporization of liquid methyl alcohol is given as \( 38 \, \text{kJ/mol} \). This means:
\[ CH_3OH_{(l)} \rightarrow CH_3OH_{(g)} \quad \Delta H = 38 \, \text{kJ/mol} \]

4. **Calculate the heat of formation of gaseous methyl alcohol:**
The formation of gaseous methyl alcohol from its elements is:
\[ C_{(s)} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow CH_3OH_{(g)} \]
Let \( \Delta H_f^{g} \) be the heat of formation of gaseous methyl alcohol.

5. **Combine the two reactions:**
The overall reaction combining the formation of gaseous methyl alcohol and its vaporization to liquid form is:
\[ C_{(s)} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow CH_3OH_{(g)} \rightarrow CH_3OH_{(l)} \]
Thus, the total enthalpy change is:
\[ \Delta H_f^{l} = \Delta H_f^{g} - 38 \, \text{kJ/mol} \]

6. **Calculate the energy required to form gaseous atoms:**
- For hydrogen: \( 2H_{2(g)} \rightarrow 4H_{(g)} \) requires \( 218 \, \text{kJ/mol} \).
- For carbon: \( C_{(s)} \rightarrow C_{(g)} \) requires \( 715 \, \text{kJ/mol} \).
- For oxygen: \( \frac{1}{2}O_{2(g)} \rightarrow O_{(g)} \) requires \( 249 \, \text{kJ/mol} \).

Total energy required to form gaseous atoms:
\[ \Delta H_{\text{atoms}} = 715 + 218 + 249 = 1182 \, \text{kJ/mol} \]

7. **Calculate the energy released from bond formation:**
- Forming 3 C-H bonds: \( 3 \times 415 = 1245 \, \text{kJ/mol} \)
- Forming 1 C-O bond: \( 365 \, \text{kJ/mol} \)
- Forming 1 O-H bond: \( 463 \, \text{kJ/mol} \)

Total energy released from bond formation:
\[ \Delta H_{\text{bonds}} = 1245 + 365 + 463 = 2073 \, \text{kJ/mol} \]

8. **Calculate the overall enthalpy change:**
Using Hess's Law:
\[ \Delta H_f^{g} = \Delta H_{\text{atoms}} - \Delta H_{\text{bonds}} = 1182 - 2073 = -891 \, \text{kJ/mol} \]

9. **Calculate the heat of formation for liquid methyl alcohol:**
\[ \Delta H_f^{l} = -891 - 38 = -929 \, \text{kJ/mol} \]

### Final Answer:
The heat of formation of liquid methyl alcohol (CH₃OH) is:
\[ \Delta H_f^{l} = -929 \, \text{kJ/mol} \]