Calculate the enthalpy of formation of acetic acid `(CH_(3)COOH)` if its enthalpy of combustion is `867 kJ mol^(-1)`.The enthalpies of fromation of `CO_(2)(g)` and `H_(2)O(l)` are -393.5 and -285.9 `kJ mol^(-1)` respectively.

Explanation

To calculate the enthalpy of formation of acetic acid (CH₃COOH) using the given enthalpy of combustion and the enthalpies of formation of CO₂ and H₂O, we can follow these steps:

### Step 1: Write the combustion reaction of acetic acid.
The combustion reaction of acetic acid can be represented as:
\[ \text{CH}_3\text{COOH} + 2 \text{O}_2 \rightarrow 2 \text{CO}_2 + 3 \text{H}_2\text{O} \]
This reaction shows that one mole of acetic acid reacts with two moles of oxygen to produce two moles of carbon dioxide and three moles of water.

### Step 2: Write the enthalpy change for the combustion reaction.
The enthalpy change for the combustion reaction (ΔH_comb) is given as:
\[ \Delta H_{\text{comb}} = -867 \, \text{kJ/mol} \]
This value is negative because it is an exothermic reaction.

### Step 3: Write the formation reactions for CO₂ and H₂O.
The formation reactions for CO₂ and H₂O are:
1. For CO₂:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \Delta H_f = -393.5 \, \text{kJ/mol} \]
2. For H₂O:
\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta H_f = -285.9 \, \text{kJ/mol} \]

### Step 4: Multiply the formation enthalpies by their coefficients in the combustion reaction.
From the combustion reaction, we have:
- 2 moles of CO₂:
\[ \Delta H_f (\text{CO}_2) = 2 \times (-393.5) = -787 \, \text{kJ} \]
- 3 moles of H₂O:
\[ \Delta H_f (\text{H}_2\text{O}) = 3 \times (-285.9) = -857.7 \, \text{kJ} \]

### Step 5: Calculate the total enthalpy change for the products.
The total enthalpy change for the products (ΔH_products) is:
\[ \Delta H_{\text{products}} = -787 \, \text{kJ} + (-857.7 \, \text{kJ}) = -1644.7 \, \text{kJ} \]

### Step 6: Use Hess's law to find the enthalpy of formation of acetic acid.
According to Hess's law:
\[ \Delta H_{\text{reaction}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \]
Here, the enthalpy of formation of acetic acid (ΔH_f) is what we are trying to find. The enthalpy of the reactants is the enthalpy of combustion:
\[ -867 \, \text{kJ} = -1644.7 \, \text{kJ} - \Delta H_f \]

### Step 7: Rearranging the equation to find ΔH_f.
Rearranging the equation gives:
\[ \Delta H_f = -1644.7 \, \text{kJ} + 867 \, \text{kJ} \]
\[ \Delta H_f = -777.7 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of formation of acetic acid (CH₃COOH) is:
\[ \Delta H_f = -777.7 \, \text{kJ/mol} \]