The enthalpy of combustion of ethyl alcohol `(C_(2)H_(5)OH)` is 1380.7 kJ `mol^(_1)`. If the enthalpies of formation of `CO_(2)` and `H_(2)O` are 394.5 and 286.6kJ `mol^(-1)` respectively, calculate the enthalpy of formation of ethyl alcohol.

Explanation

To calculate the enthalpy of formation of ethyl alcohol (C₂H₅OH), we can use the enthalpy of combustion and the enthalpies of formation of the products (CO₂ and H₂O). The combustion reaction of ethyl alcohol can be represented as follows:

\[ \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \]

### Step 1: Write the enthalpy change for the combustion reaction
The enthalpy change for the combustion of ethyl alcohol is given as -1380.7 kJ/mol. This value is negative because it is an exothermic reaction.

### Step 2: Write the enthalpy of formation for the products
The enthalpy of formation for the products (CO₂ and H₂O) is given as:
- For CO₂: ΔH_f (CO₂) = -394.5 kJ/mol
- For H₂O: ΔH_f (H₂O) = -286.6 kJ/mol

### Step 3: Calculate the total enthalpy of formation for the products
In the combustion reaction, we produce 2 moles of CO₂ and 3 moles of H₂O. Therefore, the total enthalpy of formation for the products can be calculated as follows:

\[ \Delta H_{\text{products}} = [2 \times \Delta H_f (\text{CO}_2)] + [3 \times \Delta H_f (\text{H}_2\text{O})] \]

Substituting the values:

\[ \Delta H_{\text{products}} = [2 \times (-394.5 \text{ kJ/mol})] + [3 \times (-286.6 \text{ kJ/mol})] \]

Calculating each term:

\[ = -789 \text{ kJ} + (-859.8 \text{ kJ}) = -1648.8 \text{ kJ} \]

### Step 4: Apply Hess's Law
According to Hess's Law, the enthalpy change for the combustion reaction can be expressed as:

\[ \Delta H_{\text{combustion}} = \Delta H_f (\text{C}_2\text{H}_5\text{OH}) + \Delta H_{\text{products}} \]

Rearranging to find the enthalpy of formation of ethyl alcohol:

\[ \Delta H_f (\text{C}_2\text{H}_5\text{OH}) = \Delta H_{\text{combustion}} - \Delta H_{\text{products}} \]

Substituting the known values:

\[ \Delta H_f (\text{C}_2\text{H}_5\text{OH}) = -1380.7 \text{ kJ/mol} - (-1648.8 \text{ kJ}) \]

Calculating the final value:

\[ \Delta H_f (\text{C}_2\text{H}_5\text{OH}) = -1380.7 \text{ kJ/mol} + 1648.8 \text{ kJ} = 268.1 \text{ kJ/mol} \]

### Final Answer:
The enthalpy of formation of ethyl alcohol (C₂H₅OH) is **268.1 kJ/mol**.

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