The equation represents standard heat of formation of ethanol is

Answer

To determine the standard heat of formation of ethanol (C2H5OH), we need to write the balanced chemical equation for its formation from its elements in their standard states.

### Step-by-Step Solution:

1. **Identify the Elements**: Ethanol consists of carbon (C), hydrogen (H), and oxygen (O). The standard heat of formation is defined for a compound formed from its elements in their standard states.

2. **Write the Elements in Their Standard States**:
- Carbon (C) in its standard state is solid graphite.
- Hydrogen (H2) in its standard state is a gas.
- Oxygen (O2) in its standard state is a gas.

3. **Write the Formation Reaction**: The formation of ethanol from its elements can be represented as:
\[ 2 \text{C (s)} + 3 \text{H}_2 \text{(g)} + \frac{1}{2} \text{O}_2 \text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH (l)} \]

4. **Standard Enthalpy of Formation**: The standard enthalpy of formation (ΔH°f) for ethanol can be expressed as:
\[ \Delta H^\circ_f (\text{C}_2\text{H}_5\text{OH}) = \text{Standard heat of formation of ethanol} \]

5. **Combustion Reaction**: The combustion of ethanol can be represented as:
\[ \text{C}_2\text{H}_5\text{OH (l)} + 3 \text{O}_2 \text{(g)} \rightarrow 2 \text{CO}_2 \text{(g)} + 3 \text{H}_2\text{O (l)} \]
The heat released during this reaction can be calculated using the standard enthalpies of formation of the products and reactants.

6. **Calculate the Heat of Formation**: The heat of formation can be calculated using the formula:
\[ \Delta H^\circ_{reaction} = \sum \Delta H^\circ_f (\text{products}) - \sum \Delta H^\circ_f (\text{reactants}) \]

- For the products:
- 2 moles of CO2: \(2 \times (-394 \, \text{kJ/mol}) = -788 \, \text{kJ}\)
- 3 moles of H2O: \(3 \times (-286 \, \text{kJ/mol}) = -858 \, \text{kJ}\)

- For the reactants:
- Ethanol: \(\Delta H^\circ_f (\text{C}_2\text{H}_5\text{OH})\)
- 3 moles of O2: \(0 \, \text{kJ}\) (elements in standard state)

Putting it all together:
\[ \Delta H^\circ_{reaction} = [-788 - 858] - [\Delta H^\circ_f (\text{C}_2\text{H}_5\text{OH}) + 0] \]

7. **Final Equation**: Rearranging gives us the standard heat of formation of ethanol:
\[ \Delta H^\circ_f (\text{C}_2\text{H}_5\text{OH}) = -788 - 858 - \Delta H^\circ_{reaction} \]