The enthalpy of formation of `C_(2)H_(5)OH(l)`, `CO_(2)(g)` and `H_(2)O(l)` are - 277.0, -393.5 and -285.5 kJ/mol respectively.Calculate the enthalpy change for the reaction: `C_(2)H_(5)OH(l) + 3O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l)`.

Explanation

To calculate the enthalpy change for the reaction:

\[ C_{2}H_{5}OH(l) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(l) \]

we will use the enthalpy of formation values provided for each substance involved in the reaction. The formula for calculating the enthalpy change (\( \Delta H \)) is:

\[ \Delta H = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \]

### Step-by-step Solution:

1. **Identify the enthalpy of formation values**:
- For \( C_{2}H_{5}OH(l) \): \( \Delta H_f = -277.0 \, \text{kJ/mol} \)
- For \( CO_{2}(g) \): \( \Delta H_f = -393.5 \, \text{kJ/mol} \)
- For \( H_{2}O(l) \): \( \Delta H_f = -285.5 \, \text{kJ/mol} \)
- For \( O_{2}(g) \): \( \Delta H_f = 0 \, \text{kJ/mol} \) (as it is in its standard state)

2. **Calculate the total enthalpy of formation for the products**:
- Products: \( 2CO_{2}(g) + 3H_{2}O(l) \)
- Total for \( CO_{2} \): \( 2 \times (-393.5) = -787.0 \, \text{kJ} \)
- Total for \( H_{2}O \): \( 3 \times (-285.5) = -856.5 \, \text{kJ} \)
- Total enthalpy for products:
\[ \Delta H_f \text{(products)} = -787.0 + (-856.5) = -1643.5 \, \text{kJ} \]

3. **Calculate the total enthalpy of formation for the reactants**:
- Reactants: \( C_{2}H_{5}OH(l) + 3O_{2}(g) \)
- Total for \( C_{2}H_{5}OH \): \( -277.0 \, \text{kJ} \)
- Total for \( O_{2} \): \( 3 \times 0 = 0 \, \text{kJ} \)
- Total enthalpy for reactants:
\[ \Delta H_f \text{(reactants)} = -277.0 + 0 = -277.0 \, \text{kJ} \]

4. **Calculate the enthalpy change for the reaction**:
\[ \Delta H = \Delta H_f \text{(products)} - \Delta H_f \text{(reactants)} \]
\[ \Delta H = -1643.5 - (-277.0) = -1643.5 + 277.0 = -1366.5 \, \text{kJ} \]

### Final Answer:
The enthalpy change for the reaction is \( \Delta H = -1366.5 \, \text{kJ} \).