`{:(C+O_(2)(g)rarrCO_(2)" ......(i),",DeltaH=-"393 kJ mol"^(-1)),(H_(2)+1//2O_(2)rarr H_(2)O", .....(ii)",DeltaH=-"287.3 kJ mol"^(-1)),(2CO_(2)+3H_(2)O rarr C_(2)H_(5)OH+2O_(2)" .....(iii)",DeltaH="1366.8 kJ mol"^(-1)):}`
Find the standard enthalpy of formation of `C_(2)H_(5)OH(l)`

Options

A. `"281.1 kJ mol"^(-1)`
B. `-"281.1 kJ mol"^(-1)`
C. `"562.2 kJ mol"^(-1)`
D. `-"562.2 kJ mol"^(-1)`

Explanation

To find the standard enthalpy of formation of \( C_2H_5OH(l) \), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Enthalpy Changes**:
- Reaction (i): \( C + O_2(g) \rightarrow CO_2 \)
\(\Delta H = -393 \, \text{kJ/mol}\)
- Reaction (ii): \( H_2 + \frac{1}{2}O_2 \rightarrow H_2O \)
\(\Delta H = -287.3 \, \text{kJ/mol}\)
- Reaction (iii): \( 2CO_2 + 3H_2O \rightarrow C_2H_5OH + 2O_2 \)
\(\Delta H = 1366.8 \, \text{kJ/mol}\)

2. **Write the Formation Reaction for \( C_2H_5OH \)**:
The standard enthalpy of formation of \( C_2H_5OH \) can be derived from the reaction:
\[ 2CO_2 + 3H_2O \rightarrow C_2H_5OH + 2O_2 \]

3. **Apply Hess's Law**:
According to Hess's law:
\[ \Delta H_{reaction} = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \]
For the reaction (iii):
\[ 1366.8 \, \text{kJ/mol} = \Delta H_f(C_2H_5OH) + 2 \cdot \Delta H_f(O_2) - [2 \cdot \Delta H_f(CO_2) + 3 \cdot \Delta H_f(H_2O)] \]

4. **Substitute Known Values**:
- The enthalpy of formation of \( O_2 \) in its standard state is \( 0 \).
- Substitute the values:
\[ 1366.8 = \Delta H_f(C_2H_5OH) + 0 - [2 \cdot (-393) + 3 \cdot (-287.3)] \]

5. **Calculate the Enthalpy of Formation**:
\[ 1366.8 = \Delta H_f(C_2H_5OH) - [2 \cdot (-393) + 3 \cdot (-287.3)] \]
Calculate the terms:
\[ 2 \cdot (-393) = -786 \]
\[ 3 \cdot (-287.3) = -861.9 \]
Combine these:
\[ -786 - 861.9 = -1647.9 \]
Now substitute back:
\[ 1366.8 = \Delta H_f(C_2H_5OH) + 1647.9 \]
Rearranging gives:
\[ \Delta H_f(C_2H_5OH) = 1366.8 - 1647.9 \]
\[ \Delta H_f(C_2H_5OH) = -281.1 \, \text{kJ/mol} \]

### Final Answer:
The standard enthalpy of formation of \( C_2H_5OH(l) \) is \( -281.1 \, \text{kJ/mol} \).