Given that : `C(s)+O_(2)(g)to CO_(2)(g) , Delta H = -394 kJ` and `2H_(2)(g)+O_(2)(g)to 2H_(2)O(l) , Delta H =- 568 kJ` and `C_(2)H_(5)OH(l)+3O_(2)(g)to 2CO_(2)(g)+3H_(2)O(l) , Delta H =- 1058` kJ/mole. Using the data, the heat of formation of ethanol is

Options

A. `-582 kJ`
B. 71.8 kJ
C. `-244 kJ`
D. `+782 kJ`

Answer

To find the heat of formation of ethanol (C2H5OH), 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. **Write the given reactions and their enthalpy changes:**
- Reaction 1: \( C(s) + O_2(g) \rightarrow CO_2(g) \), \( \Delta H = -394 \, \text{kJ} \) (Equation 1)
- Reaction 2: \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \), \( \Delta H = -568 \, \text{kJ} \) (Equation 2)
- Reaction 3: \( C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) \), \( \Delta H = -1058 \, \text{kJ} \) (Equation 3)

2. **Reverse the reactions as needed:**
- To find the heat of formation of ethanol, we need to reverse the reactions that produce the products of ethanol. Therefore:
- Reverse Reaction 1: \( CO_2(g) \rightarrow C(s) + O_2(g) \), \( \Delta H = +394 \, \text{kJ} \)
- Reverse Reaction 2: \( 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \), \( \Delta H = +568 \, \text{kJ} \)

3. **Add the reactions together:**
- Now, we add the reversed reactions to Reaction 3:
\[ CO_2(g) + C_2H_5OH(l) + 3O_2(g) \rightarrow C(s) + O_2(g) + 2CO_2(g) + 3H_2O(l) + 2H_2(g) \]
- The equation simplifies to:
\[ C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) \]

4. **Calculate the overall enthalpy change:**
- The overall enthalpy change for the formation of ethanol is given by:
\[ \Delta H = \Delta H_{reaction 3} + \Delta H_{reversed 1} + \Delta H_{reversed 2} \]
- Substituting the values:
\[ \Delta H = -1058 \, \text{kJ} + 394 \, \text{kJ} + 568 \, \text{kJ} \]
- Calculate:
\[ \Delta H = -1058 + 394 + 568 = -1058 + 962 = -96 \, \text{kJ} \]

5. **Final Result:**
- The heat of formation of ethanol (C2H5OH) is \( -96 \, \text{kJ/mol} \).