The heat of combustion of liquid ethanol is -327.0 kcal calculate the heat of formation of ethanol. Given that the heats of formation of `CO_(2)(g) and H_(2)O(l)` are -94.0 kcal and -68.4 kcal respectively.

Options

A. `-212.4 kcal`
B. `+212.4 kcal`
C. `-249.2 kcal`
D. `+249.2 kcal`

Answer

To calculate the heat of formation of ethanol (C2H5OH) given the heat of combustion and the heats of formation of the products, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Combustion Reaction**:
The combustion of ethanol can be represented as:
\[ C_2H_5OH + O_2 \rightarrow 2 CO_2 + 3 H_2O \]

2. **Identify the Given Values**:
- Heat of combustion of ethanol (ΔH_combustion) = -327.0 kcal
- Heat of formation of CO2 (ΔH_f CO2) = -94.0 kcal
- Heat of formation of H2O (ΔH_f H2O) = -68.4 kcal

3. **Write the Enthalpy Change Equation**:
According to Hess's law, the enthalpy change for the combustion can be expressed as:
\[ \Delta H_{combustion} = \left(2 \times \Delta H_f CO_2 + 3 \times \Delta H_f H_2O\right) - \Delta H_f C_2H_5OH \]

4. **Substitute the Known Values**:
Plugging in the values we have:
\[ -327.0 \, \text{kcal} = \left(2 \times (-94.0) + 3 \times (-68.4)\right) - \Delta H_f C_2H_5OH \]

5. **Calculate the Products' Contribution**:
Calculate the total heat of formation for the products:
\[ 2 \times (-94.0) = -188.0 \, \text{kcal} \]
\[ 3 \times (-68.4) = -205.2 \, \text{kcal} \]
Adding these together:
\[ -188.0 - 205.2 = -393.2 \, \text{kcal} \]

6. **Set Up the Equation**:
Now substitute back into the equation:
\[ -327.0 = -393.2 - \Delta H_f C_2H_5OH \]

7. **Solve for ΔH_f C2H5OH**:
Rearranging gives:
\[ \Delta H_f C_2H_5OH = -393.2 + 327.0 \]
\[ \Delta H_f C_2H_5OH = -66.2 \, \text{kcal} \]

### Final Answer:
The heat of formation of ethanol (C2H5OH) is **-66.2 kcal**.