The heat of combustion of ethanol determinal in a bomb calorimeter is -670.48 K. Cals `"mole"^(-1)` at `25^(@)C`. What is `Delta H` at `25^(@)C` for the reaction :-

Options

A. `-335.24" Kcal"`
B. `-671.08 " Kcal"`
C. `-670.48 " Kcal"`
D. `+670.48" Kcal"`

Answer

To determine the enthalpy change (ΔH) for the combustion of ethanol at 25°C, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of ethanol.
The combustion reaction of ethanol (C₂H₅OH) can be represented as:
\[ \text{C}_2\text{H}_5\text{OH (l)} + 3\text{O}_2\text{ (g)} \rightarrow 2\text{CO}_2\text{ (g)} + 3\text{H}_2\text{O (l)} \]

### Step 2: Identify the heat of combustion given.
The heat of combustion of ethanol is given as:
\[ \Delta E = -670.48 \text{ kcal/mol} \]
This value represents the internal energy change (ΔE) at constant volume.

### Step 3: Calculate ΔN (the change in the number of moles of gas).
ΔN is calculated as:
\[ \Delta N = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
From the balanced equation:
- Gaseous products: 2 moles of CO₂
- Gaseous reactants: 3 moles of O₂

Thus,
\[ \Delta N = 2 - 3 = -1 \]

### Step 4: Use the relationship between ΔE and ΔH.
The relationship between internal energy change (ΔE) and enthalpy change (ΔH) is given by:
\[ \Delta E = \Delta H - \Delta N \cdot R \cdot T \]
Where:
- R = 1.987 cal/(mol·K) (gas constant)
- T = 298 K (temperature in Kelvin)

### Step 5: Substitute the known values into the equation.
Rearranging the equation to solve for ΔH:
\[ \Delta H = \Delta E + \Delta N \cdot R \cdot T \]
Substituting the values:
\[ \Delta H = -670.48 \text{ kcal/mol} + (-1) \cdot (1.987 \text{ cal/(mol·K)}) \cdot (298 \text{ K}) \]
Convert kcal to cal for consistency:
\[ -670.48 \text{ kcal/mol} = -670480 \text{ cal/mol} \]
Now substituting:
\[ \Delta H = -670480 \text{ cal/mol} - (1) \cdot (1.987 \text{ cal/(mol·K)}) \cdot (298 \text{ K}) \]
Calculating the second term:
\[ 1.987 \cdot 298 = 592.746 \text{ cal/mol} \]
So,
\[ \Delta H = -670480 \text{ cal/mol} - 592.746 \text{ cal/mol} \]
\[ \Delta H = -671072.746 \text{ cal/mol} \]
Convert back to kcal:
\[ \Delta H = -671.072746 \text{ kcal/mol} \]
Rounding to two decimal places:
\[ \Delta H \approx -671.08 \text{ kcal/mol} \]

### Final Answer:
The enthalpy change (ΔH) for the combustion of ethanol at 25°C is approximately:
\[ \Delta H \approx -671.08 \text{ kcal/mol} \]