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` K. Cals.
B. `-671.08` K. Cals.
C. `-670.48` K. Cals.
D. `+670.48` K. Cals.

Explanation

To find the value of ΔH at 25°C for the combustion of ethanol, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of ethanol.
The combustion of ethanol (C2H5OH) can be represented as follows:

\[ \text{C}_2\text{H}_5\text{OH (l)} + 3 \text{O}_2\text{ (g)} \rightarrow 2 \text{CO}_2\text{ (g)} + 3 \text{H}_2\text{O (l)} \]

### Step 2: Identify the given data.
The heat of combustion of ethanol is given as:
- ΔE (heat of combustion) = -670.48 kilocalories per mole at 25°C.

### Step 3: Use the relationship between ΔH and ΔE.
The relationship between ΔH (enthalpy change) and ΔE (internal energy change) is given by the equation:

\[ \Delta H = \Delta E + \Delta N \cdot R \cdot T \]

where:
- ΔN = change in the number of moles of gas
- R = universal gas constant (2 × 10^(-3) kcal/mol·K)
- T = temperature in Kelvin (25°C = 298 K)

### Step 4: Calculate ΔN.
In the balanced equation:
- Moles of gaseous products (CO2) = 2
- Moles of gaseous reactants (O2) = 3

Thus,
\[ \Delta N = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - 3 = -1 \]

### Step 5: Substitute values into the equation.
Now, substituting the values into the equation for ΔH:

\[ \Delta H = -670.48 \text{ kcal/mol} + (-1) \cdot (2 \times 10^{-3} \text{ kcal/mol·K}) \cdot (298 \text{ K}) \]

Calculating the second term:

\[ -1 \cdot (2 \times 10^{-3}) \cdot 298 = -0.596 \text{ kcal/mol} \]

### Step 6: Final calculation for ΔH.
Now, substituting this back into the equation for ΔH:

\[ \Delta H = -670.48 \text{ kcal/mol} - 0.596 \text{ kcal/mol} \]

\[ \Delta H \approx -671.08 \text{ kcal/mol} \]

### Final Answer:
Thus, the value of ΔH at 25°C for the combustion of ethanol is approximately:

\[ \Delta H \approx -671.08 \text{ kcal/mol} \]

---