The value of enthalpy change `(DeltaH)` for the reaction
`C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)`
at `27^(@)C` is `-1366.5KJmol^(-1)` . The value of internal energy change for the above reaction at this tempearature will be

Options

A. `-1371.5KJ`
B. `-1369.0KJ`
C. `-1364.0KJ`
D. `-1361.5KJ`

Explanation

To find the internal energy change (ΔU) for the reaction given the enthalpy change (ΔH), we can use the relationship between these two thermodynamic quantities:

\[ \Delta H = \Delta U + \Delta N_g RT \]

Where:
- ΔH is the enthalpy change,
- ΔU is the internal energy change,
- ΔN_g is the change in the number of moles of gas,
- R is the universal gas constant (8.314 J/mol·K),
- T is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values:**
- Enthalpy change (ΔH) = -1366.5 kJ/mol
- Temperature (T) = 27°C = 27 + 273 = 300 K
- R = 8.314 J/mol·K = 0.008314 kJ/mol·K (since we want the final answer in kJ)

2. **Calculate ΔN_g:**
- The reaction is:
\[ C_{2}H_{5}OH(l) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(l) \]
- Count the moles of gaseous reactants and products:
- Gaseous products: 2 moles of CO₂
- Gaseous reactants: 3 moles of O₂
- Therefore, ΔN_g = moles of products - moles of reactants = 2 - 3 = -1.

3. **Calculate ΔN_g RT:**
\[ \Delta N_g RT = (-1) \times (0.008314 \, \text{kJ/mol·K}) \times (300 \, \text{K}) = -2.4942 \, \text{kJ} \]

4. **Substitute values into the equation:**
\[ \Delta U = \Delta H - \Delta N_g RT \]
\[ \Delta U = -1366.5 \, \text{kJ/mol} - (-2.4942 \, \text{kJ}) \]
\[ \Delta U = -1366.5 + 2.4942 = -1364.0058 \, \text{kJ/mol} \]

5. **Round the answer:**
The internal energy change (ΔU) can be rounded to:
\[ \Delta U \approx -1364.0 \, \text{kJ/mol} \]

### Final Answer:
The value of internal energy change (ΔU) for the reaction at 27°C is approximately **-1364.0 kJ/mol**.