For the reaction : `C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(g)`
if `Delta U^(@)= -1373 kJ mol^(-1)` at `298 K`. Calculate `Delta H^(@)`

Explanation

To calculate the change in enthalpy (ΔH°) for the given reaction using the provided internal energy change (ΔU°), we can use the following equation from the first law of thermodynamics:

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

Where:
- ΔU° is the change in internal energy.
- 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 1: Identify ΔU°
From the problem, we have:
\[ \Delta U° = -1373 \text{ kJ/mol} \]

### Step 2: Calculate ΔN_g
ΔN_g is calculated as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.

- Products: 2 moles of CO₂ (g) + 3 moles of H₂O (g) = 5 moles of gas
- Reactants: 3 moles of O₂ (g) = 3 moles of gas

Thus,
\[ \Delta N_g = \text{Moles of products} - \text{Moles of reactants} = 5 - 3 = 2 \]

### Step 3: Use the values in the equation
Now, we can substitute the values into the equation:

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

Substituting the known values:
- ΔU° = -1373 kJ/mol
- ΔN_g = 2
- R = 8.314 J/mol·K = 0.008314 kJ/mol·K (since we want ΔH in kJ)
- T = 298 K

Now, calculate ΔN_g RT:
\[ \Delta N_g RT = 2 \times 0.008314 \text{ kJ/mol·K} \times 298 \text{ K} \]
\[ \Delta N_g RT = 2 \times 0.008314 \times 298 = 4.964 \text{ kJ/mol} \]

### Step 4: Calculate ΔH°
Now we can plug this value back into our equation:
\[ \Delta H° = -1373 \text{ kJ/mol} + 4.964 \text{ kJ/mol} \]
\[ \Delta H° = -1368.036 \text{ kJ/mol} \]

### Final Result
Thus, the change in enthalpy for the reaction is:
\[ \Delta H° \approx -1368 \text{ kJ/mol} \]