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, we can use the relationship between change in internal energy (ΔU) and change in enthalpy (ΔH):

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

Where:
- ΔU is the change in internal energy (given as -1373 kJ/mol)
- 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 (given as 298 K)

### Step 1: Determine Δn_g
To find Δn_g, we need to calculate the number of moles of gaseous products and reactants.

- From the reaction:
\[ C_{2}H_{5}OH(l) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(g) \]

- Initial moles of gas (reactants):
- 3 moles of O2 (g)

- Final moles of gas (products):
- 2 moles of CO2 (g) + 3 moles of H2O (g) = 5 moles

Now, we can calculate Δn_g:
\[ \Delta n_g = \text{Final moles of gas} - \text{Initial moles of gas} = 5 - 3 = 2 \]

### Step 2: Substitute values into the equation
Now we can substitute the values into the equation for ΔH:
\[ \Delta H = \Delta U + \Delta n_g RT \]

Substituting the known values:
- ΔU = -1373 kJ/mol
- Δn_g = 2
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K) (since we need to keep the units consistent)
- T = 298 K

Calculating the term Δn_g RT:
\[ \Delta n_g RT = 2 \times 0.008314 \, \text{kJ/(mol·K)} \times 298 \, \text{K} = 4.975 \, \text{kJ/mol} \]

### Step 3: Calculate ΔH
Now we can calculate ΔH:
\[ \Delta H = -1373 \, \text{kJ/mol} + 4.975 \, \text{kJ/mol} = -1368.025 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta H \approx -1373 + 5 = -1378 \, \text{kJ/mol} \]