Reaction of methanol with dioxygen was carried out and `DeltaU` was found to be `-726"kJ mol"^(-1)` at 298 K. The enthalpy change for the reaction will be
`CH_(3)OH_((l))+(3)/(2)O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l)) , DeltaH=-726"kJ mol"^(-1)`

Options

A. `-"741.5 kJ mol"^(-1)`
B. `-"727 kJ mol"^(-1)`
C. `+"741.5 kJ mol"^(-1)`
D. `+"727.2 kJ mol"^(-1)`

Explanation

To solve the problem, we need to calculate the enthalpy change (ΔH) for the reaction of methanol with dioxygen using the relationship between internal energy change (ΔU) and enthalpy change (ΔH).

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction given is:
\[ \text{CH}_3\text{OH}_{(l)} + \frac{3}{2} \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + 2 \text{H}_2\text{O}_{(l)} \]

2. **Identify Given Values**:
- ΔU = -726 kJ/mol (change in internal energy)
- T = 298 K (temperature)
- R = 8.314 × 10⁻³ kJ/(K·mol) (gas constant)

3. **Calculate ΔN_g**:
ΔN_g is defined as the change in the number of moles of gas:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
- Moles of gaseous products = 1 (from CO₂)
- Moles of gaseous reactants = 1.5 (from \(\frac{3}{2} O_2\))
\[ \Delta N_g = 1 - 1.5 = -0.5 \]

4. **Use the Formula for ΔH**:
The relationship between ΔH and ΔU is given by:
\[ \Delta H = \Delta U + \Delta N_g \cdot R \cdot T \]
Substitute the known values into the equation:
\[ \Delta H = -726 + (-0.5) \cdot (8.314 \times 10^{-3}) \cdot 298 \]

5. **Calculate the Second Term**:
First, calculate \(-0.5 \cdot 8.314 \times 10^{-3} \cdot 298\):
\[ -0.5 \cdot 8.314 \times 10^{-3} \cdot 298 = -1.23 \text{ kJ/mol} \]

6. **Final Calculation of ΔH**:
Now substitute this value back into the ΔH equation:
\[ \Delta H = -726 - 1.23 = -727.23 \text{ kJ/mol} \]
Rounding to three significant figures gives:
\[ \Delta H \approx -727 \text{ kJ/mol} \]

### Conclusion:
The enthalpy change for the reaction is:
\[ \Delta H = -727 \text{ kJ/mol} \]