1. Given reaction

$CH_3OH(l)+\frac{3}{2}O_2(g) \rightarrow CO_2(g)+2H_2O(l)$

The heat measured in a bomb calorimeter is at constant volume, so it gives the change in internal energy:

$\Delta U = -726\ \text{kJ mol}^{-1}$

We need the enthalpy of combustion $\Delta H$.

2. Relation between enthalpy and internal energy

For reactions involving gases,

$\Delta H = \Delta U + \Delta n_g RT$

where $\Delta n_g$ is the change in moles of gaseous species.

3. Calculate $\Delta n_g$

From the reaction:

• Reactant gases: $\frac{3}{2}$ mol of $O_2$
• Product gases: $1$ mol of $CO_2$

So,

$\Delta n_g = 1 - \frac{3}{2} = -\frac{1}{2}$

4. Substitute values

Temperature:

$T = 27^\circ C = 300\ K$

Given:

$R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$

Now,

$\Delta n_g RT = -\frac{1}{2}\times 8.3 \times 300$

$= -1245\ \text{J mol}^{-1} = -1.245\ \text{kJ mol}^{-1}$

Thus,

$\Delta H = -726 + (-1.245)$

$\Delta H = -727.245\ \text{kJ mol}^{-1}$

5. Find $x$

Since enthalpy of combustion is written as $-x\ \text{kJ mol}^{-1}$,

$x = 727.245$

Nearest integer:

$x = 727$

6. Final Answer

$\boxed{727}$