1. Given reaction

$\mathrm{C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)}$

For a bomb calorimeter, the heat measured is at constant volume. Hence,

$q_v = \Delta U = -1406\ \text{kJ mol}^{-1}$

So,

$\Delta U = -1406\ \text{kJ mol}^{-1}$

2. Relate $\Delta H$ and $\Delta U$

We use:

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

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

From the reaction:

• Gaseous reactants = $1 + 3 = 4$
• Gaseous products = $2$

Thus,

$\Delta n_g = 2 - 4 = -2$

Therefore,

$\Delta H = -1406 + (-2)(8.3)(300)\times 10^{-3}$

Converting $RT$ into kJ:

$2\times 8.3\times 300 = 4980\ \text{J} = 4.98\ \text{kJ}$

So,

$\Delta H = -1406 - 4.98 = -1410.98\ \text{kJ mol}^{-1}$

Approximately,

$\Delta H \approx -1411\ \text{kJ mol}^{-1}$

3. Condition for equilibrium / spontaneity limit

For a reaction to just reach equilibrium,

$\Delta G = \Delta H - T\Delta S = 0$

Hence the minimum value of $T\Delta S$ needed is equal to $\Delta H$ in magnitude:

$T\Delta S = \Delta H$

Since $\Delta H$ is negative here, the asked blank is given as $(-)\,\underline{\hspace{1cm}}\,\text{kJ}$, so we report the magnitude:

$|T\Delta S| = 1411\ \text{kJ}$

4. Final answer

$\boxed{1411}$