For complete combustion of ethene,
$$C_2H_4(g) + 3O_2(g) \to 2CO_2(g) + 2H_2O(l)$$
the amount of heat produced as measured in bomb calorimeter is 1406 kJ mol$$^{-1}$$ at 300 K. The minimum value of T$$\Delta$$S needed to reach equilibrium is ($$-$$) kJ. (Nearest integer)
Given: R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$

A bomb calorimeter measures the internal energy change at constant volume, so $$\Delta U = -1406 \text{ kJ mol}^{-1}$$. To find $$\Delta H$$, we use the relation $$\Delta H = \Delta U + \Delta n_g RT$$, where $$\Delta n_g$$ is the change in moles of gaseous species.

For the reaction $$C_2H_4(g) + 3O_2(g) \to 2CO_2(g) + 2H_2O(l)$$, the gaseous moles on the product side are 2 (only $$CO_2$$; water is liquid) and on the reactant side are $$1 + 3 = 4$$. Thus $$\Delta n_g = 2 - 4 = -2$$.

Computing: $$\Delta H = -1406 + (-2)(8.3 \times 10^{-3})(300) = -1406 - 4.98 = -1410.98 \text{ kJ mol}^{-1}$$.

At equilibrium, $$\Delta G = 0$$, which gives $$\Delta H = T\Delta S$$. Therefore the minimum value of $$T\Delta S$$ needed to reach equilibrium is $$-1411 \text{ kJ}$$ (nearest integer), and the magnitude is $$\boxed{1411}$$.