In the given combustion reaction, the number of moles of gaseous products ($\Delta n_g$) is -2 (since we have 2 moles of CO₂ gas on the product side and 4 moles of gaseous reactants).

The change in internal energy ($\Delta U$) is given as -1406 kJ/mol.

We can calculate the change in enthalpy ($\Delta H$) using the equation $\Delta H = \Delta U + \Delta n_gRT$, where $R$ is the ideal gas constant and $T$ is the temperature.

Substituting the given values:

$\begin{aligned} \Delta H &= \Delta U + \Delta n_gRT \\\\ &= -1406 \mathrm{kJ/mol} + (-2) \times 8.3 \times 10^{-3} \mathrm{kJ/K/mol} \times 300 \mathrm{K} \\\\ &= -1406 \mathrm{kJ/mol} - 4.98 \mathrm{kJ/mol} \\\\ &\approx -1411 \mathrm{kJ/mol} \end{aligned}$

At equilibrium, $\Delta G = 0$, so $\Delta H = T \Delta S$, which gives $T \Delta S = \Delta H$.

Therefore, the minimum value of $T \Delta S$ needed to reach equilibrium is -1411 kJ