Increases – particles go from being in a fixed lattice to moving independently of each other.

Increases – the number of independently moving particles has increased (1 mole of gas forms 2 moles of gas).

Decreased – gone from two solutions containing free-moving ions, to one solution and a solid where the particles are in a fixed lattice.

Decreased – H⁺ and OH⁻ have gone from moving independently of each other to forming ordered H₂O molecules.

Decreased – the ions have gone from being able to move independently in solution to being in a fixed solid lattice.

Impossible to tell – although they have gone from solid and gas to gas and liquid, the number of gas molecules has decreased (8 to 5). It is more likely that entropy has decreased due to loss of gas moles, but precise values are needed.

Impossible to tell – Same number of solid particles (2 moles reactants, 2 moles products) and no change in state.

High temperatures. The reaction is endothermic (+ΔH) but increases entropy (gas produced). Therefore, TΔS must be large enough to overcome ΔH for ΔG to be negative.

Any temperature. The reaction is highly exothermic (-ΔH). Even though gas moles decrease slightly (6.5 to 6), the massive exothermicity drives the reaction at standard conditions.

Lower temperatures. The reaction is exothermic (-ΔH) but entropy decreases (gas to solid). It is feasible at low temps where TΔS is small, but at very high temps, the -TΔS term becomes positive and large enough to make ΔG positive.

High temperature. The reaction is endothermic (+ΔH) but entropy increases significantly (production of two gases). High temperature is required for TΔS to overcome the positive ΔH.