For #11 how to determine the signs?

The first reaction is exothermic as it releases heat as shown by the '+ 48 kJ'. When a reaction is exothermic the enthalpy change (ΔH) is negative as the system is losing heat. The entropy change (ΔS) is also negative as 4 moles of gas are reacting to form 2 moles of gas, reducing the disorder in the system. Since the spontaneity of a reaction is governed by ΔG = ΔH - TΔS, where ΔG must be negative for a spontaneous reaction, the reaction is only spontaneous at low temperatures as both ΔH and ΔS are negative, so as temperature increases, so does ΔG, toward 0.
The second reaction is endothermic as it absorbs heat therefore having a positive ΔH, and a positive ΔS as one mole of liquid is converted to one mole of more disordered gas. Using the same equation as before it can be shown that the reaction is only spontaneous at higher temperatures. These answers all hold true for the 3rd reaction as well (i.e. positive ΔH, positive ΔS and spontaneous at high temperatures).
Finally, reaction 4 is exothermic, therefore negative ΔH, and has a positive ΔS as 2 moles in solution are produced from one mole of solid. Therefore using the equation shown before, ΔG will always be negative regardless of temperature, and therefore the reaction is always spontaneous.
Hope this helps! If you could I would greatly appreciate you making this the brainliest answer :) If you have any questions feel free to ask.

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