Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following endothermic reaction. 2N2(g) + O2(g)2N2O(g) Clear All > 0 Hrxn < 0 Srxn = 0 Grxn > 0 low T, < 0 high T Suniverse < 0 low T, > 0 high T

∆H > 0

∆Srxn <0

∆G >0

∆Suniverse <0

Explanation:

We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.

Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.

The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.

Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.