[TEA1]

The 'reaction' where enzymes lose their activity is: Active enzyme to Inactive form

The change is endothermic and spontaneous upon heating the enzyme.  Is the structure of the active or the inactive form more ordered?

[Yggdrasil]

The change is endothermic and spontaneous upon heating the enzyme.

What does this tell you about the sign of ΔG and ΔH?  What can you infer about the sign of ΔS?

[TEA1]

G would be negative and H would be postive.  So then S would be postive but only at high temperature.  But I don't understand how this fits with order?

[Yggdrasil]

ΔG = ΔH - TΔS.  The temperature term in front of the ΔS accounts for the fact that the reaction is spontaneous only at high temperature.

ΔS is positive means that when the enzyme transitions from the active state to the inactive state, entropy increases.  How does entropy relate to order?

[TEA1]

High entropy means more chaos (disorder).  So the active form is more ordered because entropy has to increase for it to maintain its ordered structure. 

[Yggdrasil]

Yes, the active form of the enzyme is more ordered. 

Another way to think about it is to think about how many "forms" of the enzyme are likely to be active and how many are likely to be inactive?  Enzymes work because they have a specific three dimensional structure that aligns certain amino acids to aid in the catalysis of a specific chemical reaction.  Enzymes are proteins (i.e. polymers of amino acids) and these polymer can fold into many different configurations.  However, very few of these configurations have the proper alignment of amino acids in order for the enzyme to promote catalysis.  The simple fact that there are many fewer active microstates of the enzyme than inactive microstates means that the active form of the enzyme is more ordered (recall that Boltzmann defines entropy as S = k ln(Ω), where Ω represents the number of microstates corresponding to a certain state).