Equilibrium and Entropy

Hey Alex, 

I've been having trouble with some general chemistry material and I was using the Q-bank quizzes when I came upon the following question. 


14. Reaction 1 proceeds in a closed flask and reaches equilibrium. After a while, the flask beings to leak NO (g). Once the leak is fixed and the reaction has once again reached equilibrium, the entropy of the system has:

a. increased
b. decreased
c. remained constant
d. decreased by the same amount as delta g

reaction 1: 3 NO(g) --> N2O(g) + NO2(g)

The answer is a. I don't understand why. If the reaction has once again reached equilibrium, then how has the entropy changed? Is it because there was a net increase in entropy to once again reach equilibrium? How should we go about doing this?
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Initially, you may want to attack this question using ΔG = ΔH-TΔS.  ΔG = 0 at equilibrium, by definition.  There is a problem with making an assumption that ΔS will have to stay the same, though -- T will likely not stay the same (if some of the NO is leaking, the pressure inside goes down and the gas will expand to fill the container; an expanding gas will lose temperature, according to the First Law of Thermodynamics (ΔU = Q - W)).


So what can we do?  Well, let's think about LeChatelier's Principle.  When the NO escapes, the system is no longer at equilibrium.  Once we fix the leak, let's think what will happen to the system.  With a lack of NO, the system will shift left.  Remember that entropy is increased in gases (over liquids and solids) and that more moles of gas corresponds to more entropy.  This left shift, therefore, is increasing entropy as the reaction reaches equilibrium.


I hope this helps!