Kaplan Equilibrium Question

[mikeyc2040]

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I got this question incorrect on diagnostic.

The equilibrium for the reaction below can be shifted to the right by:

N2(g) + 3H2(G) <--> 2NH3(g) Change in Heat=-92kj/mol

A) decreasing volume
B) decreasing pressure
c)adding a catalyst
d)increasing temp
E)removing nitrogen

removing nitrogen would cause the rxn to move toward nitorgen to the left. catalyst isn't included in this type of problem (only concentration, temp and volume/pressure). which leaves A, B and D. I know that decreasing volume results in an increase of pressure moving toward side with least amount of moles, which is to the right. So, A is correct. But, because this is endothermic don't you count heat as a reactant. The rxn will therefore move away from the presence of heat upon adding heat. Wouldn't this shift the rxn to the left? Help?

 

[TheWiredNerv]

the change in heat is -92 kJ/ mol. The negative number tells you the reaction is exothermic so add heat to the right side of the equation. Increasing temperature would cause a shift left. You are right with choice A but I don't understand how you thought an endothermic reaction causes a shift to the left? If this reaction were endothermic an increase in temp would cause a shift right.

 

[rose786]

Do you have destroyer? There's a good memory aid in there regarding these types of problems. Just draw a triangle wherever the heat is (in this case on the right) and call him "homer." When there is an increase in temp, Homer runs away to the other side and when there is a decrease he freezes and stays on one side.

In this case: Homer is on the right, if temp increases, he runs left. If temp decreases, he freezes and stays on the right.

Another thing is this is an exothermic rxn as pointed out above.

 

[mikeyc2040]

thanks for the homer tip.. I will definitely be using this. it clears up my misunderstandings. thanks 🙂

 

[klutzy1987]

you can eliminate choice B because there are more moles on the left than on the right. Therefore if you increase the volume you will shift the equilibrium to the side that has more moles. You can eliminate D because if there is a negative delta H the raction is exothermic which would put the heat as a product. Use the homer trick in the destroyer. It makes problems like this much much easier. If you dont have the destroyer get it it is a lifesaver on the sciences.

 

[Willem]

Another username dedicated solely to praising the dat destroyer 🙁