Textbook Problem 5.33 Parts A and D

[Eliana Carney 3E]

Hey guys! I'm a little confused on parts A and D of the textbook problem 5.33.

-For part A, I don't understand how you can tell the reaction is endothermic. Is this inferred because the reaction takes place at such a high temperature, or is there a concrete way to tell that the given reaction is endothermic?

-For part D, I'm a little confused as to why adding a catalyst doesn't favor the formation of either X or X2. I know that catalysts help drive reactions forward, but does it not apply to equilibrium reactions because the reaction is proceeding in both directions simultaneously? Or is there another explanation for why adding a catalyst wouldn't affect the concentrations?

Thanks in advance!

[Stuti Pradhan 2J]

For part A, you know the reaction is endothermic because breaking a bond requires energy and in this situation, the X-X bond is being broken to form 2X.

For part D, adding a catalyst will speed up the rate of the reaction, but it will not cause the reaction to change if it was not already going to undergo that reaction. Therefore, adding a catalyst does not produce the change.

Hope this helps!

[Malakai Espinosa 3E]

Just like Stuti said, we know it's endothermic because energy/heat is required in order to break bonds. Also for D, the catalyst does not shift the equilibrium constant in any way. The catalyst's function in kinetics vs thermodynamics is not really too related!