Forward and reverse rates after pressure increase in 2NO₂ ⇌ N₂O₄ equilibrium

[mikeyman2010]

Hi, I know this is a physics/math forum, but I'd really like some help on this equilibrium question.

Consider this reaction:

2NO2 <---> N2O2

If I increase the pressure, then the system will shift to the right, and thus the forward reaction rate will increase. After awhile, a new state of equilibrium is reached, in which the forward and reaction rate will become equal again. However, the question asks me to compare the forward/reverse reaction rates of the new equilibrium system vs the initial equilibrium. Are they supposed to be increased, decreased, or stayed the same?

I have a shrewd idea, but not quite sure. Since the pressure is increased, then the concentration is increased for BOTH the reactant, and the product. then, both the forward and reverse reactions should be faster than the previous equilibrium state in the new one. Can anyone clarify or elaborate on this? Any help or idea is greatly appreciated.

 

[GCT]

First off, you can post in the chemistry subforum here at PF

Both the forward and reverse reaction rates increase, you'll just need to figure out the incremental increase of each rate relative to their initial

 

[thepatientmental]

Hi there! No problem, I'm happy to help with this chemistry question.

You are correct in your thinking that increasing the pressure will shift the equilibrium to the right, since there are more moles of gas on the right side of the equation. This means that the forward reaction will occur at a faster rate, as there are more molecules available to react.

When a new equilibrium is reached, both the forward and reverse reaction rates will have increased compared to the initial equilibrium. This is because, as you mentioned, the concentrations of both the reactant and product have increased. This means that there are more molecules colliding and reacting, leading to an increase in both rates.

It's important to note that while both rates have increased, they will still be equal at the new equilibrium. This is because the system has adjusted to the new conditions and reached a new balance between the forward and reverse reactions.

I hope this helps clarify things for you. Good luck with your studies!