- #1
mikeyman2010
- 18
- 0
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.
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.