Adding a Catalyst

[Payton Kammerer 2B]

How does adding a catalyst affect the equilibrium concentrations? A catalyst speeds up a reaction's rate, so I thought that a catalyst would favor the products, but 5.33 says that it doesn't have an effect on the eq concentrations.

5.33 Dissociation of a diatomic molecule, X2(g) -> 2 X(g) occurs at 500 K. The equilibrium state of the reaction is shown in 1 and the equilibrium state in the same container after a change has occurred is shown in 2. Which of the following changes will produce the composition shown? (a) Increasing the tempera- ture. (b) Adding X atoms. (c) Decreasing the volume. (d) Adding a catalyst. Explain your selections.
[A shows more diatomic molecules, B shows a shift to products with only two diatomics left]

The correct answer is a, and I understand why a is correct. I just don't understand why D isn't correct as well.

[Paul Hage 2G]

The catalyst wouldn't affect the equilibrium of a reaction because a catalyst speeds up both the rates of the forward and the reverse reaction. The equilibrium would only change if either the forward or the reverse reactions were sped up/favored.

[Vicki Liu 2L]

It is true that adding a catalyst speeds up a reaction, but it does not change the equilibrium concentrations because K remains a constant (unless temperature changes). Kinetics focuses only on the rate at which equilibrium is reached; the equilibrium condition itself is left unchanged.

[Hannah Romeo 1J]

A catalyst only lowers the activation energy therefore speeding up the rate of the reaction. However, catalysts are not consumed during the reaction as they are added and then regenerated. As a result, they are not written in the equilibrium expression.

[KSong_1J]

Catalysts do not affect the equilibrium constant. They create a different pathway and, since the equations for K is k over k prime, there shouldn't be a change in K

[ThomasNguyen_Dis1H]

They don't affect the equilibrium constant, it only affects the forward and reverse rates by increasing them. At equilibrium these rates are also equal to each other.

[preyasikumar_2L]

A catalyst only serves to speed up the rates of the forward/reverse reactions, but those would still be equal, and since catalysts don't change any equilibrium concentration and isn't a reactant since it can be regenerated and reused, the equilibrium constant K would not be affected. Also, K is the equilibrium constant, so it will remain the same (at the same temperature) whether there is a catalyst or not.