Adding a Catalyst

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Payton Kammerer 2B
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Joined: Tue May 01, 2018 3:00 am

Adding a Catalyst

Postby Payton Kammerer 2B » Mon Mar 09, 2020 11:32 am

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
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Joined: Thu Jul 25, 2019 12:17 am

Re: Adding a Catalyst

Postby Paul Hage 2G » Mon Mar 09, 2020 11:34 am

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
Posts: 101
Joined: Sat Aug 24, 2019 12:15 am

Re: Adding a Catalyst

Postby Vicki Liu 2L » Mon Mar 09, 2020 11:36 am

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
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Joined: Thu Jul 11, 2019 12:16 am

Re: Adding a Catalyst

Postby Hannah Romeo 1J » Mon Mar 09, 2020 12:06 pm

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.

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Re: Adding a Catalyst

Postby KSong_1J » Sat Mar 14, 2020 12:23 pm

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

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Joined: Fri Aug 30, 2019 12:17 am

Re: Adding a Catalyst

Postby ThomasNguyen_Dis1H » Sat Mar 14, 2020 12:30 pm

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.

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Joined: Fri Aug 09, 2019 12:17 am

Re: Adding a Catalyst

Postby preyasikumar_2L » Sat Mar 14, 2020 4:54 pm

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.

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