## Slow Reaction

$K = \frac{k_{forward}}{k_{reverse}}$

Maggie Elgin 2A
Posts: 19
Joined: Wed Sep 21, 2016 2:57 pm

### Slow Reaction

Im still having a hard time determining which step is the slow reaction which I understand is the reaction rate of the overall reaction. Could someone please explain how determine which is the slow and which is the fast reaction?

edward_qiao_3I
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### Re: Slow Reaction

Usually, if I'm not mistaken, this will be given to you

samuelkharpatin2b
Posts: 55
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Slow Reaction

When given a mechanism, it should provide you with which step is the slow step and which is the fast step. The slow step, sometimes called the rate limiting step, is what determines the overall rate law.

Michael Lesgart 1H
Posts: 26
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Slow Reaction

I believe that the fast step is at equilibrium and the slow step is not at equilibrium. Also do you know if you use the pre-equilibrium approach if the fast step comes before the slow step or after the slow step?

Cobie_Allen_1H
Posts: 24
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Slow Reaction

I believe you use the pre-equilibrium approach only when the fast step is first!

Mana_Sheykhsoltan_1A
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### Re: Slow Reaction

I think Professor Lavelle mentioned in Friday's lecture how the question will explicitly state which step is slow and which is fast or k, the rate constant, for each step will be given to you. By comparing the rate constants of the two steps, you can determine which step is faster and which is slower. The slowest step is the rate determining step, and thus it determines the speed at which the overall reaction proceeds. When coming up with your rate equation, you only have to look at the slowest step's reactants. Hope this helps!

Wenqian_Deng_1L
Posts: 23
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Slow Reaction

When given a two chemical reactions, and we know that one is slow and the other is fast, the problem should either give us clues to determining which is which. For example, the problem could say that the equilibrium constant (k1) of one reaction is slower than the equilibrium constant (k2) of the other reaction. If this was the case, then k1 should be smaller k2.

Zulfiqar Lokhandwala 1H
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Joined: Fri Sep 25, 2015 3:00 am

### Re: Slow Reaction

When it comes to the slow step reaction is the reaction that is reversible also the slow step reaction? So if it states that the reaction is reversible k and k(prime) would be included in the slow step reaction as well?

Xiaoman_Kang_2J
Posts: 19
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Slow Reaction

Zulfiqar Lokhandwala 1H wrote:When it comes to the slow step reaction is the reaction that is reversible also the slow step reaction? So if it states that the reaction is reversible k and k(prime) would be included in the slow step reaction as well?

If the forward the reaction is slow, then its reverse reaction should be fast.

Michael Lesgart 1H
Posts: 26
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Slow Reaction

When the question asks you to find the rate equation, what do you do when your equation requires you to know the concentration of an intermediate species? Are you supposed to replace it by looking at the fast reaction and setting the forward rate equal to the reverse rate? Or is there another method of replacing the intermediate in the equation?

Helen_Onuffer_1A
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Joined: Fri Jul 22, 2016 3:00 am

### Re: Slow Reaction

Michael Lesgart 1H wrote:When the question asks you to find the rate equation, what do you do when your equation requires you to know the concentration of an intermediate species? Are you supposed to replace it by looking at the fast reaction and setting the forward rate equal to the reverse rate? Or is there another method of replacing the intermediate in the equation?

My TA told us that intermediates are not included in rate equations, and this is so because the rate law is determined by experiment and intermediates aren't really observable. If the first step is fast and the second step is slow, you go back to the first (fast) step which involves an equilibrium. At equilibrium, the forward reaction rate is equal to the reverse reaction rate (k[R] = k'[P]) and you solve for the concentration of the intermediate (it'll be an expression). You then substitute in this expression for the concentration of the intermediate in your slow step rate law. Hope this helps!

Stephanie_Martinez3E
Posts: 9
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Slow Reaction

It should usually be given, because that rate of a reaction is usually experimentally found.