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

josmit_1D
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

how do you know if a reaction is steady-state or pre-equilibrium?

Jiyoon_Hwang_2I
Posts: 101
Joined: Sat Sep 14, 2019 12:17 am

### Re: steady state v preeq

For steady state, the first step of the reaction must be slower than the second step. On the other hand, for pre-equilibrium, the first step must be faster than the second step.

805097738
Posts: 180
Joined: Wed Sep 18, 2019 12:20 am

### Re: steady state v preeq

Jiyoon_Hwang_2I wrote:For steady state, the first step of the reaction must be slower than the second step. On the other hand, for pre-equilibrium, the first step must be faster than the second step.

how do you know which step is the fast step and which is the slow step

Andrew Liang 1I
Posts: 105
Joined: Fri Aug 30, 2019 12:18 am

### Re: steady state v preeq

I think we can determine the slow or fast step by:
1) Looking at the value of k. The smaller k is probably the slower step.

2) Looking at the activation energy for each step. The step with larger activation energy is the slower step.
Both of these can be determined by the Arrhenius equation:k = A * e^(-Ea/RT)

I'm not too sure though

Jiyoon_Hwang_2I
Posts: 101
Joined: Sat Sep 14, 2019 12:17 am

### Re: steady state v preeq

805097738 wrote:
Jiyoon_Hwang_2I wrote:For steady state, the first step of the reaction must be slower than the second step. On the other hand, for pre-equilibrium, the first step must be faster than the second step.

how do you know which step is the fast step and which is the slow step

Im not exactly sure either but sometimes they will tell you directly