## Initial Q

Samantha Chang 2K
Posts: 69
Joined: Fri Sep 28, 2018 12:17 am

### Initial Q

I was watching a YouTube video, and I got a bit confused on a topic. If you are given initial concentrations and the Kc value, and the Qinitial>Kc then do you switch the x signs for change in molarity when doing the ICE table to find the final equilibrium concentrations? If so, then why?

Elle_Mendelson_2K
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

### Re: Initial Q

I don't believe you switch the x values. Instead, if q>k then the reaction sits to the left. All that means is that the reaction favors the reactants.
I believe you get a negative value only for the c potion of the ice table usually for the reactants because that is what you have first. Then in the E section you would have the initial concentration + the change but because x is negative it would be initial concentration - change. The initial concentration of products are usually 0 so you will add a positive x

Samantha Chang 2K
Posts: 69
Joined: Fri Sep 28, 2018 12:17 am

### Re: Initial Q

I am terrible at explaining, but here is the video I watched: https://www.youtube.com/watch?v=54n1XppP-lA and I got confused with the example at 25:46 when the initial concentrations of the products were not 0.

Jeannine 1I
Posts: 73
Joined: Fri Sep 28, 2018 12:27 am

### Re: Initial Q

Samantha Chang 2K wrote:I am terrible at explaining, but here is the video I watched: https://www.youtube.com/watch?v=54n1XppP-lA and I got confused with the example at 25:46 when the initial concentrations of the products were not 0.

The reason why the initial concentrations of the products are not 0, is because they give you the initial concentrations of both the reactants and the products. All this means is that the reaction started off with product already existing, and when making an ICE table, the only difference will be that the products already exists, but the rest is all the same!(:

In other words, the products are only 0 when the question only gives you the initial concentration of the reactant. It all depends on the information given in the question.

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