## Example in class

FrankieClarke2C
Posts: 59
Joined: Fri Sep 28, 2018 12:28 am

### Example in class

When we were talking about that one example that had to do with pressure, why did we use Q instead of Kp to calculate the new pressure?

Nikki Bych 1I
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

### Re: Example in class

I'm not in lecture 2 so I may not be correct, but if he used Q then it was probably because the rxn was not at equilibrium.

Xinyi Zeng 4C
Posts: 63
Joined: Fri Sep 28, 2018 12:18 am

### Re: Example in class

We only use Kp when the reaction system is at equilibrium, other than that, we will always calculate Q and compare it with Kp (or sometimes Kc if the question is about concentrations) to see which direction (products or reactants) the reaction will move towards. Hope this helps :)

Shubham Rai 2C
Posts: 64
Joined: Fri Sep 28, 2018 12:27 am

### Re: Example in class

I agree Q is used when the reaction is not at equilibrium. The Q value then will allow us to predict the direction the reaction will proceed when compared to Kc.

Rami_Z_AbuQubo_2K
Posts: 89
Joined: Thu Jun 07, 2018 3:00 am

### Re: Example in class

Q looks at the system when it is not at equilibrium(when it ratio of products to reactants is equal to Kc). Q is compared to Kc to see whether the reaction proceeds to the right or the left. If Q<Kc then the reaction will "shift" to the right. If Q>K then the reaction will "shift" to the left. If Q=Kc, then the system is at equilibrium.

LeannaPhan14BDis1D
Posts: 57
Joined: Fri Sep 28, 2018 12:16 am

### Re: Example in class

for the example he gave, he used Q since the equation wasn't at equilibrium

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