Example in class

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FrankieClarke2C
Posts: 38
Joined: Thu Sep 27, 2018 11:28 pm

Example in class

Postby FrankieClarke2C » Fri Jan 18, 2019 5:02 pm

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 1K
Posts: 35
Joined: Thu Sep 27, 2018 11:16 pm

Re: Example in class

Postby Nikki Bych 1K » Fri Jan 18, 2019 6:27 pm

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: 51
Joined: Thu Sep 27, 2018 11:18 pm

Re: Example in class

Postby Xinyi Zeng 4C » Fri Jan 18, 2019 8:06 pm

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: 51
Joined: Thu Sep 27, 2018 11:27 pm

Re: Example in class

Postby Shubham Rai 2C » Fri Jan 18, 2019 10:39 pm

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: 77
Joined: Thu Jun 07, 2018 2:00 am

Re: Example in class

Postby Rami_Z_AbuQubo_2K » Sat Jan 19, 2019 12:39 am

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: 42
Joined: Thu Sep 27, 2018 11:16 pm

Re: Example in class

Postby LeannaPhan14BDis1D » Sat Jan 19, 2019 6:46 pm

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


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