Given:
N2(g)+O2(g) -> 2NO(g)
at equilibrium
[N]=0.3
[O]=0.3
[NO]=0.5
more NO is added bringing [NO] to 0.8
What is the final concentration of NO after the reaction reestablishes equilibrium?
When making the ICE table, I put 0.8 as the initial [NO]. After that I'm not sure on how the concentration changes. Would [NO] decrease and [N] and [O] increase? How would i go about solving for x since there is no given Kc?
Sapling HW1 Question 9
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 50
- Joined: Wed Nov 25, 2020 12:21 am
Re: Sapling HW1 Question 9
Hi, you would first have to calculate the Kc value using the given values of concentrations.
[N]=0.3
[O]=0.3
[NO]=0.5
Then you would use the equilibrium concentration again, this time using the concentrations after you add the extra NO. The Kc will be the same as what you calculated using the concentrations before you add the extra NO. The concentration of NO would decrease, and the concentrations of reactants will increase.
[N]=0.3
[O]=0.3
[NO]=0.5
Then you would use the equilibrium concentration again, this time using the concentrations after you add the extra NO. The Kc will be the same as what you calculated using the concentrations before you add the extra NO. The concentration of NO would decrease, and the concentrations of reactants will increase.
-
- Posts: 101
- Joined: Wed Sep 30, 2020 9:56 pm
Re: Sapling HW1 Question 9
The concentration of NO will decrease by 2x and the concentrations for both N2 and O2 will increase by x.
-
- Posts: 132
- Joined: Wed Sep 30, 2020 9:55 pm
- Been upvoted: 1 time
Re: Sapling HW1 Question 9
You are on the right track! Putting 0.8 for the initial concentration of [NO] is right because the ICE table is used to refer to the concentration right before a new shift. When you add more products, the chemical reaction will want to adjust to minimize the change according to Le Chatlier's Principle. So, it will shift to the left and increase the [N] and [O] concentrations.
The Kc is not given, but you can solve it. When you change the concentrations in a reaction, the value of Kc does not change because the equilibrium ratio will always stay the same (unless you change the temperature). You solve for this Kc using the original equilibrium concentrations. Then solve for the x value as usual :)
The Kc is not given, but you can solve it. When you change the concentrations in a reaction, the value of Kc does not change because the equilibrium ratio will always stay the same (unless you change the temperature). You solve for this Kc using the original equilibrium concentrations. Then solve for the x value as usual :)
-
- Posts: 104
- Joined: Wed Sep 30, 2020 9:59 pm
Re: Sapling HW1 Question 9
Hi, you can first calculate the Kc value base on the given concentration because it is already at the equilibrium. When you are setting up the ice table for every 2 mols of NO used, you gain 1 mol of N2 and 1 mol of O2 so it will be: [NO]=0.8-2x; [N2]=0.3+x; [O2]=0.3+x.
-
- Posts: 102
- Joined: Wed Sep 30, 2020 10:03 pm
Re: Sapling HW1 Question 9
You would solve for Kc using the original equilibrium values N2=O2=0.3M and NO=0.5M.
And yes 0.8 would decrease by 2x. So now you have Kc= (0.8-2x)^2/(0.3+x)^2. then you take the square root of both sides and solve for x. Then your final value of NO is 0.8-2x.
Hope that made sense!
And yes 0.8 would decrease by 2x. So now you have Kc= (0.8-2x)^2/(0.3+x)^2. then you take the square root of both sides and solve for x. Then your final value of NO is 0.8-2x.
Hope that made sense!
-
- Posts: 104
- Joined: Wed Sep 30, 2020 10:10 pm
- Been upvoted: 1 time
Return to “Equilibrium Constants & Calculating Concentrations”
Who is online
Users browsing this forum: No registered users and 14 guests