Sapling #10, WK 1

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Melody Haratian 2J
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Sapling #10, WK 1

Postby Melody Haratian 2J » Sun Jan 17, 2021 10:15 am

Hi guys,
I’m having trouble on the final part of #10 in the sapling homework. Can someone guide me through it?
The problem is:

The reaction N2O4 <—> 2NO2 is allowed to reach equilibrium in a chloroform solution at 25C. The equilibrium concentrations are 0.373 mol/L N2O4 and 2.04 mol/L NO2. Calculate the equilibrium concentrations of N2O4 and NO2 after the extra 1.00 mol NO2 is added to the 1.00 L solution.

Bella Bursulaya 3G
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Re: Sapling #10, WK 1

Postby Bella Bursulaya 3G » Sun Jan 17, 2021 10:25 am

I believe you now have the equilibrium concentrations, but then need to add 1 to NO2, so now your equilibrium concentrations become your initial, and that NO2 + 1 mole is also your initial. Then set up an ICE box and solve. However, also remember that you are adding more product, so the reaction will shift left. Therefore, you must subtract 2x from your product ICE box and add x to your reactant.

Megan Lu 3D
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Re: Sapling #10, WK 1

Postby Megan Lu 3D » Sun Jan 17, 2021 11:15 am

Hi! For the final part of this question, we can set up an ICE box.

(I) The initial concentration of N2O4 is unchanged, so we can leave it at 0.373 mol/L; the initial concentration of NO2 needs to account for the additional 1.00 mol added, so that would be 2.04 + 1.00 mol/L.

(C) Because the product NO2 has been added, the reaction will shift left as it moves towards equilibrium. Thus, we can say that the reactant N2O4 is changes +x and the product NO2 changes -2x, as there are 2 mols of NO2 per 1 mol of N2O4 in the reaction equation.

(E) Putting it all together, we get [N2O4] = 0.373 + x and [NO2] = 2.04 + 1.00 - 2x


After creating this ICE box, we can set the expression [NO2]^2 / [N2O4] equal to the equilibrium constant value from the beginning of the question, and then solve for x. After solving for x, plug the value in to get the final equilibrium concentrations. Hope this helps!

Melissa Solis 1H
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Re: Sapling #10, WK 1

Postby Melissa Solis 1H » Sun Jan 17, 2021 6:58 pm

Hey so for the final part you'll have to set up the ICE box again:

1) N2O4 is still 0.373mol/L, while the NO2 needs to be added by 1.00 mol/L : 2.04+1.00=3.04mol/L
2) Since we added more product for NO2, the reaction now shifts to the left. So in regards to the ICE box, this means that N2O4, in the changes row, will be +x and for the product, it will be -2x since it has to match the coefficient of 2.
3) Lastly, for the equation, you will need the Kc you got in the first step to equal your new equation: Kc= [NO2]^2 / [N2O4] Since this only has one product and one reactant, it can be a little tricky to use the square root. So you'll have to use the quadratic formula to find your x value.
5) After using the quadratic formula, you can plug in the x in your equation and solve for both N2O4 and NO2.

Hope this was clear!

Rachel Kho Disc 2G
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Re: Sapling #10, WK 1

Postby Rachel Kho Disc 2G » Sun Jan 17, 2021 8:35 pm

I'm also confused about the last part of this question, and I'm especially having a lot of trouble solving for x using the quadratic equation. My numbers are a bit different, so at equilibrium [N2O4] = 0.32 + x, and [NO2] = 2.89 - 2x. My K value is 11.16.

So what I have so far:
K = [2.89-2x]^2/0.32 + x = 11.16.

Then I expanded the numerator, cross multiplied, and then tried to put it in the quadratic formula, but both of my x values are positive so I'm not sure what to eliminate. When I try to plug in either, one of my final concentrations is negative. So I'm not entirely sure what I'm doing wrong, but if someone could please help walk me through it that would be great!

Charlotte Chen 3B
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Re: Sapling #10, WK 1

Postby Charlotte Chen 3B » Sun Jan 17, 2021 9:26 pm

Hey! I actually had a similar question to you, for me my x values were 0.235 and 5.78, I subbed in x=0.235 and got the right answer but I'm not sure how to know which value to eliminate. I guessed x=0.235 because it's a smaller value, but if anyone knows how in general to determine the right value I would really appreciate it!

Charlotte Chen 3B
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Re: Sapling #10, WK 1

Postby Charlotte Chen 3B » Sun Jan 17, 2021 9:35 pm

Rachel Kho Disc 2G wrote:I'm also confused about the last part of this question, and I'm especially having a lot of trouble solving for x using the quadratic equation. My numbers are a bit different, so at equilibrium [N2O4] = 0.32 + x, and [NO2] = 2.89 - 2x. My K value is 11.16.

So what I have so far:
K = [2.89-2x]^2/0.32 + x = 11.16.

Then I expanded the numerator, cross multiplied, and then tried to put it in the quadratic formula, but both of my x values are positive so I'm not sure what to eliminate. When I try to plug in either, one of my final concentrations is negative. So I'm not entirely sure what I'm doing wrong, but if someone could please help walk me through it that would be great!


For your question specifically though I've never had both my x values be negative, maybe there was an error when you calculated the K value/the expressions for the concentrations in the ICE table? If you could share the initial question I could try work through it and let you know if I got the same answer!

Kaiya_PT_1H
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Re: Sapling #10, WK 1

Postby Kaiya_PT_1H » Sun Jan 17, 2021 9:37 pm

Charlotte Chen 3B wrote:Hey! I actually had a similar question to you, for me my x values were 0.235 and 5.78, I subbed in x=0.235 and got the right answer but I'm not sure how to know which value to eliminate. I guessed x=0.235 because it's a smaller value, but if anyone knows how in general to determine the right value I would really appreciate it!


Is there a rule for this? maybe just that the concentrations still all have to end up being positive numbers?

Rachel Kho Disc 2G
Posts: 79
Joined: Wed Sep 30, 2020 9:46 pm

Re: Sapling #10, WK 1

Postby Rachel Kho Disc 2G » Sun Jan 17, 2021 9:53 pm

Charlotte Chen 3B wrote:
Rachel Kho Disc 2G wrote:I'm also confused about the last part of this question, and I'm especially having a lot of trouble solving for x using the quadratic equation. My numbers are a bit different, so at equilibrium [N2O4] = 0.32 + x, and [NO2] = 2.89 - 2x. My K value is 11.16.

So what I have so far:
K = [2.89-2x]^2/0.32 + x = 11.16.

Then I expanded the numerator, cross multiplied, and then tried to put it in the quadratic formula, but both of my x values are positive so I'm not sure what to eliminate. When I try to plug in either, one of my final concentrations is negative. So I'm not entirely sure what I'm doing wrong, but if someone could please help walk me through it that would be great!


For your question specifically though I've never had both my x values be negative, maybe there was an error when you calculated the K value/the expressions for the concentrations in the ICE table? If you could share the initial question I could try work through it and let you know if I got the same answer!



Ahhh thank you so much! I finally figured out x, but I'm pretty sure I did it wrong so I'm not sure what the proper way to use the quadratic is.

The initial problem was that the equilibrium concentrations were [N2O4] = 0.320 mol/L and [NO2] = 1.89 mol/L.


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