Large Kc for Cubic Equations
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Large Kc for Cubic Equations
Is it possible to have a large Kc in chemical formula that results in cubic equations? Or are we just staying away from that in an effort to keep complexity out of it?

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Re: Large Kc for Cubic Equations
It is possible to have large Kc values with the Kc expression being cubic. The cubic nature of the expression depends on what information is given in the problem. However, I don't think we'll get problems like this on any exams just because as you said it does get more complicated to solve since we can't use the simplification method with large Kc values, only small ones.

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Re: Large Kc for Cubic Equations
It is possible, but Lavelle stated that we don't have to solve cubic equations for this class as they are unsolvable by algebraic manipulation and you must use guesses to obtain a result.

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Re: Large Kc for Cubic Equations
It's possible, but Lavelle said we don't have to really worry about those calculations for this class since they are more complicated.

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Re: Large Kc for Cubic Equations
Hey Jake,
Based on high school chem I think it's definitely possible to have to deal with a cubic equation when you have a large K value, but like was said earlier maybe not in this class since Prof. Lavelle never went over it. Just thought it could be helpful to explain how it works. For this scenerio you'd need to make an assumption based on the fact that there will be very little reactant left, as opposed to the assumption that there will be very little product produced for a small K value. If you want an example of this all you need to do is take an example problem where we looked at a small K and flip the reaction so that you're left with a large K value (remember flipping a reaction gives you K^1, and taking the inverse of a small number gives you a really big number).
From today's lecture Prof. Lavelle gave us the reaction 2N_{2} + O_{2} ⇌ 2N_{2}O, K = 2.0 x 10^{37}. If you were to flip the reaction, you'd get 2N_{2}O ⇌ 2N_{2} + O_{2}, K = 5 x 10^{36}. Now we can look at the large K assumption. Let's say you started out with 0.9M N_{2}O. You'd set up you're ICE table like this:
2N_{2}O ⇌ 2N_{2} + O_{2}
0.9M___0M___0M
2x____+2x___+x
0.92x__2x____x
If you set up the K expression you're going to end up with a cubic equation, but the problem with this ICE table is that x represents some large value. If you remember from the small K approximation you need x to be a small number so you can ignore it when setting up your K expression (you can't just say 0.92x ≈ 0.9 for N_{2}O because x is a pretty large value here). So you want to change your ICE table to assume that basically all the N_{2}O will be converted to product, and that your change values will be the entire concentration of the reactant minus some really small number x. I'm using 0.45x to represent the "baseline" change value and multiplying it by two wherever the stoichiometric coefficients are 2.
2N_{2}O ⇌ 2N_{2} + O_{2}
0.9M_________0M_________0M
2(0.45x)__+2(0.45x)__+(0.45x)
2x__________0.92x______0.45x
This table works because now x represents a really small value. Since you still end up with a cubic expression what you want to do now is to assume that your equilibrium concentrations for N_{2} and O_{2} will be 0.9M and 0.45M, respectively (0.92x ≈ 0.9 for N_{2}, and similiary for O_{2} 0.45x ≈ 0.45). That way you end up with this K expression that can be solved for:
So basically what you're doing with the large K assumption is saying that the reaction will basically go forward to completion, and you change your ICE table method such that your x value will represent the minuscule value that you'll eliminate in the K expression. I know that was a lot but hopefully it helps with your understanding of equilibrium in general too :)
Based on high school chem I think it's definitely possible to have to deal with a cubic equation when you have a large K value, but like was said earlier maybe not in this class since Prof. Lavelle never went over it. Just thought it could be helpful to explain how it works. For this scenerio you'd need to make an assumption based on the fact that there will be very little reactant left, as opposed to the assumption that there will be very little product produced for a small K value. If you want an example of this all you need to do is take an example problem where we looked at a small K and flip the reaction so that you're left with a large K value (remember flipping a reaction gives you K^1, and taking the inverse of a small number gives you a really big number).
From today's lecture Prof. Lavelle gave us the reaction 2N_{2} + O_{2} ⇌ 2N_{2}O, K = 2.0 x 10^{37}. If you were to flip the reaction, you'd get 2N_{2}O ⇌ 2N_{2} + O_{2}, K = 5 x 10^{36}. Now we can look at the large K assumption. Let's say you started out with 0.9M N_{2}O. You'd set up you're ICE table like this:
2N_{2}O ⇌ 2N_{2} + O_{2}
0.9M___0M___0M
2x____+2x___+x
0.92x__2x____x
If you set up the K expression you're going to end up with a cubic equation, but the problem with this ICE table is that x represents some large value. If you remember from the small K approximation you need x to be a small number so you can ignore it when setting up your K expression (you can't just say 0.92x ≈ 0.9 for N_{2}O because x is a pretty large value here). So you want to change your ICE table to assume that basically all the N_{2}O will be converted to product, and that your change values will be the entire concentration of the reactant minus some really small number x. I'm using 0.45x to represent the "baseline" change value and multiplying it by two wherever the stoichiometric coefficients are 2.
2N_{2}O ⇌ 2N_{2} + O_{2}
0.9M_________0M_________0M
2(0.45x)__+2(0.45x)__+(0.45x)
2x__________0.92x______0.45x
This table works because now x represents a really small value. Since you still end up with a cubic expression what you want to do now is to assume that your equilibrium concentrations for N_{2} and O_{2} will be 0.9M and 0.45M, respectively (0.92x ≈ 0.9 for N_{2}, and similiary for O_{2} 0.45x ≈ 0.45). That way you end up with this K expression that can be solved for:
So basically what you're doing with the large K assumption is saying that the reaction will basically go forward to completion, and you change your ICE table method such that your x value will represent the minuscule value that you'll eliminate in the K expression. I know that was a lot but hopefully it helps with your understanding of equilibrium in general too :)

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 Joined: Wed Sep 30, 2020 10:05 pm
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Re: Large Kc for Cubic Equations
Having a large Kc is always possible, even with cubic equations. We will not have those on the exams, since we won't be able to use the approximation concept Dr. Lavelle addressed in class today. I think it is more important to focus on a very small K, since Dr. Lavelle mentioned that we are able to use the approximation calculation for cubic and quadratice equations if we have a very small K. I don't think he would give us a large K with a cubic equation since that wasn't the main goal of his reasoning for showing us that cubic problem. He wanted us to learn that approximations can be applied to what we have already learned, which mainly consists of quadratic problems.

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 Joined: Wed Nov 18, 2020 12:26 am
Re: Large Kc for Cubic Equations
Hi jake,
Dr. Lavelle stated in his lecture that while it is quite possible to have a large K value when doing these types of calculations, we will not have to do it in 14B. he said that he won't require us to simplify cubics, so this means we will be expected to recognize a small K value and use the approximation concept to eliminate the possibility of a messy cubic equation.
pierce
Dr. Lavelle stated in his lecture that while it is quite possible to have a large K value when doing these types of calculations, we will not have to do it in 14B. he said that he won't require us to simplify cubics, so this means we will be expected to recognize a small K value and use the approximation concept to eliminate the possibility of a messy cubic equation.
pierce
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