Cubic approximation
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Cubic approximation
On the first example from lecture 4, the value of x is approximated to be very small, in order to make the cubic equation solvable. Why do we approximate this, does this have to do with the fact that very few products are being formed? Or is there some relationship between K_{C} and x? Also, does this approximation apply when K_{C} is very large ( >10^{4} )?

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Re: Cubic approximation
We approximate this because the K value is very small, therefore you're correct, it would signify that there are very little products being formed . Since the numerator is equal to (2x)^2, the x value must be very small. We would pose the x as 0 when calculating the concentrations of the reactants since you're subtracting such a small number from the initial concentration. The approximation as Dr. Lavelle said, was only to be applied if the K value is less than 10^3, and we can't apply it to larger numbers. I think he also said he wouldn't give us a problem where we would actually have to solve the cubic approx too.

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Re: Cubic approximation
Yes, because the K_{c} value is so small, we can approximate that the change taking place is also so small that it is negligible. However, if K_{c} is very large we would not be able to make a similar approximation. In fact, the change would be predicted to be very large and would have a considerable effect that we cannot ignore.

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Re: Cubic approximation
We approximate when Kc is really small because that means that x must also be really small, so subtracting/adding it to a larger number is insignificant. This does not apply when Kc is large because then it would be significant. Also, Professor Lavelle mentioned that we don't have to solve cubic equations in this class so that should be a good hint or reminder to approximate when you get a cubic equation.

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Re: Cubic approximation
I would say that you are correct from my understanding. The value is so small that it does not really affect the end product within our means of accuracy. Adding it in would just unnecessarily complicate the equation. However, for larger numbers I would say that you should not do this as larger numbers have a greater affect on the end product. I would not use it if it significaly affected the sig figs that I was using. Hope this helps!

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 Joined: Wed Sep 30, 2020 9:38 pm
Re: Cubic approximation
Yes, when the K value is very small it means that there are very few products being formed and so the change is negligible. However, if K was very large it would not apply because many products would be formed so the change would definitely be something we want to notice. I think this ties in very well to Dr. Lavelle's example with the million dollars. If K was very small and someone with a million dollars gave away ten dollars they would not see that large of a difference and we could say they still have approximately a million dollars. However, if K was very large and someone with a million dollars gave away a hundred thousand dollars they would notice a difference and we would no longer be able to approximate that they have a million dollars.
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