Lecture 4 Question: Cubic equation

[MHarrold_1E]

Is it possible to get a cubic equation when the K>10^-3, if so, what do we do from there? Also, if K<10^-3 and we do not get a cubic function, do we still omit -x/+x?

[Pranav Kadiyala 1A]

From my understanding, you would not get a cubic with a Kc > 10^-3. Those would have to be solved through either guess and check, or graphing.

If you get a quadratic with kc< 10^-2, I believe you can assume the change is insignificant again (especially if it is egregiously small, like 10^-20). However, if it is something like 10^-4, solving the quadratic all the way through without approximating would get you a closer answer. I believe the Sapling HW contains an example of this (Problem 3, I think).

[Sid Panda 3A]

Is it possible to get a cubic equation when the K>10^-3, if so, what do we do from there? Also, if K<10^-3 and we do not get a cubic function, do we still omit -x/+x?

I think it is possible to get a cubic equation when the K>10^-3. I'm not sure what Dr. Lavelle would recommend, but what I usually do is graph the equilibrium constant and try to see where it hits the zero points (those zero points are the x values).

For example, if you have something like. X^3/( (2-x)^2 * x) = 45. I would move the 45 over to the left side of the equation and solve for the zero's of this equation by graphing it. There is a function on your calculator that helps you calculate zero values.

For your third question, I would say that you should omit +x/-x because you should try to keep those values as much as possible, unless they are a part of a cubic function with a K<10^-3 (which is not what is the case in your question here).

[Samantha Pedersen 2K]

Dr. Lavelle mentioned in lecture today that we won't be expected to solve cubic equations in Chem 14B, so I think that means we won't get any problems in Chem 14B where we have a cubic equation with K > 10^-3. If K < 10^-3 and you have a quadratic function, then you can still make the assumption that allows us to omit +x/-x, but I would still recommend keeping the +x/-x and solving the quadratic because it will give you a more accurate answer. I hope this helps!