How to make ICE box

[Nick Fiorentino 1E]

I understand the initial and equilibrium rows in the ICE box but how do you figure out the change row? That is, how do you figure out what the change in concentration will be?

[Eileen Si 1G]

By looking at the balanced chemical equation. If the molecule does not have a stoichiometric coefficient, then the change is simply x. If the molecule does has a stoichiometric coefficient, then the change is the coefficient * x. Also, if the molecule is a reactant, then the change is negative, but if the molecule is a product, then the change is positive.

[Asha Agarwal 1E]

If you are given the equilibrium concentration of a product, you can use stoichiometry to find the corresponding amounts of reactant used, and that becomes your change.

[Angus Wu_4G]

Unless you are given the precise change, you would just use x as an placeholder. For reactants, the change will be negative x, and for products the change will be positive x. Be sure to also account for the coefficients in the balanced chemical equation, if there is a coefficient of 2, then your change needs to be correspondingly 2x. After you assign the x values, you can substitute the x into the equilibrium constant equation, and set it equal to a known K value, then solve for x.

[Justin Seok 2A]

Since change is often unknown in problems using the ICE box, we usually attribute a variable to the amount of change in products and reactants, oftentimes x. Using stoichiometric analysis, like in an example equation 2 A + 3 B <--> C, the change would be -2x for A, -3x for B, and +x for C. Then use the variable x in the equilibrium constant and solve for x.

[Kavya Immadisetty 2B]

Allow the change to be x (ie. on the reactants side, the change would be -2x if the stoichiometric coefficient is 2). Add the change to the initial value to get the equilibrium expression. Plug in the expressions into a K equation and if the value of K is given, algebra can be used to solve for x aka the change value.

[KBELTRAMI_1E]

Unless you are given the precise change, you would just use x as an placeholder. For reactants, the change will be negative x, and for products the change will be positive x. Be sure to also account for the coefficients in the balanced chemical equation, if there is a coefficient of 2, then your change needs to be correspondingly 2x. After you assign the x values, you can substitute the x into the equilibrium constant equation, and set it equal to a known K value, then solve for x.

Are there ever instances where more advanced math than algebra is used?

[Angus Wu_4G]

Unless you are given the precise change, you would just use x as an placeholder. For reactants, the change will be negative x, and for products the change will be positive x. Be sure to also account for the coefficients in the balanced chemical equation, if there is a coefficient of 2, then your change needs to be correspondingly 2x. After you assign the x values, you can substitute the x into the equilibrium constant equation, and set it equal to a known K value, then solve for x.

Are there ever instances where more advanced math than algebra is used?

I don't believe so, most that we will need is probably just algebra and the quadratic equation. I think soon we will also learn that for specific types of K values there's an assumption that we can use that will allow us to skip the quadratic equation and just use simple algebra, depending on how big the K value is. As for the rest of the course, I think there could potentially be more advanced math like calculus when we cover other topics but I wouldn't be sure.

[Joanne Lee 1J]

When you look at the balanced equation, the number of moles of each compound will tell you the change. For example if there is two moles of water the change would be 2x. The positive and negative depends on what compound you start out with. If you are given the amount of reactant and water is a product, 2x is added (positive) but if you are given the amount of product and water is a product 2x is subtracted (negative).

[Sally Qiu 2E]

The change would be x if you aren't given the equilibrium concentrations. Unless there are coefficients, then it would be corresponding coefficient times x.

[805097738]

the moles represent the amount of X that will change. If the molecules go into the eq the change is -, and if the molecules come out of the eq the change is +

[Jasmine Fendi 1D]

To add onto the comments above, make sure you go back to the problem to see what you need to solve for. After you find x, you may need to multiply it by 2 if the concentration you are looking for is 2x

[Maya Beal Dis 1D]

Are pure liquids and pure solids always left out of the ICE box equations?

[Uisa_Manumaleuna_3E]

Pure liquids and solids are always left out. And to find the change column, its just -x/+x with the molar ratios considered

[Cooper Baddley 1F]

Always leave out the solids and bases they will not be changed enough to be included.

[Nicholas_Gladkov_2J]

I understand the initial and equilibrium rows in the ICE box but how do you figure out the change row? That is, how do you figure out what the change in concentration will be?

What is in the change row is determined by the stoichiometric coefficients of the balanced chemical equation. If in the initial molar concentration (the I of the ICE box) there is a zero anywhere, then the change of that column (the C of the ICE box) will be +X or +2X or etc. (it's the same as the coefficient).

[KBELTRAMI_1E]

Unless you are given the precise change, you would just use x as an placeholder. For reactants, the change will be negative x, and for products the change will be positive x. Be sure to also account for the coefficients in the balanced chemical equation, if there is a coefficient of 2, then your change needs to be correspondingly 2x. After you assign the x values, you can substitute the x into the equilibrium constant equation, and set it equal to a known K value, then solve for x.

Are there ever instances where more advanced math than algebra is used?

I don't believe so, most that we will need is probably just algebra and the quadratic equation. I think soon we will also learn that for specific types of K values there's an assumption that we can use that will allow us to skip the quadratic equation and just use simple algebra, depending on how big the K value is. As for the rest of the course, I think there could potentially be more advanced math like calculus when we cover other topics but I wouldn't be sure.

Wait what other situations would we use the ICE box for? Other than for calculating the equilibrium constant

[Angus Wu_4G]

You would also use an ice box for weak acid/base calculations. These acids and bases will also have their own K values.