ICE Chart Question

[Hasmik Dis 2F]

Does anyone know why when creating ICE charts, and we have a decrease of "-2x" we still have to include the "2" as an exponent when writing out the constant equation?
For example, if my product is -2x, and whenever it's time to calculate either K or for X, we would then do products over reactants. The products in this case would be (2x)^2. However, isn't this double-counting the coefficient of 2? We already incorporate it when saying -2x, so why do we include it again in the exponent?

[Laura 3l]

You would still have to include the coefficient 2 as an exponent so it would be [-2x]^2 in your equilibrium equation.
I know it seems like it is taking it into account two times but I'm honestly not sure exactly why...

[Madison Wong 3H]

You still include the coefficient in calculation for the equilibrium constant. I think this is because the ()^2 accounts for the coefficient in the equilibrium constant calculation while the -2x matters in relation to the molar ratios of other reactants and products.

[sophie esherick 3H]

You would still have to include the coefficient 2 as an exponent so it would look like [-2x]^2 in your K equation. I'm not entirely sure why we include both but I'm assuming it is because the -2 in -2x matters since it relates to molar ratios of the reactants and other products (whatever x is, since there are two moles of it in the balanced equation you would multiply that x concentration by 2). The ^2 I think is used to acknowledge the coefficient within the equilibrium equation but again I'm not 100% sure.

[Aditya Sundaram 3D]

Does anyone know why when creating ICE charts, and we have a decrease of "-2x" we still have to include the "2" as an exponent when writing out the constant equation?
For example, if my product is -2x, and whenever it's time to calculate either K or for X, we would then do products over reactants. The products in this case would be (2x)^2. However, isn't this double-counting the coefficient of 2? We already incorporate it when saying -2x, so why do we include it again in the exponent?

The 2 is essential to finding the relative proportions and calculating the molar concentration differences from initial to equilibrium constants.

[Nick Saeedi 1I]

Yes it is also included in the exponent.

[Jason Knight - 1F]

When creating ICE charts, even when we decrease by [-2x] we still have to include the ^2 in the equilibrium concentration. For example, in this situation our equilibrium expression for a acid/base with a molar coefficient of 2 will look like [2x]^2.

[Mursall M 2A]

You would still need to include the exponent because the 2x on the inside is only representing the concentration. So when you calculate K, you just treat the 2x as a concentration, and plug it into your equation.

[Becca Nelson 3F]

Lets say your equation is O2 +2H2 -> 2H2O and you start with only 1 mol hydrogen gas and oxygen.

For every two water molecule made you will loose 2 hydrogens and 1 oxygen, so your ICE table would have -x for change for oxygen, -2x for hydrogen and +2x for water. In your equilibrium it would be 1-x and 1-2x. You still square the hydrogen and water because they have the coefficients of 2.

your final K would be (2x)^2/([1-x][1-2x]^2). Hope this helps!

[Kimiya Aframian IB]

Does anyone know why when creating ICE charts, and we have a decrease of "-2x" we still have to include the "2" as an exponent when writing out the constant equation?
For example, if my product is -2x, and whenever it's time to calculate either K or for X, we would then do products over reactants. The products in this case would be (2x)^2. However, isn't this double-counting the coefficient of 2? We already incorporate it when saying -2x, so why do we include it again in the exponent?

Hi! The reason that we need to do -2x or +2x is because there are two moles so the change must be at this ratio as well. Then, we need to square it because when calculating our K value, the concentration (or partial pressure) but be raised to the power of its coefficient. Hope this helps!

[Andy Hon 3E]

You would leave it in because the -2x gives you the equilibrium concentration. You use that equation to plug it into the equilibrium constant equation. You would still have to square it or raise it to its stoichiometric coefficient.

[Claudia_Danysh_2B]

Yes, we still include 2 in the K equation!!

[Liam Bertrand 3]

You would still take into account the exponents.

[Chinmayi Mutyala 3H]

2x is considered to be the concentration of that compound. When you calculate K, you do each concentration to the power of it's stoichiometric coefficient so it would be (2x)^n. The exponent would be on the 2 as well.

[Michael Cardenas 3B]

You would still use the exponents because they are a part of the equation for Q, which is separate from the ice chart.

[Charmaine Ng 2D]

Yep, you would still take them into consideration for the exponents! :))

[Jacob Schwarz-Discussion 3I]

Yes you still do take it into consideration for the exponents. This is because the -2x accounts for the molar concentrations whereas the ^2 exponent is related to the equilibrium concentration.

[Lorena_Morales_1K]

Yes you still do take it into consideration for the exponents.

[Sara Sandri 2B]

Coefficients matter for both what you subtract (-2X) and what power your raise the concentration 2. I was confused on this as well!

[tholz11]

You have to account for it "twice," once by writing in 2x and then a second time by squaring the concentration in the K expression.

[Constance Newell]

You have to do both, I can't explain why but I guess the concentration has some sort of exponential effect on the constant

[Britney Tran IJ]

yeah you have to consider the coefficients for the exponents

[Diana Aguilar 3H]

Hi! I believe you still have to take into consideration the coefficients for the exponents when writing the constant equation.

[Naomi Hernandez-Ramirez 1J]

even when you decrease by -2x you still have to include the ^2 in the equilibrium concentration

[Audrey Banzali-Marks 1A]

You have to include the 2 in 2x because it's fully counting the concentration of the associated reactant or product. For example, if you have Cl2 <-> 2 Cl, you have to include K = [2x]^2/[Cl2-x] because if you wrote K=[x]^2/[Cl2-x], it would only count half of the product's concentration. The formula for K is K=[product concentration]^coefficient/[reactant concentration]^coefficient, so you won't be double counting if you include the 2 and square it because the coefficient-power part is unrelated to the concentration. I hope that helps!

[Naomi Christian 1E]

Yes, you would still have to include the 2. It is because this coefficient relates to the change in molar concentration, and is not related to the stoichiometric coefficient. If you don't you will get the wrong answer.

[Michelle Argueta 1E]

Hello! I believe that you still have to use the "2" in consideration of the exponent as it is still the coefficient of one of the reactants and/or products and therefore important when solving for the equilibrium constant. The squaring of the concentration when solving for the equilibrium constant is important to include in order to get the accurate K.