HW 1 #2

[mya warner 2C]

At a certain temperature, 0.940 mol SO3 is placed in a 3.00 L container.
2SO3(g)<-->2SO2(g)+O2(g)
At equilibrium, 0.180 mol O2 is present. Calculate Kc.
I'm having some problems with what to do with this question after completing the ICE box. If someone could explain the process to me that would be great!

[Clayton Dinh 3C]

After your ICE box, you would just find the molarity of each molecule and plug it into the Kc equation!

[Aliza Hacking 1A]

Hi! so for this one, I had slightly different numbers, but I think for the ICE box the initial SO3 concentration would be .94 mol/3 liters, and the other 2 initial concentrations would be zero. Then for the change, you know from the givens that the concentration of O2 went up by .180 mol/3 liters (to get moles/liter concentration), and based on the ratios of the chemical equation, the SO2 went up by double that concentration, and SO3 went up by 3/2 of that amount of O2. After completing that ice box, you would just plug those values into the Kc equation (making sure to raise the concentrations to their appropriate powers based on their coefficients), and you should get the right value! Make sure products are on top and the reactant is on the bottom :)

[LaurenAkason2A]

Hi! After you find the molar concentrations, you just need to plug the values into your P/R equation to find the Kc. Don't forget to raise the numbers to the right powers or else you'll get the wrong answer! You also need to make sure you converted your answer to molarity and are not giving it in moles. Hope this helps!

[Phoebe Ko 3E]

After you set up the ICE box, you would plug the concentrations of each species into the equation Kc=[O2]*[SO2]^2/[SO3]^2. This question is tricky because the question gives you moles instead of mol/L, so make sure that the concentrations in the ICE box are all mol/L, for example 0.940 mol/3 L SO3, instead of just the given moles in the problem. Also, make sure each species is raised to the correct exponent (should correspond to their respective coefficient in the reaction) in the Kc equation! I hope this was helpful!

[Samantha Melin 2E]

After calculating your ICE box, it should become clear that your x value is 0.150 mol. From there, plug your x value into the your equilibrium row in your ICE box to find the number of moles of each substance at equilibrium and plug these values into your Kc equation. Remember to change moles to concentrations and use the proper powers based on stoichiometric coeffecients. Hope this helps!

[Madelyn_Rios_2c]

After completing the ICE box, change all moles to molarity (mol/L) and then plug it into the kc equation.

[405566265]

what are people referring to with "ice box"?

[mya warner 2C]

it is the Initial molar concentration, Change in molar concentration, and Equilibrium molar concentration table that we learned in lecture!

[Kitana_Garcia_3A]

what are people referring to with "ice box"?

I: Initial Molar Concentration, C: Change in Molar Concentration, and E: Equilibrium Molar Concentration. I hope this helps :)

[Kitana_Garcia_3A]

I am a bit lost, can someone explain how we get our values from the ICE box, do we use the quadratic equation to solve for x?

[mya warner 2C]

I am a bit lost, can someone explain how we get our values from the ICE box, do we use the quadratic equation to solve for x?

so for this problem you get the values for the ICE box from what the problem initially tells you. The initial values that are given to you go under the 'I' category and for the other R/P that aren't given an initial value, it is assumed to be 0. Then for the change category you are either adding or subtracting X from the initial value (keeping in mind that if there is a coefficient in front of the compound then that same coefficient is in front of X). Finally for Equilibrium, you simply write out initial+/- the change. For example 2SO3 (R) had an initial of 0.94M in my problem, so the change is -2X and the equilibrium is 0.94-2X. Hope this helps!

[EvaHeinrichs3C]

Hi! so for this one, I had slightly different numbers, but I think for the ICE box the initial SO3 concentration would be .94 mol/3 liters, and the other 2 initial concentrations would be zero. Then for the change, you know from the givens that the concentration of O2 went up by .180 mol/3 liters (to get moles/liter concentration), and based on the ratios of the chemical equation, the SO2 went up by double that concentration, and SO3 went up by 3/2 of that amount of O2. After completing that ice box, you would just plug those values into the Kc equation (making sure to raise the concentrations to their appropriate powers based on their coefficients), and you should get the right value! Make sure products are on top and the reactant is on the bottom :)

Do you multiply the O2 concentration by 3/2 to find the change in SO3 concentration because the stoichiometry instructs you to multiply by 2 and it is O3 instead of O2?

[Claire_Sabol_2G]

I am a bit lost, can someone explain how we get our values from the ICE box, do we use the quadratic equation to solve for x?

Adding on to Mya's explanation about the ICE Box, because we know O2 begins at a concentration of 0 and ends in equilibrium at the given value (for me, this value was 0.190 mol), we know that the change in O2 was +0.190mol. This value acts as our "x" value, so we can figure out the concentrations at equilibrium for SO3 and SO2 using this logic. If O2 is increasing as a product, this means SO2 must also be increasing; thus, the reactant (SO3) must be decreasing. Do not forget to factor in stoichiometry constants as well into the change portion of the ICE box (as Mya noted). As well, make sure to convert your final answer from moles to M using the provided value for L :)