Achieve HW #2

[Shannon Clark 1F]

Hi,
I'm having trouble understanding exactly what to do after we form the ICE table.

Here is the question:
At a certain temperature, 0.740 mol SO3 is placed in a 2.00L container.
2SO3(g)↽−−⇀2SO2(g)+O2(g)
At equilibrium, 0.180 mol O2 is present.

My ICE chart is:
0.37 0 0
-2x +2x +x
0.37-2x 2x 0.09

What do I do with the x? I know I'm supposed to calculate Kc with [SO2]^2[O2]/[SO3]2, but do I leave the x in when I put in the values??? very confused, any help is welcome!

[benkw_3I]

Hi! Not sure if this helps but I did the problem by thinking of it conceptually. Initially there is only SO3 and at equilibrium there is 0.180 mol 02. Therefore S03 reacted to 0.180 02 and some amount of S02. For every molecule of 02 produced 2 molecules of S03 are produced so the concentration of S02 is 2 * 0.180 mol. By similar logic, for every 02 produced 2 S03 molecules are used. Therefore the concentration of S03 is reduced by 2 * 0.18 mol. So the total concentration of S03 is 0.740 - (2* 0. 18).

[Lexi Tempera 3B]

Hi Shannon,

Whatever your final concentration of O2 is X. Use this to solve your E equations to find the final concentrations. Then, plug the E concentrations into the Ka equation:

[O2][SO2]^2 / [SO3]^2

Voila!

[Brianna Cronyn Lec3E]

For me, I used that information to see that O2 is equal to x and SO2 is equal to 2x at equilibrium, for their concentrations at equilibrium. The amount of moles of O2 at equilibrium is given, 0.190 if I am not mistaken. So, that divided by the volume is x, which will get you the concentration for O2. That times 2 will get you the concentration of SO2. Now, with all of the concentrations at equilibrium, you can solve for K.

Hope this helps!

[Shannon Clark 1F]

Hi Shannon,

Whatever your final concentration of O2 is X. Use this to solve your E equations to find the final concentrations. Then, plug the E concentrations into the Ka equation:

[O2][SO2]^2 / [SO3]^2

Voila!

This helped, thank you so much! I solved the problem correctly, but just for future knowledge, why is the equilibrium concentration of O2 equal to x?

[Saebean Yi 3E]

Hi Shannon,

Whatever your final concentration of O2 is X. Use this to solve your E equations to find the final concentrations. Then, plug the E concentrations into the Ka equation:

[O2][SO2]^2 / [SO3]^2

Voila!

This helped, thank you so much! I solved the problem correctly, but just for future knowledge, why is the equilibrium concentration of O2 equal to x?

In this problem, the equilibrium concentration of O2 equals x because in the beginning (initial), there were no O2. If you wrote out the ICE table, the "initial" for O2 would be 0, the "change" would be +x (because look at the equation, O2 only has a coefficient of 1), so naturally, the "final", which is the equilibrium constant of O2 should be x.

One important thing to note though, is that in this problem (at least how I did it) my ICE table was set up so the values were all in moles, not molarity (which is what concentrations are in). So technically, the equilibrium concentration of O2 would not exactly equal x, but x (moles) divided by the volume.