11.11

Moderators: Chem_Mod, Chem_Admin

Helen Shi 1J
Posts: 78
Joined: Sat Jul 22, 2017 3:00 am

11.11

Postby Helen Shi 1J » Sat Nov 25, 2017 8:03 pm

A 0.10-mol sample of pure ozone, O3, is placed in a sealed 1.0-L container and the reaction 2 O3(g)∆3 O2(g) is allowed to reach equilibrium. A 0.50-mol sample of pure ozone is placed in a second 1.0-L container at the same temperature and allowed to reach equilibrium. Without doing any calculations, predict which of the following will be different in the two containers at equilibrium. Which will be the same? (a) Amount of O2; (b) concentration of O2; (c) the ratio [O2]/[O3]; (d) the ratio [O2]3/[O3]2; (e) the ratio [O3]2/[O2]3. Explain each of your answers.

Can someone explain why (c) the ratio [O2]/[O3] would not be equal if there are .1 mol of O3 as opposed to .5 mols.

Curtis Tam 1J
Posts: 105
Joined: Thu Jul 13, 2017 3:00 am

Re: 11.11

Postby Curtis Tam 1J » Sat Nov 25, 2017 10:12 pm

I did that problem too and got confused. I think it's cuz you want to focus on the ratio of O2 raised to its nonzero coefficient to O3 raised to its nonzero coefficient. This ratio would never change because this produces the value Kc but that doesn't necessarily mean that the ratio of O2 to O3 would remain the same too. This is because the ratio of O2 to O3 doesn't equal Kc.

Tess McDaniel 1F
Posts: 16
Joined: Fri Sep 29, 2017 7:04 am

Re: 11.11

Postby Tess McDaniel 1F » Sun Nov 26, 2017 4:11 pm

I was also confused on this one but I think its simply because of the powers that the concentrations are raised to. Obviously when calculating Kc we have to raise the concentrations to the power of their coefficient and so larger concentrations will be exponentially larger than smaller concentrations rather than varying directly. Like Curtis said Kc itself doesn't change but the ratio is that of the concentrations raised to their powers not just the ratio of concentrations. Hope that helps!


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 1 guest