5I.23 7th Edition

Moderators: Chem_Mod, Chem_Admin

klarratt2
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

5I.23 7th Edition

Postby klarratt2 » Tue Jan 15, 2019 4:46 pm

I've been working on this problem and can't figure out what I am doing wrong. I set up the ICE table and tried to solve for Kc, but I keep getting the wrong value. Can someone break this problem down for me?

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

Re: 5I.23 7th Edition

Postby Matthew Tran 1H » Tue Jan 15, 2019 8:10 pm

You have to use the balanced chemical reaction in order to solve this problem, which is CO(g) + 3H2(g) <--> CH4(g) + H2O(g). I'm assuming that you remembered to divide the moles of CO and H2 by the volume of the container to get the initial concentrations of CO and H2. The initial concentrations of CH4 and H2O are 0 since they are not given. Since you know that there are 0.478 mol CH4 at equilibrium, using the balanced chemical equation, you can infer that there are 0.478 mol H2O at equilibrium because of the 1:1 ratio (remember to divide by volume!) and that the moles of CO decreased by 0.478 (1:1) and the moles of H2 decreased by 3*0.478 (3:1). After that it's all hard numbers so you can just plug them into the expression for Kc, assuming that you raised the concentrations to their appropriate powers.


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 1 guest