5I.23

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905373636
Posts: 62
Joined: Thu Aug 01, 2019 12:15 am

5I.23

Postby 905373636 » Mon Jan 06, 2020 10:24 pm

Has anyone solved 5I.23?

It involves CO(g) + 3H2(g) -><- CH4(g) + H2O(g) and we know the initial [] of CO is 0.2 mol/L and H2 is 0.3 mol/L (the products are 0) and that [CH4] at equilibrium is 0.478, but we don't know anything about [H2O] at equilibrium. How do we solve for Kc with that unknown? I tried making an icebox, but I keep getting multiple unknown variables.

Thanks!

Charisse Vu 1H
Posts: 101
Joined: Thu Jul 25, 2019 12:17 am

Re: 5I.23

Postby Charisse Vu 1H » Mon Jan 06, 2020 10:37 pm

Hi! When you make your ICE table, the x value that you use to add and subtract to and from the reactants/products is actually equal to .478 M, which was given in the original problem because it is given that the equilibrium concentration (the E section of the ICE table) is 0.478 M.


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