HW 5I23

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005321227
Posts: 90
Joined: Sat Sep 07, 2019 12:15 am

HW 5I23

Postby 005321227 » Mon Jan 13, 2020 10:35 pm

A reaction mixture consisting of 2.00 mol CO and 3.00 mol H2 is placed in a reaction vessel of volume 10.0 L and heated to 1200. K. At equilibrium, 0.478 mol CH4 was present in the system. Determine the value of Kc for the reaction CO(g) 1 3 H2(g) ∆ CH4(g) 1 H2O(g) at 1200. K.

Can someone explain how to do this step by step? I'm a little confused since I assumed we'd have to use the ICE box, however we are not given a K value in order to solve it, but the concentration of H2O is not given either.

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: HW 5I23

Postby nicolely2F » Mon Jan 13, 2020 11:35 pm

We know that we have 0.478 mol CH4 at equilibrium. The balanced equation is given, which means we can tell the concentration of H2O is 0.478 M since it has a proportion of 1:1 to CH4. From there, we can build the ICE table.

CO 3H2 CH4 H2O
2 3 0 0
-x -3x* +x +x
2-x 3-3x .478 +x
*This has to be 3x because H2 is consumed at a rate 3 times faster than the other reactant.

We know x = 0.478 because (again) it is the equilibrium concentration for CH4. And because CH4 has a 1:1 proportion to H2O, we know that 0.478 mol H2O is formed. From there, we can discover the number of moles for CO and H2:
CO = 2-x = 1.522
H2 = 3-3x = 1.566
Before plugging all these numbers into Kc, recall that these moles are mixed into 10L of water, thus the molarities are actually 0.1522, 0.1566, 0.0478, 0.0478.
Thus, Kc = (0.0478)(0.0478) / (0.1522)(0.1566^3) = 3.88


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