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April 16, 2014

April 16, 2014

Posted by **bkue** on Monday, May 14, 2012 at 11:28pm.

At equilibrium there are .6 mol of H2.

Determine the concentrations of NH3, H2, and N2 at equilibrium.

2NH3-> 3H2+N2

^ all gas

- chem, find concentrations -
**DrBob222**, Tuesday, May 15, 2012 at 12:03am.........2NH3 ==> N2 + 3H2

I.........2.2.....0......0

C.........-2x......x.....3x

E.......2.2-2x.....x......3x

The problem tells us that there is 0.6 mol H2 at equilibrium. So 3x = 0.6 mol which means x = 0.2 mol. (Note that I did NOT change H2. It still is 0.6 mol. For proof of that 3x = 3*0.2 = 0.6. I write this only because I helped a student last week with this and she thought I had changed H2 but I didn't.)

That gives you N2 (x) and NH3 = 2.2-2x.

Then divide mols each by 2 L to obtain concn of each.

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