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A $1.00-\mathrm{L}$ vessel at $400^{\circ} \mathrm{C}$ contains the following equilibrium concentrations: $\mathrm{N}_{2}, 1.00 \mathrm{M} ; \mathrm{H}_{2}, 0.50 \mathrm{M} ;$ and $\mathrm{NH}_{3}, 0.25 \mathrm{M}$ . How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 $\mathrm{M}$ ?

0.33 $\mathrm{mol}$

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

Carleton College

Rice University

University of Kentucky

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this question says that a one liter vessel at 400 degree Celsius contains the following equilibrium concentrations one Moeller of End, 2.5, mauler of H two and $20.25 H three and asked us how many moles of hydrogen gas must be removed from the vessel to increase the concentration of measured in to 1.1. The first thing I did was figure out the equilibrium constant with respect to concentration, that is, the concentration of N H three squared the product divided by the concentration of n two times concentration beach too cute, plugged in each of the values that gave us unsolved and found that the equilibrium constant with respect to concentration is 0.5. So we use that in the moment. Then I went back and filled in part of a nice table for this reaction here the initial concentrations that gave us and it says that we want the final concentration of end to to be 1.1 Moeller and it's changed could be represented by X. What sex and simple subtraction will tell us that X must equal 0.1. I'm ignoring H two for now because we're changing its concentration by physically removing some. But we can look at the product and h three here you can write exchange as minus two X. And since we just found that we know that this equilibrium, then concentration must then be 05 Mueller, we're at age three. Okay, Now let's rearrange this equilibrium constant and get, uh, H two by itself. So h two cubed equals concentration of N h three squared by the by the concentration of into times. Casey, all I did was divide both sides by Casey and multiple. You both sides by H two constellation of H two cubed. We can plug in all of our parts. Now we're gonna use these new equilibrium mounts. It's a 0.5 squared for an H three and 1.1, um, times 0.5 for the construction of end, too. And the concentration of, uh, and sorry for the equilibrium constant. And if you saw this, if you saw this, you will see that just double checking here. You'll see that the concentration of each too cools, um equals point one seven. We rounded a bit different, but 0.17 and so if this is the equilibrium concentration of each too and its original was 0.5. The change, which it says how much we need to remove is going to equal 0.5 minus 2.17 equals zero point three three moles of each to have to be removed in order for the equal of concentration of end to to be 1.1 Mueller.

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