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Calculate the equilibrium concentrations of NO, $\mathrm{O}_{2},$ and $\mathrm{NO}_{2}$ in a mixture at $250^{\circ} \mathrm{C}$ that results from the reaction of 0.20 $\mathrm{M} \mathrm{NO}$ and 0.10 $\mathrm{M} \mathrm{O}_{2}$ . (Hint: $K$ is large; assume the reaction goes to completion then comes back to equilibrium.)$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad K_{c}=2.3 \times 10^{5} \mathrm{at} 250^{\circ} \mathrm{C}$

0.0034 $\mathrm{M}$

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

University of Central Florida

Rice University

Drexel University

University of Maryland - University College

Lectures

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

04:15

Calculate the equilibrium …

02:34

At $2000^{\circ} \mathrm{C…

02:07

So here we've got an equilibrium expression that were given some starting concentrations for materials and were asked to calculate with the concentration of each species is gonna be once this reaction reaches equilibrium. So if we take a look at this closely, the values that were given a beginning are that the mill aridity of nitrogen oxide, his point to Mueller and that the concentration of oxygen gases 0.1 Mueller is our initial concentrations. But one thing that we should take into account with this problem is how large this casey value is. Casey values huge. It's the number of times 10 to the fifth. So what we can assume from that is instead of doing this as these values being are starting equilibrium, we should consider these, um, this reaction going totally to completion and actually consider it as if the reaction we're starting from the products and proceeding with reverse to their reactions. So to figure out our starting amount of nitrogen dioxide, we actually have to figure out which of these re agents are going to be limiting to ensure that we choose the right polarity value. So we're just gonna do some very quick straight geometry here. So point to O molds for leader uh, nitrogen oxide times a conversion factor of two moles of enter, too, for more. I know, for two moles of mm, Sorry. They're 1 to 1 ratio. So if this reaction, we're totally to go to completion, we would wind up with point to Mueller and over to. And if we look at this in terms of oxygen, what one No more oxygen for leader. Two moles of nitrogen oxide for one more a vote, too. And we also get point to a mall. So this is a case where the amounts of these that we have in our reaction vessel we're going to be equal so and this because they both produce the same amount instead of using This is our initial for a change to reach equilibrium, we can assume that we're starting with neither of these to start with 0.2 Moeller nitrogen dioxide. So now the steps are going to be very much the same to what we're used to do him. There's going to be some amount of change in the species. The easiest thing to do is to represent those as changes in X using the strike Yama trees in the equation. So here we're going to lose two moles of nitrogen dioxide for everyone. Mole of auction that we gain and every two moles of nitrogen oxide that we gain. This is our change, and then our final concentrations therefore going to be two x for the nitrogen oxide X for oxygen and point to minus two X for the nitrogen dioxide. So now the same way that we do in all of our problems, we're gonna take these values and take them as our concentrations and plug them into our equilibrium expression. So we have Our Casey is equal to 2.3 times 10 to the fifth, and we know that that's equal to the concentration of nitrogen dioxide squared on top, which here we're going to substitute out for 0.2 minus two x in that whole terms gonna be squared and we're going to divide that by the concentration of nitrogen oxide squared and the concentration of oxygen. No one important thing that we can do in this problem is because it's equally remiss so high we can assume that the shift backwards if we were to start with all nitrogen dioxide and shipped back towards towards the reactions, we can assume that this number here is gonna be very small compared to point to. So we're just going to assume that that number is zero, which is gonna leave us with 0.2 On top of this fraction, we simplify this bottom again, bottom a bit we're gonna have for X squared times X, which is for X cubed. Now, if we take these away through, we do some quick arithmetic and and, uh, rearrange a little bit, we will end up with this. Where excuse he put 1.74 times 10 to the negative seven. Finally, it would take this forward. We find that X is equal to 3.4 times 10 to negative third. So what that means is that the concentration that's going to be equal to the concentration of oxygen, the concentration of nitrogen oxide is gonna be two times that. So we're going to have 7.8 times 10 to the negative. Third is equal to the concentration of nitrogen oxide. And really, if you think about how small this value is in comparison to our starting concentration for Reggie and dioxide. Concentration of nitrogen dioxide really isn't going to change. Um, if you want to be very specific about it, it will drop it down to about 1.98 or so. 1492 Um, but again said, assuming that that change was negligible was very accurate. So these are the final concentrations for these species in this reaction, that equilibrium.

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