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Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl_oduced when asample of NOCl with a pressure of 10.0 atm comes to equilibrium according to this reaction:$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \qquad K_{P}=4.0 \times 10^{-4}$

.209$\mathrm{~atm}$

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

University of Central Florida

University of Maryland - University College

Brown University

Lectures

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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.

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this question asks us to calculate the pressures of all species that equilibrium in a mixture of n, f, c, l, n o and seal to produced when a sample of N o. C. L. With the pressure of 10 atmospheres comes equilibrium. According to this reaction, here's that reaction to an O. C. E o forms to Eno and seal to the equilibrium. Constant respect to pressure is four times 10 to the negative forth. We have initially 10 atmospheres pressure from an O. C. L. And none from any of the products. So let's finish our ice table. We'll have minus two x here because of this too plus two x here. Because of this too. We are close to X and plus X here because there's no term in front of seal too. And we could express the equilibrium Sze as 10 minus two x here just to X and here just X All right. Now that we have an expression to describe the equilibrium pressures, we can plug them into three equilibrium constant equation. For this reaction, we would have the partial pressure from Anna squared times a partial pressure from seal too divided by the partial pressure from N O. C. L. Squared again. Those squares come from the twos out in front of those two, um, substances. So let's plug in our values. We had two ex squared times X divided by 10 minus two x squared. We can set that equal to the equilibrium constant it gave us, which is four times 10 to the negative forth. And this one is a little bit tough to solve by hand. You would, Ah, I just expanded this to 100 minus 40 x plus four x squared and I simplified this to four x cubed. Multiply the denominator over and then simplified. It's that I got a cubic function that looked like zero equals four x cubed minus 40.16 x squared plus 0.16 x minus points here four. And, uh, you can tell that by hand, or you can use a cubic function calculator. But if you did, either way, you should get X equals 0.209 that we could plug this X back into each of these functions that we use to describe the equilibrium. And we would find that the pressure pressure's would now be, uh, 9.5 heat atmospheres for N S E L. It would be 0.419 atmospheres. The round is a little bit off. I kept all of the digits that were in this value when I calculate each of these but that rounding 0.419 atmospheres for N O. And just 00.209 atmospheres for seal, too.

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