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For which of the reactions in Exercise 13.15 does $K_{c}$ (calculated using concentrations) equal $K_{P}$ (calculated using pressures)?

a) equalb) equal c) not equald) not equale) not equalf) not equalg) not equalh) not equal

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

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So for this problem, we have to take a look back at problem 13 15 and used the equations that they list there to determine whether each of the equations in question is going to have a K P that is equal to a Casey. So to do this problem. One thing that you should recall is that there's a pretty straightforward relationship between K P and K C. And that is that the K P is equal to the Casey, multiplied by the product of the gas, constant times, the temperature raised to the power of the change in number of active species within the equation. That's what I mean by that is Delta N is going to include, um, the number of moles of gases, iniquitous species. What is not going to include are solids and liquids. And that's because in terms of your equilibrium, the solids and liquids are not going to be taking up an appreciable amount of space and the reaction and they're not going to be, um, they just don't get considered the same way in the equations. So because of that, we're looking for a scenario where K P is equal to K C and for KP to be equal to K. C. We look at this a little bit more closely. The only way that we can have KP equal to K C is ever changing the number of moles a zero, because if we have zero here instead of don't end. But that does is it makes this whole term one. And that's where we can say that K P is equal to Casey. So for each of the equations listed in 13 15 we're really just going to look at the change in moles of gaseous and a curious species on the left, and compare that to the change in moles of gaseous and a curious species on the right. So for a we have molecule of methane gas interacting with chlorine gas, and that's an equilibrium with methyl chloride, also a gas and gaseous hydrochloric acid. So for this equation, we have one mole gas, two moles of gas ed equilibrium with one mole of gas to most of gas. What that means is that the change in end is zero, and this is an example of an equation where K P is going to equal or Casey so for the rest of the equations. In this problem, we're just going to run through middle it You look back the equations and look at the changes in the number. Like the specific questions equations themselves is going to run through the changes in moles as they pertain to each of these. So for part B, we have one mole nitrogen gas, plus a mole of oxygen gas Is that equilibrium with two moles of N 02 are nitrogen oxide. So that means, is that our delta and a zero? This is another scenario. Where are KP is equal to or Casey? For 13. 15 c. We're looking at two moles of sulfur dioxide gas interacting with one mole of after getting gas. And that's an equilibrium with two moles of sulfur dioxide gas. So in this case are changing and is negative one because we have three moles on the left and two moles on the right. And in that sense it means that our KP is not going to be equal to our Casey for letter D For the letter d, we have a soul. So no moles on the left, at equilibrium with another soul and a gaseous species on the rape. So that means that our delta and is one and r k p is not equal to our Casey. Moving on to e. We have a full of gas interacting with five moles of gas and they're an equilibrium with a solid. The change of number of models of gas for this is actually gonna be negative. Six. And what that means is that the KP is not going to be able to the Casey. Same thing for F one mole of gas interacting with two moles of gas changing end one. KP and Casey are not equal looking at G one mole of gas with two moles of gas at equilibrium with one mole of gas and two moles of liquid. I remember the liquids don't get included in our species. So we write that is zero. That means the change in and is negative too, and that our KP in our case he cannot be equal. Finally, for letter h, we have a solid no moles on the left is going to be an equilibrium with another solid and five moles of water vapor. It means you're changing and it's five and get her. KP is not equal to our Casey. That's what we've learned from this. Is that part a part B from 13 5 are the only scenario here. KP is. When did you be equal to Casey? And again? That's because the only scenario where you have a K p equal that Casey is where the number of moles of gas and a quick of species on the left and the right are balanced.

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