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A quantity of 0.20 mole of carbon dioxide was heated to a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached:$$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$Under these conditions, the average molar mass of the gases was $35 \mathrm{g} / \mathrm{mol} .$ (a) Calculate the mole fractions of $\mathrm{CO}$ and $\mathrm{CO}_{2}$. (b) What is $K_{P}$ if the total pressure is 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

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(a) $y_{\mathrm{CO}_{2}}=0.44, y_{\mathrm{CO}}=0.56$(b) $K_{p}=7.8$

Chemistry 102

Chapter 14

Chemical Equilibrium

Carleton College

University of Maryland - University College

University of Toronto

Lectures

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.

00:54

In chemistry, chemical equilibrium (also known as dynamic equilibrium) is a state of chemical stability in which the concentrations of the chemical substances do not change in the course of time due to their reaction with each other in a closed system. Chemical equilibrium is an example of dynamic equilibrium, a thermodynamic concept.

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A quantity of 0.20 mole of…

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A quantity of 0.29 mole of…

07:44

The reaction $\mathrm{CO}_…

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12:58

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The reaction CO2(g) + C(s)…

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$\mathrm{CO}_{2}$ is passe…

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The amount of carbon dioxi…

This is a fairly challenging problem. It requires you to think of equilibrium in a slightly different way in terms of fractions and average molar masses. And then you need to understand the concept of average Miller Mass being a weighted average based upon those fractions. So let's start with the equation that's given to us. We have solid carbon dioxide reacting. I'm so sorry. Solid carbon reacting with gaseous carbon dioxide, one mole of each of them producing two moles of carbon monoxide. Then we know that at equilibrium there's an average molar mass of 35 grams per mole. If the average Mueller mass at equilibrium is 35 grams per mole, then that should be equal to the fraction that is carbon dioxide multiplied by the molar mass of carbon dioxide, plus the fraction that is carbon monoxide multiplied by the molar mass of carbon monoxide. So where did we get the fraction than that is carbon dioxide and the fraction that is carbon monoxide? Well, if we start out with 0.2 moles of carbon dioxide, then as the reaction proceeds, it will decrease by X X is just an unknown amount as it approaches equilibrium so the moles, then of carbon dioxide at equilibrium will be 0.2 minus X. The moles of carbon mine oxide at equilibrium will be two X, assuming we have no moles of carbon monoxide, which is the way the problem was described, Um, when the reaction began, then we produced twice Aziz many moles of carbon monoxide as there is a decrease in moles carbon dioxide. So the fraction then will be the molds that we have an equilibrium divided by the total bowls at equilibrium, the total moles at equilibrium will be the moles of carbon monoxide. Sorry, moles of carbon dioxide 0.20 minus X plus the moles of carbon monoxide, which is two x. So that's how we get our denominator. The moles of carbon dioxide that air created at equilibrium, divided by the total moles of gas. It equilibrium the most of carbon monoxide at equilibrium, divided by the sum which is the total moles of gas at equilibrium. Then we have a fair amount of algebra to do. If we multiply both sides of the equation by this denominator will end up getting 35 x plus seven will be equal to and then we'll distribute this 44 through here and will take the 28 multiply it by two x and we'll get 8.8 minus 44 X plus the 56 X coming from this contribution over here. Then we'll go ahead and collect our like terms and solve for X. And we get X equal. 2.78 Okay, now we're not done. That's just X in order to solve for the fraction that is carbon dioxide we need to solve for this fraction here and then for the fraction that is carbon monoxide. We need a plug in X to this ratio here, so we'll do that for carbon dioxide. It'll be the 0.20 minus X, divided by the point to zero minus x plus two x or essentially just plus X, and we get a fraction of about 0.44 We'll do the same thing with carbon monoxide, and we get a fraction of about 0.56 then to solve four K P. If the total pressure is 11 atmospheres, then KP is going to be equal to the pressure that is carbon monoxide squared, which will be the fraction that is carbon monoxide multiplied by the total pressure and then divide that by the fraction that is carbon dioxide multiplied by 11 and we get a K P value of 7.8.

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