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A quantity of 1.0 mole of $\mathrm{N}_{2} \mathrm{O}_{4}$ was introduced into an evacuated vessel and allowed to attain equilibrium at a certain temperature$$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$The average molar mass of the reacting mixture was $70.6 \mathrm{g} / \mathrm{mol} .$ (a) Calculate the mole fractions of the gases. (b) Calculate $K_{P}$ for the reaction if the total pressure was 1.2 atm. (c) What would be the mole fractions if the pressure were increased to 4.0 atm by reducing the volume at the same temperature?

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(a) $y_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.535, y_{\mathrm{NO}_{2}}=0.465$(b) $K_{p}=0.49$$(\mathrm{c}) y_{\mathrm{NO}_{2}}=0.29, y_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.71$

Chemistry 102

Chapter 14

Chemical Equilibrium

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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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One mole of $\mathrm{N}_{2…

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A container with a volume …

for this problem we have into a floor going to Tuen OAS are equilibrium. And if we have the average Mueller math of the mixture at equilibrium, which it didn't specify but should have at equilibrium of 70.6 in the fraction that is N 02 multiplied by its smaller mass, plus the fraction that is into +04 which will be the one minus the fraction that is an +02 because there's just the two of them multiplied by its Moeller Mass should give us the 70.6, so we could then do a little bit of algebra, rearrange it and solve for X value, which is 0.465 which is the fraction that's n 02 And then we define the fraction that is into a forest one minus X. So it's going to be 10.535 Now, knowing these two fractions, we should be able to solve four K p. KP is going to be equal to the concentration or the pressure that is, ah and oh too squared because of the coefficient to in the balanced chemical reaction, which I should probably right here as a reference close to two you know to. So KP is going to be equal to its pressure squared, divided by the pressure of the other one. And if the total pressure is 1.2 than the pressure that is, an 02 will be the total pressure multiplied by the fraction that is a no to the point for 65 and then we square it. And then we'll divide by the pressure that is into 04 which will be the total pressure multiplied by its fraction. And we get a K P value of 0.485 The next part of the question asked this to determine what the fractions would be if the total pressure was now four atmospheres instead of 1.2 atmospheres. So to solve for this we used the KP expression and the KP value, which we just calculated to be 0.485 and set K p equal to the pressure that is an 02 which would be the total pressure now multiplied by its fraction, which we don't know X squared. Well, then, divide that by the pressure that is into a four, which will be the total pressure multiplied by its fraction, which will be one minus X. We can. Then you solver, calculator, um, do the longhand, algebra and solve for X and we get a value of 0.29 And that'll be the fraction that is ano, too, as that's what we defined X to be. And then the fraction that is into a four will be one minus that or 0.71

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