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At $25^{\circ} \mathrm{C},$ the equilibrium partial pressures of $\mathrm{NO}_{2}$ and $\mathrm{N}_{2} \mathrm{O}_{4}$ are 0.15 atm and $0.20 \mathrm{atm},$ respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

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$$\begin{array}{l}P_{\mathrm{NO}_{2}}=0.10 \mathrm{atm} \\P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.090 \mathrm{atm}\end{array}$$

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.

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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.

04:15

The partial pressures of a…

02:36

The equilibrium constant, …

02:08

An equilibrium mixture con…

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An equilibrium mixture co…

01:50

04:09

At $25^{\circ} \mathrm{C}$…

03:16

01:52

The total pressure for a m…

So since we have been given an equilibrium value for both the product in reaction, partial pressure, the first thing we can dio for this problem it's find the KP before we change the volume it all. So we were told that the partial pressure of the product was your 0.15 We can raise that to two. That's what we see here. Then we can divide that by what are partial Pressure of reactions is which is 0.2 zero. So I find KP initially here to the 0.11 and this KP will hold true for the reaction going forward even if we change the volume. So we know that we are taking the volume and we're going to multiply our initial volume by two. And what this means for the pressure, however, is that we are going to divide by two or take half of the pressure that we had before. So that means that our new initial, where when we're going to find the new equilibrium, personal pressure, our new initials are going to be 0.15 divide by two, which is equivalent to 0.7 fine a t m and then to the same for our reacted into 04 0.2 divided by two. It is 0.1 zero a t m. So now that we have our new starting initial partial pressures after we have changed the volume, we can go ahead and put this into in ice table, which is an initial change that at equilibrium excuse a different color here. So we have 0.1, which is what we found here. And we have zero 0.75 which is what we found here. So we have no zero for the initial because we already were at an equal of room value value before we change. Ah, state of the system itself and that was we made the volume two times as large. So we know that this is going to be minus X plus two x 0.7 five plus two eggs here for the product and 0.1 minus x here for the reacted. So since we already calculated KP when we were given the equilibrium values before we change the volume, we know that the cave he stays the same. We can go ahead and use the KP to make a new equilibrium equation to solve four our X value. So we're gonna first plug in our products and then plug in our reactant. But then, if you simplify this in Seoul for X, you find X to be 0.1 to, and now we're able to find the partial pressure of our product and reactant at our new equilibrium. So partial pressure of of our reactant is going to be 0.1. We had minus X. Gonna go ahead and seven or that X with our found X value of 0.12 And that is 0.9 a. T him. Now that we're gonna do the same thing for our product, we're gonna take what we have initially 0.75 plus two times what we found to be ex near 20.12 and that is going to be equivalent to 0.1 a. T. M.

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