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At $25^{\circ} \mathrm{C}$ and at 1 atm, the partial pressures in an equilibrium mixture of $\mathrm{N}_{2} \mathrm{O}_{4}$ and $\mathrm{NO}_{2}$ are $\mathrm{P}_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.70 \mathrm{atm}$ and $\mathrm{P}_{\mathrm{NO}_{2}}=0.30 \mathrm{atm} .$(a) Predict how the pressures of $\mathrm{NO}_{2}$ and $\mathrm{N}_{2} \mathrm{O}_{4}$ will change if the total pressure increases to 9.0 $\mathrm{atm}$ . Will theyincrease, decrease, or remain the same?(b) Calculate the partial pressures of $\mathrm{NO}_{2}$ and $\mathrm{N}_{2} \mathrm{O}_{4}$ when they are at equilibrium at 9.0 $\mathrm{atm}$ and $25^{\circ} \mathrm{C}$

a) increase.b) 8.0 $\mathrm{atm}$ and 1.0 $\mathrm{atm}$

Chemistry 102

Chapter 13

Fundamental Equilibrium Concepts

Chemical Equilibrium

Aqueous Equilibria

Carleton College

University of Toronto

Lectures

00:41

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.

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.

05:27

At $25^{\circ} \mathrm{C},…

04:15

The partial pressures of a…

01:50

The equilibrium constant, …

So here we've got an equilibrium between die nitrogen tetroxide and major jintai oxide and were asked if we raise the pressure of this reaction vessel that's closed and has these two species together, How is that going to affect the pressures of each of the gases? And then if we increase the pressures, that equilibrium, the total pressure is nine atmospheres. Um, what are the final partial pressures of each of these species going to be? So, for starters, report A If we increase the pressure, the pressure of both of these. So we increased p total. I should say that means that the pressure, the pressure of each gas is also going to increase now, even though this might elicit some swing in the equilibrium. Ultimately, the total pressure in the reaction vessel the equilibrium shift is not going to compensate entirely for the increase in pressure. So there will be at least a small increase in each of the gases. Partial pressures. So then report be, uh, we assume that the total pressure, the equilibrium is nine. Whereas to figure out um, what the partial pressures of these things are in that equilibrium vessel. So one thing that we're not given in this problem is the value of K P. So we're going to have to do that using the starting equilibrium values here. So we know that KP is going to be equal to the partial pressure of nitrogen dioxide squared in our numerator. So 0.3 squared, divided by the partial pressure of Diane Nitrogen tetroxide on what we find that is, that K P is equal to 0.129 So there's another important relationship here. If we think about our new reaction equilibrium is that the partial pressure of nitrogen dioxide, plus partial pressure of die nitrogen tetroxide? He's gonna be equal to nine. And it's important because it makes it a lot easier to some values indoor new equilibrium expression, for example. I can rewrite this to say that the partial pressure of nitrogen during nitrogen tetroxide is equal to nine, minus the partial pressure of nitrogen dioxide. So if we take this over to our reaction equation or back to this equilibrium expression, we consider this now under our new conditions, we know that this is going to be maintained to the 0.1 to 9 to maintain because we're not changing changing temperature. We know that this is now going to be equal to put some placeholders in here. The partial pressure of nitrogen dioxide divided by nine minus the partial pressure of nitrogen dioxide. And if we rearrange this and solve all the way through, what we could end up with is a quadratic equation, which is equal to the partial pressure of di nitrogen tetroxide squared that should be squared in the top of this equation. I messed up plus 0.129 parts of pressure nitrogen dioxide minus 1.161 So now we can use the quadratic formula and go through, calculate with these final concentrations, or be well what these final partial pressures will be. And if we do that, we find that the partial pressure of nitrogen dioxide is one which by rule means that the partial pressure of nitrogen die. Nitrogen tetroxide is eight. So to start out with this problem, what we did is we took the equilibrium concentrations at the equilibrium we were given use that to find r. K. P used the simple relationship between the two gases that were given and then plug them in directly into equilibrium expression to finally find our final partial pressure is using this new equilibrium

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