consider-the-reaction-2-mathrmcogmathrmo_2g-rightleftharpoons-2-mathrmco_2g-suppose-the-system-is-al

Question

Consider the reaction
$$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)$$
Suppose the system is already at equilibrium, and then an additional mole of $\mathrm{CO}(g)$ is injected into the system at constant temperature. Does the amount of $\mathrm{CO}_{2}(g)$ in the system increase or decrease? Does the value of $K$ for the reaction change?

Stephanie L · Accepted Answer

in this question, we&#x27;re told that the following reaction is already at equilibrium. And whereas what happens to the amount of see you to carbon dioxide when Moore CEO carbon monoxide is added to the system Now according to a shot alias principle, if we increase the amount of his species in a chemical reaction, the reaction will shift away from it to consume it unless it&#x27;s a solid or a liquid. In this case, CEO is a reactive and a gas. So if we add one mole of carbon monoxide, were essentially increasing the amount of carbon monoxide in the container. Now because we increased the amount of carbon monoxide, the reaction is going to shift away from it to consume its in order to reach equilibrium again. Now, because the reaction is shifting to the right, everything on the left of it is in a decrease. Yes, with the amount of CO. Two will decrease, and on the right side of it, everything will increase. So the amount of Sio two will increase as we&#x27;re producing it, E. So see you, too. Amount increases. And that&#x27;s gonna happen until the system reaches equilibrium again. Now, because this reaction is done under constant temperature. This means that the equilibrium constant does not change, and the only thing that affects the value of the equilibrium constant is temperature. And in this case, since we&#x27;re not changing it, the 11 constant will not change. So the reaction will shift to the rights until the same ratio of products to reactions is a cheek, and that&#x27;s when it will stop because it will be at equilibrium.

Ronald Prasad · Answer

threat the equilibrium to end to old gas plus two gas in equilibrium with four and oh, gas. And we&#x27;re gonna assume that we&#x27;re told that the system is at equilibrium and then add equilibrium. We&#x27;re gonna add an additional mole of into Oh, so the polarity of N two is going to increase. So I&#x27;m gonna graft this here. Let&#x27;s do a concentration versus time graph. And I&#x27;m just gonna graft them, lady, even to o so at equilibrium, the polarity of into it would be constant. And we&#x27;re gonna do our change at this time here. Change would be increased the concentration of into Oh, so the change goes up there and then listed as principal wouldn&#x27;t want to compensate this increase by shifting the equilibrium towards the rate So many draw dotted line. This is the original equilibrium Compensation was listed. High&#x27;s principal would look something like this here. Overall, this is an overall increase in the mill aridity of into oh, even with list highest principle compensating for the increase by shifting towards the right, it will not go back to its original equilibrium position so overall, by adding in extra night die nitrogen monoxide, the into oh shows an overall increase when it reestablishes equilibrium. Now, now we look at what happens to the equilibrium. Constant, uh, equilibrium concert is unchanged as we&#x27;re, um, changing the concentration concentration does not effect equilibrium constant. Uh, as temperature is constant, the only factor that will change the value of the equilibrium constant is temperature. And since temperature is constant, the equilibrium value or the value of the equilibrium constant is unchanged.

Carina Carlos · Answer

the equation for the equilibrium constant KP is equal to you, partial pressure of the products divided by the partial pressure of the reactant. Since given this equation, let&#x27;s take a look. That&#x27;s our reaction. So we have carbon monoxide and oxygen, and this is going to be an equilibrium with carbon dioxide. We&#x27;re told that this reaction is exo thermic, so we add heat to the product side. The reason that we&#x27;re adding heats on the product side is because he is being released. If he is being released, then we can treat it as if it were a, um, a product itself. So given this, let&#x27;s consider what will happen if we increase the temperature. When we increase the temperature, we are essentially increasing heat, and again we&#x27;re pretending that he&#x27;d Z product. If we&#x27;re increasing the concentration of a product, then you want to shift the reaction left towards theory reactant to counterbalance this increase and restore equilibrium. This is going to be as a result of Luschek Taylor&#x27;s principle. So if we know what our reaction is shifting to the left, we know we&#x27;re increasing the concentration of our reactant since and decreasing the concentration of our products. This is going to be directly related to their partial pressures. If you have a greater concentration of reactant, since then, they&#x27;re going to contribute more to the pressure in the reaction mixture. So these increases again. We&#x27;re going to increase the partial pressures the reactant and decrease the partial pressure of the products. So looking back at our equation, if we increase our numerator and decrease our denominator, I&#x27;m sorry you have that backwards. We are increasing our denominator and decreasing our numerator. So if this occurs, we&#x27;re going to be overall decreasing for equilibrium constant. So that is how he can justify why the equilibrium constant is going to decrease.

Jorge Villanueva · Answer

you can Clearly a question about 16 talks about the relationship between BP and H. Since you look at your equation, you see Dr Ancient Grave for so okay. If you increase the temperature when you&#x27;re h is available for your KP value also increase.

Carina Carlos · Answer

in the problem. We are given that our Delta H is equal to 200 six killing Jewell&#x27;s. So if we know that bill to H is a positive number, we know our reaction is endo thermic. If our reaction is under thermic, we can treat heat as a reactant. So if we follow the shuttle ears principle, if we increase temperature for increased heat, it&#x27;s almost as if we&#x27;re increasing the concentration of a reactant that will shift the reaction to the rights, which is the forward reaction. And if we do that, we will be decreasing the concentration of our reactant since while simultaneously increasing the concentration of the products, remember that in our equation for ideal guesses, we know that pressure and concentration which will be en over V so pressure and concentration are directly related. So if we decrease the concentration, we&#x27;re decreasing the pressure of the reactant since and increasing the pressure of our products. Why is that important thinking to what the question is asking about? KP KP is in a very simplified way, the partial pressure of the products divided by the partial pressure of the reactant. It&#x27;s so if we increase temperature we are increasing the numerator while simultaneously decreasing the denominator. This is going to results in the KP that is larger than it originally. Waas. So the enter to the question is that KP is going to increase if we increase the temperature.

Stephanie L · Answer

So for this question, were given the following equilibrium reaction. And we&#x27;re told that it&#x27;s already at equilibrium and were asked what will happens into this reaction if we do the following perturbations. So for a what happens if we add B right and being is a solid now because solids and liquids do not show up in the equilibrium, constant expression, there will be no effects, so the reaction will stay at equilibrium. Now, the part b were as what will happen to me equilibrium if we remove see gas as its forms. Now, if we remove, see yes, right which is a product according to lash Achilles principle and the system gonna wanna undo the change. And so because we&#x27;re moving, see, the system is going to shift to produce See again you know, the reestablish equilibrium. So in this case, the reaction is gonna shift to the right and for part C were asked what will happen if the volume of the system is the decrease. So if the container sizes decrease by a factor of two, all right, so if we decrease the volume and in this case the factor of two doesn&#x27;t matter, because this is just a qualitative question. And so if we decreased the buying when you think about what happens to the pressure of the system I and if we decrease the volume, the total pressure increases. These are the pressure of all the gases and the container will increase. And because of that, the Excalibur is gonna want to shift to reduce the pressure. And in order to do that, it will shift to the side with les miles of gas. Since we know that, yes, we know that the moles of gas and the pressure are directly proportional. So in this case, if we look at the left side, we have two moles of gas. This by looking at the Stoke Geometrical efficient uh, the gas one and then on the right side, we have four moles of gas since we have one well coming from C and three moles coming from the deep. So in order to reduce the pressure is going to shift towards the left where we have a less molded gas when we have two moles of gas chefs left and then the final perturbation in part D is what happens if the temperature is increased So if you increase the temperature now to consider that we need to look at the facts that this reaction is Endo thermic nice and still th is confident. And so that means that heat is absorbed so heat is used up as a reaction proceeds. So we want to think about heat as a reactant because it&#x27;s consumed as a reaction and happens now. If we increase the temperature, we want to think about it. As Addie heats, these were increasing the amount of heats. So now, according to the shuttle, it&#x27;s principal write just for, like any of the species analysts, solid or liquid. If we increased the amount of heat, the reaction isn&#x27;t a shift to consume it. Maize is gonna shift over A from it. And so here the reaction is going to shift to the rights

Problem

For an equilibrium involving gaseous substances, …

Problem 30 Easy Difficulty

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