when-a-reaction-system-has-reached-chemical-equilibrium-the-concentrations-of-the-reactants-and-pr-2

Question

When a reaction system has reached chemical equilibrium, the concentrations of the reactants and
products no longer change with time. Why does the amount of product no longer increase, even though
large concentrations of the reactants may still be present?

Anatole Borisov · Accepted Answer

so it&#x27;s answers. Question. Don&#x27;t you have to consider the reaction generic and be being an equilibrium with a B? And so we were told his at equilibrium. The concentration of all of these is constant. Concentration was put out like that. And so the question is, how come it&#x27;s constant, even if there&#x27;s a huge amount of A and B Francis still available, that that&#x27;s possible now? What happens is A and B are still gonna be reacting. They&#x27;re suddenly reacting to form a beat together. But the thing that&#x27;s often overlooked is that, and at equilibrium, there&#x27;s still a B that can break apart and do the reverse reaction and become a plus B. And so, even though the concentrations aren&#x27;t changing because the rate of this reaction is equal to the rate of that reaction rate, one equals rate, too. So that means the amount of a plus B and A B are gonna stay the same. Um, there&#x27;s still a reaction happening in both directions. That&#x27;s why it&#x27;s called a dynamic equilibrium and chemical systems, because the actual molecules are still reacting in both directions. Is just that it? When you look at the system, it appears that nothing is changing. There&#x27;s no change in concentration or volume. And so um yeah, that&#x27;s that&#x27;s how we can explain the fact that even though there&#x27;s a huge amount of A and B available at equilibrium is the concentration of A B isn&#x27;t going up because A B is still reacting to become a plus B at this.

Samantha Grieco · Answer

so we know that for an equilibrium constant K, the value will always be the same for a system at equilibrium unless we change the temperature. So if I change the concentration of the reactant since or products in a system in equilibrium, the value of K does not change. What does happen is the concentration of your reactant and your products will change accordingly in order to maintain the same value of K so lush. Aaliyah&#x27;s principle says that if a system of equilibrium is stressed, it will work to relieve that stress and reach a new equilibrium. So the question isn&#x27;t really why doesn&#x27;t k change? What we&#x27;re really looking at is when we change the concentration of, let&#x27;s say, the concentration of X if I decrease the concentration of X with shot. Leah&#x27;s principle says that the system is going to shift equilibrium to the right. So if I&#x27;m removing X, I need to replace it means I need to make more acts. So what will happen is the concentration of X will increase as well as why, because why is also a product of equilibrium is now favoring this forward reaction because I&#x27;m favoring the forward reaction. What happens is the concentration of A and B decrease. Now the amount that the concentrations in your products increase is going to be in the same ratio, that the concentration of your reactant swill decrease. And so it&#x27;s not that the system will shift because we shot Lee, I says. So the shift. The system shifts because K must remain the same. So if I change the concentration of one of my reactions or products, I have to shift the concentration of the others in order to maintain the ratio between them. Remember, this fraction is really just the ratio between the concentration of reactant or products versus react INTs, and so changing one is going to force the others to change so that the overall ratio stays the same so the K value will not change unless you change the temperature

David Collins · Answer

the rates of nearly all reactions, except for zero order reactions are going to decrease as the reactant become products, because the rate of the forward reaction is a function of the reactant concentrations. And as the reaction proceeds, the reacted concentrations decrease as expressed by the differential rate law. As the concentrations decrease, the rate will decrease.

Steven Swee · Answer

So this problems all about equilibrium and the conditions we have here is that I really have a changing conditions. And the problem says that the Ford reaction is increased or the rate of her four reactions increased more than the rate of the reverse reaction. So in that case that we changed the reaction, conditions were driving the reaction forwards. Okay, so we&#x27;re driving the reaction for over Dr Lee Reaction forward. Then we&#x27;re gonna end up having more products here, right? Because if the four reaction is greater in Bangor, it&#x27;s faster than the reverse reaction. Then equilibrium will be established if, and only if when we have, um, more products than that of the reactant. Then at least go into this changing conditions so you can write your general Que que is equal to see your law mass action formula. But if you joined the reaction forward and we have more products than you have more proxy right and fuel reactions, that means Katie increases if you drive the reaction for them. We also have more products here than the concentration of the products also increases and the concentration of price increases. Then the relative concentration of the reactant CE we&#x27;ll end up decreasing. So this is a result that we have. If we came in reaction conditions and drive the reaction forward, you see the equilibrium constant increasing becomes a concentration of progress is increasing and the concentration of the reactant decreasing.

Anatole Borisov · Answer

So whenever, um, reaction reaches equilibrium, the rate of the Ford reaction rate forward is equal to the rate of the reverse reaction. What that means is like, for instance, we have a plus B, just two random molecules reacting to form Maddock a B. This in the equilibrium reaction, the rate of a plus B becoming a B will be the same as of a B breaking apart to form a plus B once it reaches equilibrium. And so what that means is there is always me the same fruity and a pose. Differences. There&#x27;s always the same amount leaving as there is convenience. The concentration of a plus B is going to be constant. And that&#x27;s what that&#x27;s an important fact. Remember that equilibrium? It appears as if there&#x27;s no change in the amount of multi. So there there is really no net change. But the specific

Charotte Mcadam · Answer

so the concentration off products on reactant are constant at equilibrium. So, however, we do need to remember that this doesn&#x27;t mean that our products and reactant are the same. They just remain constant. So for some reactions, most off the reactors will be converted into products. However, if we take a look at other reactions, small amounts of reaction reactant will be converted into products.

Problem

Ammonia, a very important industrial chemical, is…

Problem 116 Medium Difficulty

When a reaction system has reached chemical equilibrium, the concentrations of the reactants and
products no longer change with time. Why does the amount of product no longer increase, even though
large concentrations of the reactants may still be present?

Answer

Related Courses

Introductory Chemistry

Related Topics

Discussion

Top Chemistry 102 Educators

Lizabeth T.

Nadia L.

Stephen P.

Allea C.

Chemistry 102 Courses

Liquids - Intro

Comparison Of Gases Liquids And Solids

Recommended Videos

Watch More Solved Questions in Chapter 17

Video Transcript

Steven S. Zumdahl, Donald J. DeCoste

Introductory Chemistry

Lizabeth T.

Nadia L.

Stephen P.

Allea C.

Liquids - Intro

Comparison Of Gases Liquids And Solids

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Introductory Chemistry

Lizabeth T.

Nadia L.

Stephen P.

Allea C.

Liquids - Intro

Comparison Of Gases Liquids And Solids

Steven S. Zumdahl, Donald J. DeCosteIntroductory Chemistry

Lizabeth T.

Nadia L.

Stephen P.

Allea C.

Steven S. Zumdahl, Donald J. DeCoste

Introductory Chemistry