The Equilibrium State - Online Tutor, Practice Problems & Exam Prep

Chemical reactions often reach a state of equilibrium where reactants and products coexist, indicated by equal rates of the forward and reverse reactions. The equilibrium constant, K=[products][reactants], helps determine the favored direction of the reaction. If K > 1, products are favored; if K < 1, reactants are favored. Understanding these concepts is crucial for predicting reaction behavior and calculating equilibrium concentrations.

The Equilibrium State

When it comes to chemical reactions, we're going to say that many of them never go to completion. This means that reactants do not completely convert into products. As a result, reactant concentrations do not go down to 0. We're going to say that these types of chemical reactions reach a chemical equilibrium in which the forward and reverse reactions are happening. What this means is, for example, we have, 3 moles of H₂ gas reacting with 1 mole of N₂ gas to produce 2 moles of ammonia gas as a product.

What happens is initially, all we have are reactants. Over time, those reactants will break down in order to recombine to form this product. We're going to say moving in this forward direction to make the product, we have associated with it our rate constant in the forward direction. When enough time has passed, the products have built up to a sufficient number, it will begin to break down and move in the reverse direction to reform our reactants. That has associated with it the rate constant for the reverse direction.

Now when it comes to these rate constants, they're tied to the rate. Here in our forward rate and our reverse rate, we're going to say initially again we only have reactants, so we're going to start out with our forward rate in which reactants break down to form products. Realize here that over time this forward rate will decrease. That's because we're having fewer and fewer reactants that can break down, so the forward rate is slowing down. At the same time, we have no reverse rate initially because there are no products, but once our products reach a sufficient amount, it'll begin to break down and reform our reactants.

When that starts to happen, we start to get a reverse rate. Over time, both of these rates will reach a number in which they are equal. When the rates become equal in both the forward and reverse direction, that is when equilibrium is established. Now with equilibrium, we have our capital K variable here which is our equilibrium constant. The equilibrium constant is equal to your rate constant in the forward direction divided by your rate constant in the reverse direction.

It is also equal to products over reactants. Now remember, in the latter part of chemistry, gen chem when you guys took it, remember what's associated with our equilibrium constant is our equilibrium expression. Here we ignore solids and liquids. We're going to have here NH32 / H23 * N2.

So this would represent our equilibrium expression or equilibrium equation. Now, if we had actual values for the products and reactants, we could plug them in and find a numerical value for our equilibrium constant K. By knowing what that value is, we can determine which side of the equation is favored. Is the product side favored, meaning that our reaction will move more in the forward direction to make more products, or is it favored in the reverse direction? Now if we're going to talk about that, we have to take a look here at these 3 images.

Now in the first image, realize here we have our reactant A, our product B. We start out with some reactants initially, no products initially. Over time our reactants break down to form products. Realize here because of that our reactants over time are decreasing while our product is increasing. Around the 5-minute mark, we can see that their amounts reach a plateau where they stop changing.

It is at that exact moment at 5 minutes where equilibrium is established. Here again remember K equals products over reactant so that's B over A. At equilibrium, it looks like B is equal to around 8, A is equal to around 1. So K here is 8. We're going to say here when K is more than 1 or greater than 1, that means that the forward direction is favored and products are favored because if you're favored in the forward direction, that means you're moving in the forward direction more so to help make more products.

In this one here, we can see that our reactants are still decreasing, products are still increasing. But at the end when equilibrium is established, we still have more reactant than we do product. Here we'll say this is around 6 it looks like and then B here looks like it's around just over 4. In this situation, you can see that your K is less than 1 and that means that the reverse direction is more favored and that means your reactants are favored. So again, the magnitude for K, whether K is equal to or greater than 1, it means that products are favored.

If it's less than 1, that means reactants are favored. In this last image at 5 minutes, they both meet at around 5. So K here equals 1 which means that we are at ideal equilibrium. So when K equals 1, that's when we're at ideal equilibrium. So it's equal to 1.

So no direction is favored and products and reactants are both favored. Just realize that when it comes to the equilibrium of a chemical reaction, it's moving both in the forward and reverse direction. Because the rate of the forward direction and the rate of the reverse reaction gets to a value that's equal after x amount of time, at that moment chemical equilibrium can be established. As we go deeper and deeper into chemical equilibrium, we'll talk about additional topics such as equilibrium concentrations of reactants and products at the end, ice charts of course, as well as other important topics. But fundamentally remember that equilibrium cannot be established until the rate of the forward and the rate of the reverse reactions gets to an equal value.

Equilibrium State Calculations 1

Here we're told for the following chemical reaction, nitrogen gas plus oxygen gas produces 2 moles of nitrogen monoxide gas. We're told that our equilibrium constant Kc is 3.7 times 10 to the negative 5. Our rate constant for the forward direction is 2.5 times 10 to the negative 3, and our rate constant for the reverse reaction is 67.57. Now they're saying the addition of a catalyst increases the forward rate constant to 1.8 times 10 to the negative one. What is the new reverse rate constant after the addition of the catalyst?

So remember, the equation is our equilibrium constant Kc equals our rate constant in the forward direction divided by our rate constant in the reverse direction. Now when it comes to our rate constants, they can be affected by a few factors. They can be affected by the addition of a catalyst. They can be affected if we increase or decrease our reactant concentration. They can be affected by increasing the surface area of our reactants. They can also be affected by a change in temperature. The addition of a catalyst here is the direct factor in changing my forward rate constant from 2.5 times 10 to the negative 3 to 1.8 times 10 to the negative one. Now when it comes to our equilibrium constant though, so capital K, the only factor that can affect its value, making it different, would be temperature. Because we never mention a change in temperature here, that means that our equilibrium constant Kc will remain the same value.

It's still going to be 3.7 times 10 to the negative 5 even after the addition of a catalyst. Here, the rate constant in the forward direction we're told is 1.8 times 10 to the negative one. And all we're doing here is solving for the rate constant in the reverse direction. So we're just solving for KR there. All we have to do is isolate KR, so multiply both sides by KR.

So multiply both sides by KR. So KR times my equilibrium constant equals the rate constant in the forward direction. And now, we just need to divide both sides here by our equilibrium constant to isolate the rate constant in the reverse direction. So when I do that now, KR from my reverse direction equals 4.86 times 10 to tthh3ree as my final answer. So just remember the factors affecting the rate constant are these four.

Whereas to affect our equilibrium constant, all that we can do is affect temperature and that will cause a change in my equilibrium constant. We'll talk in greater detail on if this increases or decreases my equilibrium constant when we discuss Le Chatelier's principle. Also, what you should keep in mind in terms of this question is the addition of a catalyst not only increased the rate of the forward reaction but also affects the reverse reaction. So catalyst affects both the reverse and forward directions in a chemical reaction. Now that you've seen this, go on to the next video and take a look at example 2.

If you want, you can attempt it on your own. But again, if you get stuck, don't worry. Just come back and see how I approach example 2.

Equilibrium State Calculations 1

Consider the following reactions at 25 degrees Celsius. So in reaction 1, we have 2 moles of NO decomposing to produce N₂ gas and O₂ gas. Here it has a K value of 1×10³⁰. Reaction 2, we have 2 moles of water breaking down to give us 2 moles of hydrogen gas and 1 mole of oxygen gas. It has a K value of 5×10^-82.

And then finally, we have 2 moles of carbon monoxide plus 1 mole of oxygen gas combining to give us 2 moles of carbon dioxide gas. Here it has a K value of 3×10⁹¹. Now here it asks which compound is most likely to dissociate and give O₂ gas as at 25 degrees Celsius. So, we have a compound that's going to dissociate or break up to produce O₂. O₂ has to be a product. When we look at our 3 equations, automatically the third equation is out because in this one, O₂ is a reactant, not a product.

Now, we have narrowed down the answers to either reaction 1 or reaction 2. Since we want to create O₂ as a product, realize here that our equilibrium constant or K value looks at the magnitude of our equation. Basically, the larger your K value, the more products you can produce. Here in reaction 1, this equilibrium constant value is really large, which means that we will produce a great amount of product.

So we'll produce a lot of O₂ gas. What helps to produce that O₂ gas? Nitrogen monoxide breaking up helps to produce that O₂ gas. The answer here would be option B. Reaction 2 doesn't work because here its equilibrium constant is much smaller than 1, which means that reactants are much more greatly favored.

We're not going to produce much product at all. Very little O₂ would be produced. That's why H₂O gas could not be an answer. Remember, this all hinges on the fact that K, your equilibrium constant, equals products over equilibrium constant equals products over reactants. If your K is larger than 1, then products would have to be a much larger number than reactants.

That's what makes reaction 1 the best choice and that's why NO gas is the correct compound which will break up to produce the most amount of O₂.

Write the equilibrium expression for the following reaction.

4 NH₃ (g) + 3 O₂ (g) ⇌  2 N₂ (g) + 6 H₂O (l)

• A.
• B.
• C.
• D.

Equilibrium State Calculations 2

When reactions 1 and 2 below are added together, the result is reaction 3. We're asked to find the equilibrium constant for reaction 3, which is K₃. Now, the way we approach this question is the way we would approach a question if we're dealing with Hess's Law. Our goal is to isolate this equation here. And the way we isolate that equation is by manipulating the other equations that have known equilibrium constant values.

All right. We're going to go 1 by 1 and see where we find each one of these compounds. Here we need 1 mole of nitrous acid as a reactant. If we look up above at the reactions 1 and 2, here we have 1 mole of nitrous acid as a reactant. It matches exactly with the 1 mole that I need here.

That tells me that I cannot manipulate reaction 1 at all, so I leave it as is. Next, I need 1 mole of hydroxide ion as reactant, and here it goes here. Again, it matches perfectly, which means I don't do anything to this equation. I don't manipulate it at all. And then we can see here that my 1 mole of NO₂^- as a product that I need is right here.

And my 1 mole of water as a liquid as a product, I need 1 mole. Here I have 2 moles of it as products, but then I also have 1 mole of it over here as reactants. What's going to happen is that this one's going to completely cancel out with 1 from here, leaving me with one left. Everything that hasn't canceled out would come down. So this would come down to give me this.

This also cancels out with this. This would come down to give me this. This would come down to give me this, and this would come down to give me this. I've canceled out everything that I can cancel out, and I've isolated the equation that I need. All we say at this point is to find K₃, we would say K₃ equals K₁ times K₂.

So we'd multiply 4.50 times 10 to the negative 4 times 1.00 times 10 to the 14. When we do that, it gives me a K₃ value of 4.50 times 10 to the 10. This one was pretty simple. We didn't have to manipulate the equations at all. All we had to do was cancel out intermediates, which are compounds that look alike, except one is a reactant, one is a product and then just multiply your equilibrium constants to find the missing one.

In the bottom example, we will have to do some types of manipulations to equations to get the desired final equation. In those cases, we'll see what happens to our equilibrium constants when we manipulate the equations as a whole. So, guys, keep in mind the simple things that we did here as we move on to the next example in the next video.

Equilibrium State Calculations 3

Here we're asked what the equilibrium constant for the reaction is. The reaction we're trying to isolate is 1 mole of ammonia plus 1 mole of hydronium ion produces 1 mole of ammonium ion plus 1 mole of water as a liquid. We have to manipulate these two equations because they have known equilibrium constant values. We're going to go 1 by 1 and see where we find each one of these compounds from the two equations below. I need 1 mole of NH₃ as a reactant.

Here goes our 1 mole of NH₃ as a reactant. It matches up perfectly with the one I need to isolate here. It means I cannot manipulate that equation at all. I leave it as is. Next, I need 1 mole of H₃O⁺ as a reactant.

If I look at the 2 equations below, I have what? I have 2 moles of H₃O⁺ as products. I need to do 2 changes. I first need to reverse the reaction so that it becomes a reactant like I need it to be here. I also need to divide by 2 because here there's only 1 mole of H₃O⁺ but here we have 2.

All right. So we're going to reverse. And when we reverse it, we're going to get 2 OH^- aqueous + 2 H₃O⁺ aqueous gives us 4 H₂O liquid. Now, when I reverse the reaction, that's going to give me the reciprocal of my equilibrium constant K. So now my K_c becomes 1 over 1.0×10-28.

Now this is gone because it's been manipulated. Again, we have H₃O⁺ as a reactant like I want but I need only 1 mole of it. What I have to do here is I'm going to have to divide by 2. When it comes to your equilibrium constant K, we tend to associate things in terms of multiplication. If I'm going to divide by 2, it really means that you're multiplying by 1 half.

We have to do it in terms of multiplication because that will help us understand what K is going to be. All right. So again, I have to divide by 2 because I need only 1 mole of H₃O⁺ like I have 1 mole of H₃O⁺ up here. So now I multiply it by a half. So that's OH^-, aqueous, + H₃O⁺, aqueous, gives me 2 H₂O liquid.

Now, whatever you multiply by becomes the power. So I multiply by one half, so it becomes 11.0×10^-281/2. Since I multiplied by half that becomes the half power. Now half power is the same thing as square root. So you could also say it just means the square root of 1 over 1.0×10-28.

So that also works. All right. So now this is gone because it's been manipulated to this final equation. Now I have 1 mole of H₃O⁺ as reactant just like I have 1 mole of H₃O⁺ here as reactant. All right.

Next, I need 1 mole of NH₄⁺ as a product, so it matches up perfectly here. And I need 1 mole of H₂O liquid as a product. Here I have 2 moles of H₂O liquid here as products. And I have 1 mole of H₂O liquid as reactants. They look alike except one's a product, one's a reactant.

That is not allowed. So this one's going to cancel out with 1 from here leaving us with exactly 1. Who else can we cancel out? Well, here we have 1 mole of OH^- as reactant. Here we have 1 mole of OH^- as product.

They're going to cancel out. And then we look to see what's left. So what we have left is NH₃, H₃O⁺, NH₄⁺ and 1 mole of H₂O. Those compounds that are left help to recreate this equation that I need to isolate up here. So I know I've done it correctly.

Now, if we say that this is K_C1 and this represents K_C2, then K_c3 here would just be K_c1 times K_c2. So that's 1.8×10-5 times So when we take the square root of 1 over 1.0×10-28, that's going to give you 1.0×1014. So that's K_c2. So plug that in. So at the end, we're going to get as our answer 1.8×109, which gives me option A as my answer.

So remember, when it comes to questions like this, we need to isolate the equation with the unknown equilibrium constant value by manipulating the other equations that have known equilibrium constant values. When you reverse a reaction, we get the reciprocal for k. When you multiply by a number, that becomes the power. If we're dividing by anything, we have to understand it in terms of multiplication. So that way we can apply it to the rule of it becoming the power.

So if I divide by 2, it's really multiplying by a half, which means half becomes my power. If I divide by 3, then that would mean I'm multiplying by 1 third and 1 third becomes the power, so on and so forth. Once we've isolated the equation that we needed, we've multiplied the known equilibrium constant values to get the unknown one that we're looking for. Remember these fundamental steps and you'll always be able to calculate the missing equilibrium constant for any given equation.