16. Chemical Equilibrium
Chemical Equilibrium
Most chemical reactions never go to completion where all the reactants are converted into products. Instead these reactions reach a chemical equilibrium.Â
Chemistry & Equilibrium
1
concept
Understanding Chemical Equilibrium
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Hey, guys, in this brand new video, we're going to take a look at the equilibrium state and figure out when exactly do some reactions reach equilibrium? Now, first, we're gonna say that most chemical reactions never go to completion When we say the word completion, all that really means is that our reactant since do not completely convert into products a reaction that goes to completion. Every amount of reactant that's their gets broken down in order to build up to make our products. But we're going to say that this doesn't happen with all chemical reactions, so there's always gonna be some amount of reactant left. Now, that amount could be a very small amount or could be a very large amount. It all depends on the chemical reactions. Now, since are reacting remount is always gonna be some amount. We're gonna say reaction concentrations do not do not go down to zero Now we're going to stay. Instead, these chemical reactions reach a state of chemical equilibrium in which the reaction moves in the forward and the reverse direction. Remember, we talked about this. We see that these reactions have double arrows. One going in the forward one going in the reverse, we're gonna say because they can go in the Ford or the reverse direction, they're called reversible reactions. Here we have a typical reaction. We have reacted a going in the four direction to give us product be and just realize we're going in the four direction here with this forward facing arrow. And we use K one to signify this forward direction. Now, in the same way, the reaction could go in the opposite direction where some of our product actually breaks down itself in order to rebuild some of the reactant that was lost. When we're going in the reverse direction, we use K minus one. The minus signifies that we're going in the reverse direction. Now, if we're gonna look at this plot here, you can see on this plot that are Blue Line represents our reacted A Our red line represents our product be. And what you should take from this chart is this We're gonna be losing reacting, so we lose reactant to make products. So that's the most common theme that we have to take from this image. And what you should realize here is eventually product amount will stop increasing and reacting amount will stop decreasing and they'll reach a plateau for both of them. So they're amounts are gonna be held constant after a while. It's at this five minute mark where they become constant in their concentrations. So we're gonna say equilibrium happens at that exact moment. Okay, so that's how we're able to tell. A reaction has reached chemical equilibrium. The reactant and the product concentrations or amounts have been held constant now reactions air still going on on a molecular level. But for every amount of reacting we lose. It's OK because our products are going in the reverse direction to make up for what we just lost. So in that way, they're leveled. Maintain a same number over time. Okay, so we're gonna say once our amounts plateau, that's when equilibrium is reached. Now, knowing that, I want you guys to take a look at this practice question on the bottom and try your best to answer it on your own. Once you figure out the answer, come back to our next video and click on the explanation. But after answering and see if your answer matches up with my own good luck on this question, guys,
A reaction reaches equilibrium once the rate of the forward reaction equals the rate of the reverse reaction. Once at equilibrium there is no net change in the concentration of reactants or products.
Whereas the rate constant, k, is affected by factors dealing with the rate of reaction, the equilibrium constant, K, is affected by only temperature.
2
Problem
ProblemWhich one of the following statements does not describe the equilibrium state?
a. While at equilibrium, a dynamic process is still occurring.
b. The concentration of the reactants is equal to the concentration of the products.
c. The concentration of the reactants and products reach a constant level.
d. At equilibrium, the net concentration of all species is not changing.
e. All are true.
A
Choice A
B
Choice B
C
Choice C
D
Choice D
E
Choice E
The Equilibrium Constant K
Associated with any reaction at equilibrium is the equilibrium constant K. Its numerical value determines if reactants or products are more greatly favored within a reaction.
3
concept
The Equilibrium Constant Concept 1
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Hey, guys, In this brand new video, we're gonna take a look at the equilibrium. Constant. Now we're going to say that the variable we usedto illustrate our equilibrium. Constant is the variable K. We're gonna say K is a number equal to the ratio of products to react. It's at a given temperature. Why do I say at a given temperature? Because we're gonna say that temperature directly effects are equilibrium, constant K increasing the temperature will increase our K value. Decreasing our temperature will decrease r k value. Now we're gonna say it's a ratio of products to react in. So we're gonna say here that K equals products, overreact INTs. Now we're gonna say that K is important because it it's magnitude tells us how far to the left or to the right are. Chemical reaction is at a given temperature. So we're gonna say if K is greater than one than products, our favorite overreact. It's and think about it. This makes sense because we said K equals products. Overreactions. So let's say our products are 10 on our reactant are one k equals 10 definitely greater than one. We're gonna say when products are favorite over our reactant. That means we're making more products. How do we make more products by going in the forward direction so the four direction would be favored now in the opposite way if K is less than one? So let's say we have. In this case, products are one, but our reactions air 10 we have a K less than one. So in this case, if K is less than one than react INTs our favorite over products, which means our reaction is heading in the reverse direction so the reverse direction would be favored. So I just remember these two differences when K is greater than one when K is less than one Now, what if we say Kay was equal toe one? We know that this is products, overreact INTs. So that means that our products and reactant would have to be equal. Let's say they're both 10. 10/10 is one. So when K equals one, we're gonna say both are reacting in our product. Amounts are equal to one another. Now we're gonna say the equilibrium constant K takes into account all the states of matter except to it doesn't look at solids, and it doesn't look at liquids. It ignores those two states for matter
The equilibrium constant K is a ratio of products to reactants. It only deals with gaseous or aqueous compounds.
4
example
The Equilibrium Constant Example 1
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so it looks at it only cares about really gasses and also acquis a quiz Just mean that the solvent is water. So a compound is dissolved in water. It would be a quiz. Now, knowing this, let's take a look at at the following example it says right, the equilibrium expression for the following reaction. Here we have to end 205 Acquis gives us four and 02 acquis plus O two. Acquis. It's important to look at the phases because remember, if it's a solid or liquid, we ignore it. Now we're gonna say here K equals products, overreact INTs So we're gonna have our products is n 02 on. Remember, you have to use the coefficients in these calculations. The number in front of No. Two is a four, so that four will become the power. So it's gonna be no to to the fore times 0202 just has a coefficient of one in front of it, which we don't have to show divided by and 205 again, the coefficient in front of end 12 5 is too. So that becomes a power. So we would say that this represents our equilibrium, expression or are equilibrium equation Same thing. Equilibrium, expression, equilibrium, equation. Now that we've seen this one, let's look at B okay, for be we're gonna say here, we're gonna ignore this compound because it's a solid and we're gonna ignore this compound because it's a solid here. We're gonna say K equals products. Overreacting. It's like before, but here's the thing we're gonna say we don't have any products available, but you have to put something on top. We're gonna say it's equal toe one solids and liquids are ignored, and in place of them, we're gonna put one. We're going to say this because things such as pressure don't really affect solids and liquids like they do gasses and acquis compounds. That's why they're equal toe one because their numbers being held constant. Okay, so we're gonna replace solids and liquids with the number one on on the bottom. We'd have 02 now differently your professor could give you this is we know there's a one here, so this could also be or in verse one. So be aware of both types of answers. Both are correct. Both are saying the same thing in verse. One just means one over whatever it ISS for the next one. Again, we ignore solids. The only thing that we don't ignore is this gas. So here K equals products. Overreact INTs, so just equal x e cute because of the three over one. But we don't need to put the one because anything over one is the same exact thing. So that would be our equilibrium expression. So it was pretty simple. But remember the differences in B. N. C. We ignore the solids and liquids. Now that you've seen this, I want you guys to attend to answer a question that follows this one. The practice one. Here. I want you to tell me who's greater in amount is a products or is a reactant from that. You have to remember, what do we say about K when it's greater than one who's favored when it's less than one who's favored when it's equal toe? One who's favored remembering that will be a great way for you to approach this problem. Good luck, guys.
5
Problem
ProblemState which is greater in amount:reactants or products, based on the given equilibrium constant, K.
A
a) products, b) products, c) reactants
B
a) reactants, b) products, c) reactants
C
a) products, b) reactants, c) equal
D
a) reactants, b) products, c) equal
6
Problem
ProblemThe decomposition of nitrogen monoxide can be achieved under high temperatures to create the products of nitrogen and oxygen gas.
6 NO(aq) ⇌ 3 N2(aq) + 3 O2(aq)
a) What is the equilibrium equation for the reaction above?
b) What is the equilibrium expression for the reverse reaction.
A
B
C
D
7
Problem
ProblemThe equilibrium constant, K, for 2 NO (g) + O2 (g) ⇌ 2 NO2 (g) is 6.9 x 102.
What is the [NO] in an equilibrium mixture of gaseous NO, O2, and NO2 at 500 K that contains 1.5 x 10 –2 M O2 and 4.3 x 10 –3 M NO2?
A
4.15 x 10-4 M
B
1.34 x 10-3 M
C
2.04 x 10-2 M
D
1.80 x 10-6 M
Kp and Kc
When dealing with equilibrium units in terms of pressure the equilibrium constant of Kp must be used, but if the equilibrium units are in molarity then we must use the equilibrium constant of Kc.
8
concept
The Different Types of Equilibrium Constants
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Hey, guys, In this new video, we're gonna take a look at the different types of equilibrium. Constants. Now, one thing we're gonna realize by going through chemistry one in chemistry to is that K is a very, very popular variable. We have it everywhere. We have our rate constant K. We have actually a K for acids and bases. We call them K a N K B K A. Is called our acid association constant, whereas K B is called Are based association Constant. Now we're talking about different types of equilibrium. Constants. So K is really, really popular. And when it comes to calculations in chemistry now we've talked about the equilibrium constant. Up to this point, we've said that it's equal toe products. Overreact. It's and we should realize, is that we have different types of equilibrium constants based on the units that we're using. Now we're going to say when dealing with gas is, we usually use the equilibrium constant K p. We're gonna say KP uses the partial pressure unit off atmospheres, and we're gonna say when dealing with a quiz salutes remember, acquis means that water is dissolving my compound. So when we're using Equus salutes, we tend to use the equilibrium constant Casey. Now Casey uses the concentration unit off polarity E. Remember, Polarity is moles over leaders. And also remember, polarity is the same thing as concentration and the same thing as Saul. You bility. So we've seen that we have K, p and K C now, and we're going to say that these two are related together. By this equation, KP equals Casey rt to the Delta end here. We're going to say that our is called our gas constant and it equals 0.8 to 06 leaders times atmospheres over moles Times K. Now remember, there's another are variable out there. It's 8.314 Jules over moles times K. But we only use that one when we're dealing with energy, velocity or speed. In this case, we're not dealing with energy, velocity or speed. We're dealing with equilibrium concentrations. So that's why we're using this particular are we're gonna say that t represents our temperature in Kelvin and then we're gonna say, most importantly, is our delta and exponents. Delta End equals moles off gas as products minus moles off gas again. Remember gas as react INTs. So we started out with the broad idea of an equilibrium constant equilibrium, constant K. But now we've delved in deeper to K, P and K C.
Kp and Kc are related to one another by the following equation below:
9
example
For the reaction above, the Kc = 4.9 x 10-9 at 25oC
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now knowing the connection between them. Let's take a look at the following example and try to answer the question. So in this question, it says for the following reaction to a solid plus three. B gas gives us to see gas. This equation has a Casey value off 4.9 times 10 to the negative nine at 25 degrees Celsius. Now we're saying which of the following statements is true, so I'm gonna take a look at the first one. It says the reaction is favored in the in the Ford direction. Remember, we've seen this type of terminology before. When is it favorite In the four directions? It's on Lee favorite in the four direction when K is greater than one. But we can see that based on the Casey, Value K is a lot smaller than one, so the four direction would not be favored. Next, we're gonna say the concentration of the products is greater than the concentration of the reactant. Again, this would Onley happen when K is greater than one. So this is also out. See, the reaction is favored in the reverse direction. This is true when K is less than one. And here are K. C is times 10 to the negative nine. So it's incredibly small, So the reverse direction is highly favored, so C is definitely true. Now let's take a look at the last one D. It says the value of KP will be larger than the value of K C. Now we know that the equation that connects them is k p equals K. C are tee to the Delta end. We know that, Casey, I told you, is 4.9 times 10 to the negative nine. Our is our gas constant, which doesn't change. Our temperature is 25 degrees Celsius. So we add 2 73 15 to that to get to 98. 15 Calvin when it's times to 98.15. Calvin, remember, we have to figure out what Delta and is here. So remember Delta and is gonna look at our e r products as gas. We have two moles of gasses products minus this Here is a solid. We don't look at solids. We're only looking at the gas minus three moles of gas so delta and is minus one. So when we punch that into our calculators we're gonna get back to 0.0 times 10 to the negative 10 and you can see that KP is actually smaller than Casey, So D is also false. Hopefully, guys were able to remember the relationship for our equilibrium. Constant its products overreact. INTs. When K is greater than one, the four direction is favored, and our products are greater in amount than react. It's when K is less than one than the reverse is favored, and our reactant will be greater than our products. Now that we've seen how to do this one, I want you guys to attempt to do this one on your own. So in this question, I want you to figure out what Casey is. I give you all the equilibrium amounts for each compound so we can use products overreacting to help us get to Casey. See if you guys can answer this, then come back and click on the explanation button and see what I do to solve the same exact question. Good luck, guys.
10
Problem
ProblemMethane (CH4) reacts with hydrogen sulfide to yield hydrogen gas and carbon disulfide, a solvent used in the manufacturing rayon and cellophane. What is the value of Kc at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm methane, 0.15 atm hydrogen sulfide, 0.30 atm carbon disulfide and 0.10 atm hydrogen gas?
CH4 (g) + 2 H2S (g) ⇌ 4 H2 (g) + CS2 (g)
A
9.90 x 10-5
B
6.7 x 10-4
C
9.90 x 10-7
D
0.0067
Determination of the value for Δn can help us determine if Kp is greater than, less than or equal to Kc.
11
Problem
ProblemIn which of the given reactants is Kp greater than, less than and equal to Kc?
a) SO3 (g) + NO (g) ⇌ SO2 (g) + NO2 (g)
b) P4 (s) + 5 O4 (g) ⇌ P4O10 (s)
c) 4 NH3 (g) + 3 O2 (g) ⇌ 2 N2 (g) + 6 H2O (g)
A
a) Kc > Kp, b) Kc < Kp, c) Kc = Kp
B
a) Kc = Kp, b) Kc < Kp, c) Kc > Kp
C
a) Kc < Kp, b) Kc = Kp, c) Kc > Kp
D
a) Kc = Kp, b) Kc > Kp, c) Kc < Kp
12
Problem
ProblemGiven the hypothetical reaction 2 A (s) + ? B (g) ⇌ 3 C (g), Kp = 0.0105 and Kc = 0.45 at 250 degrees Celsius. What is the value of the coefficient of B?
A
4 moles
B
3 moles
C
2 moles
D
5 moles
Hess's Law and the Equilibrium Constant K
The old concept of Hess's Law can be applied to our new concept of the Equilibrium Constant K.
13
example
The equilibrium constant K for the first reaction is 6.83 x 10-12 at 1000 K
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Hey, guys, In this new video, we're gonna take a look at Hess's law. At least kind off. Now, we're gonna say that when it comes to Hess's law, we learned that the total entropy change. So total delta h of a reaction by taking into account each individual step. Now, we're gonna do the same exact thing. Except instead of using Delta H, we're gonna use our equilibrium, constant K. So they pay very close attention to what I do here for these two particular examples. Now we're gonna say the equilibrium constant for the reaction. This reaction here is 6.83 times 10 to the negative, 12 at 1000 Kelvin. Now I'm asking you to find a new K for this new reaction. Now, what you should realize here is that both of these reactions are the same. They both have CO two. They both have CEO. They both have 02 for the same exact equation. The only difference here is I changed and manipulated the one on the top so that it now looks like this one. So what we have to figure out is what changes that I do and what effect will that have on the equilibrium? Constant K. So we're gonna say first Co two was a reactant, but now it's a product. These guys were products. Now they're reacting. So it looks like I reverse the reaction and just realize when I reverse a reaction, K becomes the reciprocal. So K was 6.83 times 10 to the negative 12 while now it becomes 1/6 0.83 times to the negative 12 it becomes the reciprocal of whatever it waas. Now, that's one change that I did. The other change that I did. It looks like I multiplied by four. Because here this CO two has a four. Now, this CEO has a four now. And when we do four times a half the two councils out with this four and that's how we got this to here. Now, when you multiply by a number, that number becomes the power. So this is gonna become to the fourth. So it's gonna become 1/6 0.83 times 10 to the negative, 12 to the fourth. Okay, so when you multiply by number, it becomes a power. So what we're gonna do here is just solve for what's inside 1st. 1/6 0.83 times, 10 to the negative 12. That gives us 1.46 times 10 to the negative 11. Well, actually to the positive 11. And now that's to the fourth. So that's going to give me 4.60 times to the 44 don't. So that will be our answer here. So remember, if you reverse the reaction, then we're gonna say that becomes the reciprocal one over whatever. When we multiply by a number, that's gonna become the power you multiply by five, it becomes power five. Let's say we divided by a number. Let's say we divided by three. Well, just remember, if you're dividing by three, that is the same as multiplying by one third. And if you're multiplying by one third, then one third becomes the power, let's say you have to divide by two. If you're dividing by two, that's the same thing as multiply by a half. And in that way half becomes the power. Okay, so that's how we get around the dividing aspect of this
Reversing a reaction gives us the reciprocal of K, while multiplying a reaction by a value makes it the power for K.
14
example
Calculate the rate constant, Kc, for the reaction below
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So here this one's a lot more complicated. What we're gonna say here is we need to find the K value Casey value for this equation here. We're gonna call this K C four. This will be K C 12 and three. Since we're gonna need some room to do this, guys, I'm gonna take myself out of the image so we have more room to work with. What? So our best strategies for a question like this is, since we're trying to isolate that equation above, we have to look and see. Where do we see each one of these elements? So what we need to do first is find h gas and we're gonna notice that we see h gas right here. Now we have to manipulate this one in red, so it matches up with this one in blue. So what to changes? Do I need to do to it first? The H in red is a product, but we needed to be reacted. So we need to reverse the reaction once we reverse it. We also need to realize that there's a two in front of this H But over here in the blue box there, Isn't it a two. So we need to divide by two. And we just said dividing by two is the same thing as multiplying by half. So we have to reverse it and multiply by a half. Remember, dividing by two is the same thing is multiplying by half. So now our equation becomes H gas gives me half H two gas. When we reverse it, Que becomes 1/11 0. and because we multiplied by a half, half becomes the power. And just remember, half power really means square root. Okay, we're going to say that this right here got manipulated so we no longer care about it. And remember, we did all of this so that the red H gas that we have matches up with the H gas in the blue box. We're basically trying to manipulate these three equations so that they look like the compounds in the blue boxes above next we're gonna see where do we see br br gas here in the blue box? It's a reactant, but over here it's a product also here in the blue box, it doesn't have a coefficient in front of it. But here it has a two So the two changes are the same We need to reverse it So it becomes a reacted to and we need to divide by two which is the same thing as multiplying by a half. So this becomes BR gas gives me half br to gas So we reversed it So it becomes the reciprocal and we multiplied by a half so half becomes the power Remember half powers, the same thing A square root Finally hbr where do we see HBR? Luckily, we c h b r and only one place We see it right here as a product So they're both products so that's good. So there's no need to reverse but there's a two in front of that HBR So what do we need to do? We need to divide by two or same thing is multiplied by a half So this is gonna become half h two gas plus half br to gas gives me hbr Since I multiplied by half this is to the half power half powers, the same thing a square root. Okay, What we're gonna do next is we're gonna cross out intermediates things that look alike but our but is a product in one equation and are reacting in another. So we're gonna say here this h two is a product. This H two is a reactant. So they're going to cancel out this half Br two is a product, but this half br two is a reactant. They cancel out whatever doesn't get canceled. Out comes down. So what did you canceled out? H gas didn't get canceled out be our gas didn't get canceled out hbr gas and get canceled out and look, the equation we just isolated matches up with the equation above. Now all we have to do is we're gonna say K c four for missing equation equals K C one times k c two times k c three. So, basically, all we have to do here is we have to multiply each one of these right here. So multiply this times this times this multiplying all three of them will give us a K C four value of 4526. Hopefully, you guys were ableto follow along as we did this problem. Just realize the equation with the missing Casey value is the one we have to look at And we just looked compound by compound, look to see Where do we find them? In the given equations below When when were able to isolate them, you would just multiply all their K values together. And that will give us the overall K value for this equation we're looking for in the first place. So I just remember the tips that we did. If you reverse it, K becomes a reciprocal. If you multiplied by a number that becomes the power. If you divide by number, then it becomes one over that number as a power. So if you divided by five would be multiplying by 1/5. If you divided by four, it be like multiplying by 1/4. Those would become the powers. Just remember those small little principles, and you'll be able to get these questions correct.
15
example
Hess's Law & Equilibrium Constant K
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Hey, guys. So here we're gonna apply the thermal chemistry concept of Hess's law and applied toe are variable equilibrium, constant K. So here they tell us that the value for KP is 0.39 for the following reaction. Now they're asking is what is going to be the new KP for this new equation at the same temperature. Realize that this equation here is this same equation given to us down here. Except this second equation has been manipulated to be in this form. So let's take a look at the first equation and realize what the changes that occurred. One change that occurred waas. We reversed the reaction. No. Two was a reactant. Now it's a product and tool for is a product. Now it's a reactive. So we reversed the reaction. And remember, when you reverse the reaction, we don't switch the sign of K, we do the inverse of K. So here KP was 0.39 because we reversed, it becomes 0.39. The next thing we need to realize is that we divided this by three. So that's how it became two thirds and one third. But instead of saying dividing by three, we can say we multiplied by one third. So again, dividing by a number means you're multiplying by one over that number. Now, why do we need to talk about it in terms of multiplication? That's because whatever number we multiply our equation by becomes the power. So if you multiply by one third, that means you're taking Kate to the one third power and you're one third power just means your cube root. So you're gonna do one divided by 10.39 take the cube root of that number for some of you. You may see this on your calculator. You may see this button. So to do cube root you would hit you had three than that button, and then you would do parentheses one divided by 10.39 close parentheses for others. You may not have that button. So instead, what you should do is you should do 1. to the one third, so raise it to that power now, whichever one you do, what you get as your answer at the end would be 1. as the new K for your equation So remember, if you are are reversing the reaction, you get the reverse sign or not the riverside. If you reverse your reaction, get the reciprocal Sorry. So if you reverse your reaction, you get the reciprocal. If you multiply by a number, it becomes the power. Remember, divided by number means that you're multiplying by one over that number. So multiplication creates the power that that you've seen this attempt to do the next. Following example. If you get stuck, don't worry, just come back and take a look at how I approach that question. But let me give you guys some hits on how to do this. So just realize we have to isolate the K of this equation. And to do that we manipulate thes equations here. So if you're familiar with Hess's law, you should be able to do this. And remember what happens when you reverse the reaction or multiplied by a number. Once you've isolated each of these equations, you would multiply each one of these cases to get this missing K here, and that all seems too complicated. Just come back and take a look at how I do it
16
example
Hess's Law & Equilibrium Constant K
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Hey, guys. So here we have to isolate this equation. So what we're gonna do is we're gonna go one by one and see what we find each one of these compounds. So ch four, where do we see ch four? Here's ch four and matches perfectly with the one that we see up here. Which means we do not manipulate or touch this equation next. 02 Or do we co two? We find it in two places, which means that we should not do anything. If you find it in more than one place, skip it and move on. Co two, Where do we see co two? We see CO two right here again. They match perfectly with each other, which means we should not attempt to touch this equation either. And then finally, we have two moles of H 12 liquid. Here goes H two a liquid. But we have two issues. First, we needed to be a product like it is here and also we need two moles of it. So what to changes? Do we need to do well, we're gonna have to reverse this reaction in order to make each to a liquid a product and we're gonna have to multiply by two. So that has two moles of it. So when we do that, we get to H 20 gas gives me to H liquid. Remember, when we reverse the reaction, we get the reciprocal of K. So it's gonna become 1/6 0.12 times 10 to the negative four and I multiplied by two. When you multiply by a number, it becomes the power when you punch this into your calculator. So first do one divided by this number 6.12 times 10 to the negative four. Make sure this is in parentheses. When you do it, it should give you 1633. And then you square that answer which gives you 2.66991 times 10 to the six. So now this equation is gone because it's been manipulated to this new one same thing with cat and then we cross out our intermediates soon cancels out Well, we have two moles of H 20 gas which cancels out with these two moles of H 20 gas as products. So this here is called formaldehyde, which cancels out with this formaldehyde, which is a product, Remember, intermediates look alike, except one will be a reactive and the other will be a product. Now, those are the only intermediates. Everything else comes down. So ch four comes down and then these two old tools are on the same side. When they're on the same side, they don't cancel out. They add up this co two gas comes down and then are two moles of H 20 liquid come down. So we isolated the equation we needed so to figure out this K C four. Now all we do is we multiply K C one times k c. Two times, Casey three which we found earlier. When you multiply all three together, that will give you your K up 1.29 times 10 to the six. So this will be our answer. And this will be the missing equilibrium constant for our equation.
Additional resources for Chemical Equilibrium
PRACTICE PROBLEMS AND ACTIVITIES (64)
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- Ethene 1C2H42 reacts with halogens 1X22 by the following reaction: C2H41g2 + X21g2 Δ C2H4X21g2 The followin...
- The diagram shown here represents the equilibrium state for the reaction A21g2 + 2 B1g2 Δ 2 AB1g2. (a) Assumin...
- For the general, single-step reaction A1g2 + B1g2 ∆ AB1g2, Kc = 4.5 * 10-6, which of the following statements ...
- Suppose that the gas-phase reactions A ¡ B and B ¡ A are both elementary processes with rate constants of 4.7 ...
- The equilibrium constant for the dissociation of molecular iodine, I21g2Δ2 I1g2, at 800 K is Kc = 3.1 * 10-5. ...
- Write the expression for Kc for the following reactions. In each case indicate whether the reaction is homogen...
- Write the expressions for Kc for the following reactions. In each case indicate whether the reaction is homoge...
- Write the expressions for Kc for the following reactions. In each case indicate whether the reaction is homoge...
- Which of the following reactions lies to the right, favoring the formation of products, and which lies to the ...
- Which of the following statements are true and which are false? (c) As the value of the equilibrium constant i...
- If Kc = 0.042 for PCl31g2 + Cl21g2 Δ PCl51g2 at 500 K, what is the value of Kp for this reaction at this temp...
- Write an expression for the equilibrium constant of each chemical equation. a. SbCl5(g) ∆ SbCl3(g) + Cl2(g)
- Find and fix each mistake in the equilibrium constant expressions. b. CO(g) + Cl2(g) ⇌ COCl2(g) K = [CO][Cl2]/...
- Find and fix each mistake in the equilibrium constant expressions. a. 2 H2S(g) ⇌ 2 H2(g) + S2(g) K = [H2][S2]/...
- When this reaction comes to equilibrium, will the concentra- tions of the reactants or products be greater? Do...
- Ethene (C2H4) can be halogenated by this reaction: C2H4(g) + X2(g) ⇌ C2H4X2(g) where X2 can be Cl2 (green), Br...
- At 1000 K, Kp = 1.85 for the reaction SO21g2 + 12O21g2 Δ SO31g2 (c) What is the value of Kc for the reaction i...
- H2 and I2 are combined in a flask and allowed to react according to the reaction: H2(g) + I2(g) ⇌ 2 HI(g) Exam...
- A chemist trying to synthesize a particular compound attempts two different synthesis reactions. The equilibri...
- Consider the equilibrium N21g2 + O21g2 + Br21g2 Δ 2 NOBr1g2 Calculate the equilibrium constant Kp for this r...
- Calculate Kc for each reaction. a. I2(g) ⇌ 2I(g) Kp = 6.26 * 10^-22 (at 298K)
- Calculate Kc for each reaction. b. CH4(g) + H2O(g) ⇌ CO(g) + 3 H2(g) Kp = 7.7x10^24 (at 298 K)
- Calculate Kp for each reaction. a. N2O4(g) ⇌ 2NO2(g) Kc = 5.9x10^-3 (at 298 K)
- Calculate Kp for each reaction. b. N2(g) + 3H2(g) ⇌ 2NH3(g) Kc = 3.7x10^8 (at 298 K)
- The equilibrium 2 NO1g2 + Cl21g2 Δ 2 NOCl1g2 is established at 500 K. An equilibrium mixture of the three gase...
- The equilibrium 2 NO1g2 + Cl21g2 Δ 2 NOCl1g2 is established at 500 K. An equilibrium mixture of the three gase...
- Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl31g2 + Cl21g2 Δ PCl...
- Consider the reaction: N (g) + 3H (g) ⇌ 2NH (g) Complete the table. Assume that all concentrations are equilib...
- Consider the following reaction: H2(g) + I2(g) ⇌ 2 HI(g) Complete the table. Assume that all concentrations ar...
- Consider the reaction: 2NO(g) + Br2(g) ⇌ 2NOBr(g) Kp = 28.4 at 298K In a reaction mixture at equilibrium, the ...
- Consider the reaction: SO2Cl2(g) ⇌ SO2(g) + Cl2(g) Kp = 2.91*10^3 at 298 K In a reaction at equilibrium, the p...
- Consider the interconversion of A molecules (red spheres) and B molecules (blue spheres) according to the reac...
- At 900 K, the following reaction has Kp = 0.345: 2 SO21g2 + O21g2 Δ 2 SO31g2 In an equilibrium mixture the p...
- The following picture represents the equilibrium state for the reaction 2 AB ∆ A2 + B2. Which rate constant i...
- Identify the true statement about the rate of the forward and reverse reaction once a reaction has reached equ...
- At 373 K, Kp = 0.416 for the equilibrium 2 NOBr1g2 Δ 2 NO1g2 + Br21g2 If the pressures of NOBr(g) and NO(g) ...
- At 218 C, Kc = 1.2 * 10-4 for the equilibrium NH4SH1s2 Δ NH31g2 + H2S1g2 Calculate the equilibrium concentra...
- At 80 C, Kc = 1.87 * 10-3 for the reaction PH3BCl31s2 Δ PH31g2 + BCl31g2 (b) If the flask has a volume of 0.25...
- At 80 C, Kc = 1.87 * 10-3 for the reaction PH3BCl31s2 Δ PH31g2 + BCl31g2 (a) Calculate the equilibrium concent...
- For each of the equilibria in Problem 15.56, write the equi-librium constant expression for Kp and give the eq...
- At 25 C, the reaction CaCrO41s2ΔCa2+1aq2 + CrO4 2-1aq2 has an equilibrium constant Kc = 7.1 * 10-4. What ar...
- The vapor pressure of water at 25 °C is 0.0313 atm. Cal- culate the values of Kp and Kc at 25 °C for the equil...
- Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO2(aq) + CO(aq) ⇌ HbCO(a...
- For each of the following equilibria, write the equilibrium constant expression for Kc. Where appropriate, als...
- Which of the following reactions yield appreciable equilib- rium concentrations of both reactants and products...
- The value of Kc for the reaction 3 O21g2 ∆ 2 O31g2 is 1.7 * 10-56 at 25°C. Do you expect pure air at 25 °C to ...
- An equilibrium mixture of N2, H2, and NH3 at 700 K con- tains 0.036 M N2 and 0.15 M H2. At this temperature, K...
- The value of Kc for the reaction of acetic acid with ethanol is 3.4 at 25°C: CH3CO2H1soln2 + CH3CH2OH1soln2 ∆ ...
- The following equilibria were measured at 823 K: CoO1s2 + H21g2 ΔCo1s2 + H2O1g2 Kc = 67 H21g2 + CO21g2 ΔCO1g2 ...
- In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (c) What is the value ...
- In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (b) By using data in A...
- In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation ...
- For the reaction A2 + 2B ∆ 2 AB, the rate of the for- ward reaction is 18 M/s and the rate of the reverse reac...
- Consider a general, single-step reaction of the type A + B ∆ C. Show that the equilibrium constant is equal to...
- Which of the following relative values of kf and kr results in an equilibrium mixture that contains large amou...
- Consider the reaction of chloromethane with OH- in aque- ous solution: CH Cl1aq2 + OH-1aq2 ∆kf CH OH1aq2 + Cl...
- Refining petroleum involves cracking large hydrocar- bon molecules into smaller, more volatile pieces. A simpl...
- The equilibrium constant Kc for the gas-phase thermal decom-position of cyclopropane to propene is 1.0 * 105 a...
- Consider the sublimation of mothballs at 27 °C in a room having dimensions 8.0 ft * 10.0 ft * 8.0 ft. Assume t...
- Consider the sublimation of mothballs at 27 °C in a room having dimensions 8.0 ft * 10.0 ft * 8.0 ft. Assume t...
- Ozone is unstable with respect to decomposition to ordi-nary oxygen: 2 O31g2 ∆ 3 O21g2 Kp = 1.3 * 1057 How man...