Skip to main content
Loading Explore guides…
Chemical Equilibria · Acedra
Home
Explore
Chemistry
Chemical Equilibria
Sign in to save
Chat
If YouTube asks you to sign in, open the video on YouTube.
Watch on YouTube
Reload player
Transcript
64 segments
Copy
Hide video
0:00
Welcome to Unit 7 review. In this unit we will be learning about equilibria, which are systems in which there is both a forward reaction and a reverse reaction. Specifically we will discuss details regarding equilibria and reversible reactions, reaction quotients and equilibrium constants, equilibrium concentrations and associated calculations, le chatelier’s principle, solubility equilibria, the common ion effect, pH,
0:30
and free energy of dissolution. Let’s start out very simply by defining equilibria and their characteristics. Many chemical reactions that we have studied so far can be considered essentially unidirectional. There are reactants, and these get used up to form products, after which the reaction is over. But many systems do not operate like this. There exist chemical equilibria, in which there is a forward reaction and reverse reaction,
0:57
and both are happening constantly. In such a dynamic equilibrium, it appears that there is no activity, but it is only because the rates of the forward and reverse reaction are the same, so it’s as though nothing is changing. This is the case with dinitrogen tetroxide, or N2O4, which will decompose to form nitrogen dioxide, NO2, which can then recombine to form the reactant again. We can monitor this reaction because the reactant is colorless while the
1:27
product is brown, so as the reaction initially proceeds in the forward direction it will become darker, until it reaches equilibrium. At equilibrium, concentrations will become constant, and the forward and reverse rates of reaction will become constant as well as equivalent to one another, since the forward and reverse reactions must be occurring at precisely the same rate in order for the concentrations to remain constant. In order for such an equilibrium to occur, the
1:57
reaction must be reversible, and we will denote this with the equilibrium arrows, which illustrate motion in both directions. All reactions are technically reversible, but for some, one direction is preferred by such an astronomical amount that we regard it as essentially unidirectional. Some are easily reversible under certain conditions, so those are the ones that we will discuss in terms of an equilibrium. It is therefore important to always keep in mind
2:25
that in a chemical equilibrium the forward and reverse reactions never stop, they simply result in a constant concentration, which gives the illusion of inactivity. Equilibria exist in the way of physical processes too, like phase changes, such as the equilibrium between liquid and vapor for any substance in a closed container. Some particles are vaporizing and some particles are condensing at any given time, but at equal rates, so the system appears inactive. The same goes
2:55
for the dissolution and precipitation of ionic solids, both processes can be happening at once. Now that we understand what chemical equilibria are, we will want to be able to look at the concentrations of substances involved in reversible reactions and be able to predict how these concentrations will affect the equilibrium of the system. Given this general equation for a chemical equilibrium, we can write something called the reaction quotient expression,
3:23
symbolized by Q. Q will be equal to the concentrations of the products each raised to the power of their stoichiometric coefficients, over the concentrations of the reactants each raised to the power of their stoichiometric coefficients. So for the following equilibrium involving N2O4 and NO2, the reaction quotient, or Qc, since we are using solution concentrations, will be as follows. Note that NO2 concentration is squared because of the coefficient of 2 from
3:53
the equilibrium. It is important to understand that when a reaction begins, Q will be zero, since there will not yet be any products, and it will begin to increase as products are formed. But Q will eventually become constant once equilibrium is reached, as concentrations will become constant at that point, and once that happens Q becomes equal to the equilibrium constant of the reaction at that temperature. This is symbolized as Kc, and the equilibrium constant expression
4:23
takes the same form as the reaction quotient expression, except that it is a constant value, since it is derived from constant concentrations. We should be able to calculate either Qc or Kc for any set of concentrations, we simply plug them into the reaction quotient expression. If they are equilibrium concentrations, we get Kc, if they are nonequilibrium concentrations, we get Qc. Let’s note that these expressions will only contain aqueous and gaseous species,
4:54
as it doesn’t make sense to discuss a solid or pure liquid in terms of concentration.
5:01
The magnitude of Kc will tell us the yield of a reaction once it reaches equilibrium. If Kc is very large, much greater than one, it means this fraction is very large in the numerator, so products dominate the reaction mixture. If Kc is very small, much smaller than one, it means this fraction is very large in the denominator, so reactants must dominate the reaction mixture. Once we know the Kc for a reaction, we can predict which way the equilibrium will shift
5:33
given a reaction quotient for some nonequilibrium concentrations, since the reaction quotient must eventually become equal to Kc. If Q is bigger than K, it will shift left until equilibrium is reached, meaning the reverse reaction is favored. If Q is smaller than K, it will shift right, meaning the forward reaction is favored. We should also note that a reaction can begin from reactants, products, or from a combination of both. No matter what the circumstance,
6:02
the system will move so as to attain equilibrium. Now we will have to be able to perform calculations regarding equilibrium concentrations and the equilibrium constant or reaction quotient. Let’s use the decomposition of ammonia as an example to see how this works. Say we start with some amount of ammonia and allow it to decompose. We want to be able to keep track of all the concentrations as they change. As a result of the decomposition, the nitrogen
6:31
concentration increases by 0.11 molar. What would be the increase in hydrogen concentration? Well according to these coefficients of one and three, with every nitrogen molecule that forms, three hydrogen molecules form. That means whatever the change in concentration for nitrogen, hydrogen concentration will change triple that amount. So we can multiply 0.11 by three and see that hydrogen concentration will increase by 0.33 molar. Likewise, for every molecule of
7:02
nitrogen that forms, two molecules of ammonia have to decompose, due to this 2 to 1 ratio. So ammonia concentration will change twice as fast as nitrogen concentration changes, and with opposite sign. So in order for nitrogen concentration to increase by 0.11 molar, ammonia concentration must have decreased by 0.22 molar. Now let’s instead say that we did not know any specific numbers. We could still discuss relative changes in concentrations by doing algebra. Instead of saying
7:36
0.11 molar for nitrogen, let’s just say X. Then the change for hydrogen would be 3x and the change for ammonia would be -2x. So as you can see, all we are doing is applying the stoichiometry of the reaction. Now that we know how to assign these terms, we are ready to learn how to calculate parameters of an equilibrium algebraically. The most basic calculation we can do is to calculate Kc given the equilibrium concentrations of all the substances involved, or similarly,
8:07
to calculate a missing equilibrium concentration given Kc and the other equilibrium concentrations. Let’s try each of these now. First, let’s calculate Kc, perhaps for the decomposition of ammonia. This is quite simple, because we already know how to write the equilibrium constant expression, which is the same as the reaction quotient expression. That would look like this. If we happen to know all these concentration values from experimental data, we would simply
8:35
plug them into this expression to calculate Kc. But sometimes, we may have to calculate one or more of these concentrations. To do this we must learn how to use an icebox. For these, ice stands for initial, change, and equilibrium, which represents the initial concentration of each substance, the change in concentration, and the equilibrium concentration. Let’s see how we can construct an icebox for the following equilibrium, if we are given these initial concentrations.
9:08
First let’s construct the icebox, with rows for initial, change, and equilibrium. Then, we can plug our initial concentrations into the first row here. The reactants were given, and there isn’t any product at first so that will be zero. We learned how to assign variables for the second row already, we do this according to the stoichiometry. If x is the increase in product, since some of that must form, -x will be the change in each reactant, because it
9:37
is a 1 to 1 to 1 ratio for these substances. We add initial and change to get equilibrium concentrations. But beyond the algebra, we also know the equilibrium concentration of I2, so if we set this expression equal to that value, that will allow us to solve for x, which in turn allows us to assign all the equilibrium concentrations. Now we just plug all these values into the reaction quotient expression and do the arithmetic to get Kc, which will be 776. So as we can see, if we
10:13
are given all the equilibrium concentrations, or a way to find all the equilibrium concentrations, we simply plug them into this expression to calculate Kc for any equilibrium. What if instead we already know Kc, and want to use this to calculate one of the equilibrium concentrations that we don’t know? To do this, we can take the Kc value and other known concentrations, and plug them into this expression. Take for example the following
10:41
equilibrium. Let’s say we know the following value for Kc, and we also know the equilibrium concentrations of nitrogen and oxygen. What will be the equilibrium concentration of NO? First we must certainly be able to write the correct reaction quotient expression. This will be NO concentration squared, because of the molar coefficient of two, over nitrogen concentration times oxygen concentration. Then we can rearrange this to solve for NO concentration. We simply
11:11
have to bring the reactant concentrations up to the other side, and then take the square root of both sides. Now we are ready to simply plug in all the values we know and do the arithmetic. And this should be the equilibrium concentration of NO, 3.6 x 10-4. Let’s try one more for practice. For this equilibrium, here is the Kc as well as the equilibrium concentrations for nitrogen and hydrogen. What will be the equilibrium concentration of ammonia? Well
11:42
let’s first write the correct expression. Then we will need to solve for ammonia concentration, so let’s rearrange and take the square root. Then we can plug in what we know, and 1.53 molar is what we should get. Again, if we must do some algebra to attain one or more concentrations, we will simply consult the stoichiometric ratio. Now let’s learn how to calculate equilibrium concentrations given some nonequilibrium concentrations and the Kc. To do this,
12:12
we will need to follow four steps. First, we need to know the direction in which the equilibrium will move, in order to equilibrate. We can do this by comparing Qc to Kc. If Q is smaller, it will move to generate products. If Q is larger, it will move to generate reactants. This is important because we need to know which substances are going to decrease in concentration and which will increase in concentration in order to do the math correctly. Next, we can algebraically determine
12:42
the relative changes needed to reach equilibrium, which we have learned how to do, like for this equilibrium where the stoichiometry dictates that as one mole of each reactant is used up, one mole of product is generated. We can then plug initial concentrations and the algebraic relative changes into an icebox to find the equilibrium concentrations, whatever those might be. Next, we solve for the equilibrium concentrations by plugging these algebraic terms into the
13:11
equilibrium constant expression and solving for X, which allows us to get quantitative answers for all of the equilibrium concentrations involved. Let’s say for example we are looking at this equilibrium. Given the following Kc and initial concentration of PCl5, what are the equilibrium concentrations for all the substances? Well let’s build up our icebox. We know that these are our initial concentrations. For the change in concentration, since there is
13:40
only reactant to start, we will be moving in the forward direction, so PCl5 will be decreased by X, and both the products will increase by X. This is due to the stoichiometry of the equation. From these, we can write the equilibrium concentrations as follows. Now we plug all of these into the equilibrium constant expression and see what we get. So looking at the expression, let’s put in everything we have, the Kc, X for the two products, and 1.2 minus X for the reactant.
14:12
Now we just do some algebra. Let’s multiply both sides by this denominator, and then distribute this number across the parenthetical. Then let’s just get everything on one side and equal to zero. That leads to this equation. Remembering our basic algebra we can factor this to give x plus 0.4 times x minus 0.3, which means the solutions for x are negative 0.4 and positive 0.3. Since the negative value would give nonsensical results, as we can’t have negative amounts of substances,
14:46
let’s take the positive value and plug it in to our existing expressions for the equilibrium values. And these are what we will get. With some of the more tedious calculations out of the way, let’s now look at how equilibria can be manipulated by introducing Le Chatelier’s principle. Let us consider a system that has reached equilibrium, but then experiences some kind of stress that causes the system to shift
15:14
away from the perfect balance of equilibrium. This would mean that K and Q would no longer be equal. We want to be able to predict how such a system will react, and we can make that prediction using Le Chatelier’s principle. Le Chatelier’s principle says that if a system at equilibrium experiences a stress it will move in a direction so as to relieve that stress and return to equilibrium, re-establishing a Q = K relationship. It can only do one of two things
15:44
as the result of experiencing a stress. It can either shift right, producing more products, or it can shift left, producing more reactants, and we need to be able to predict the direction in which the equilibrium will move depending on the type of stress that the system is experiencing. Let’s examine the types of stress that a system can experience. The first would be a change in concentration of one of the reactants or products. If there is balance, that balance will be thrown
16:11
off if we add or remove some amount of one of the substances involved. Let’s say we add more reactant. The way that the system can restore balance is by shifting right to use up some of the extra reactant. If we remove reactant, the system will shift left to make more of it. Likewise, if we add more product, the system will shift left to get rid of some of it, and if we remove product, the system will shift right to regenerate some of what was lost.
16:40
The second type of stress involves a change in pressure for the system. If the equilibrium involves gases, a change in pressure could have a substantial effect, provided that the forward and reverse reactions may change the number of gaseous particles in the system. This can be discerned from the balanced chemical equation. Let’s say we are looking at the following equilibrium. There
17:02
are three moles of gaseous particles on the left, and two moles on the right. Let’s now say that we take the vessel containing this equilibrium and we cause it to compress. As the volume goes down, the pressure will go up, since particles are still moving the same speed as before and will therefore hit the sides more often. As a result, the system will make an effort to relieve some of the additional pressure by shifting the equilibrium towards the side with
17:28
fewer particles. In this case, that would be the right side. Favoring the side with fewer particles will help relieve some of the additional pressure that has developed. On the other hand, if the vessel expands and the pressure drops, the equilibrium will favor the side with more gaseous particles in an attempt to regain some portion of the lost pressure. Once again, this only has an
17:53
impact if the number of moles of gaseous reactant and the number of moles of gaseous product are not the same. If they were to be the same, such as with this equilibrium, there would be no ability to alleviate additional pressure one way or another as a result of shifting the equilibrium, so a change in pressure will not cause any shift. The third stress that can be placed on a system is
18:17
a change in temperature. Unlike concentration and pressure, changing the temperature will actually change the value of the equilibrium constant, but even so, we can qualitatively predict the direction the equilibrium will shift when the temperature increases or decreases. The key is to look at the change in enthalpy of the reaction. Let’s examine this reaction here, which has a delta H listed. We can see that delta H is negative, which makes this an exothermic
18:45
reaction, meaning that heat is released. Since heat is released we can essentially consider it a product of the reaction. So let’s rewrite the equation with the word heat listed as a product. When written this way, we can simply treat heat as another substance involved in the reaction, and a change in temperature will cause the equilibrium to shift just as though we were altering the concentration of a substance. For this one, if the temperature went up, there would be more heat,
19:14
so the equilibrium would shift left, to use some of the excess heat to drive the reverse reaction and balance things out. If the temperature went down, the equilibrium would shift right, to regenerate some of the heat lost. Of course the equilibrium concentrations will also change when the equilibrium shifts, so this must be taken into account, and we can describe the shift in equilibrium in terms of the increasing or decreasing concentrations of these substances
19:43
as a result of the temperature change. Looking at one more example, the change in enthalpy for the forward reaction is positive which makes it endothermic, which means heat can be treated as a reactant, since heat must be absorbed for the reaction to occur. If we cool down the system, which way will the equilibrium shift? It must shift left, to regain some of the heat lost. So we should now be able to use Le Chatelier’s principle to predict the direction an equilibrium
20:12
will shift when there is a change in concentration, pressure, or temperature. As we mentioned earlier, in addition to chemical reactions, processes like dissolution of ionic solids qualify as equilibria as well, so let’s see how solubility fits into this concept. We have probably learned that some ionic compounds are water soluble, and will completely disperse in solution, due to the ion-dipole interactions they will make with water molecules. We have
20:42
also probably learned that other ionic compounds are water insoluble, and will not dissolve at all, remaining completely in the solid phase. Beyond this, if the components of such a compound find each other in solution, they will precipitate, and leave solution completely to form a solid. This is actually a slight oversimplification, and we will now require a more sophisticated and quantitative understanding of solubility, so let’s revisit this concept.
21:09
It’s true that many ionic compounds are completely soluble, but the reality is that even the ones that we call insoluble will dissolve to a minuscule degree. Whatever the degree to which dissolution occurs for a particular substance, we can communicate this using the solubility product, abbreviated as Ksp. Let’s look at a substance like silver chloride, which we label as water insoluble. As we said, if there is an excess of it in water, it will dissolve to a tiny extent and
21:39
cause a dynamic equilibrium. Some ions dissolve as other ions rejoin the lattice. Just as with any other type of equilibrium, there must be an equilibrium constant to describe it, and that’s the solubility product. As we might guess, this will simply be equal to the product of the two ion concentrations in solution, each raised to the power of their stoichiometric coefficients, which in this case is simply one and one. Notice that we do not include the solid itself in this
22:09
expression, as we do not include solids in equilibrium expressions. Soluble salts will have Ksp values of greater than one, while salts that we regard as insoluble will have very tiny Ksp values. The smaller the constant, the fewer the ions that will be present in solution, and therefore the less soluble the substance must be. And compounds that are extremely water insoluble will have Ksp values on the order of ten to the negative thirty, negative fifty, or even smaller.
22:41
Let’s make sure we can write these solubility products for various equilibria. Take something like calcium carbonate. First, we must write out the complete equilibrium. Upon dissociating, this will form a calcium ion and a carbonate ion. Therefore, the solubility product will be equal to the product of these two ion concentrations. If we have something like magnesium hydroxide, it’s the same thing, except that dissociation produces one magnesium ion and two hydroxide ions,
23:11
which means that the hydroxide ion concentration will be squared in the solubility product expression. We can even do this for something more complicated, like apatite, which is a mineral. This will dissociate into five calcium ions, three phosphate ions, and a hydroxide ion, so the solubility product expression will look like this. In this way, a number of ionic compounds that we previously deemed insoluble should actually be categorized as slightly soluble. If we have a way
23:44
of measuring ion concentrations in solution, we can calculate the Ksp value for a given substance. Let’s say we have a saturated solution of milk of magnesia, which is magnesium hydroxide. This is slightly soluble and dissolves according to the following equilibrium. Say we measure the resulting magnesium ion concentration as being 3.7 x 10-5 molar. What will be the solubility product
24:11
for magnesium hydroxide? First let’s write the Ksp expression for this substance, which we derived earlier. Notice that hydroxide concentration is squared due to the coefficient in the equilibrium. Now if this is the concentration of the magnesium ion, then the hydroxide ion concentration must be precisely double, since there are two hydroxide ions for every magnesium ion in this substance. So
24:36
all we need to do is plug these two concentrations into the expression, evaluate, and we will get 2.0 x 10-13 for Ksp, which will be a unitless constant, just like other equilibrium constants. We can also go the other way around, and predict the concentrations of ions that will result in solution when dissolving a substance with a given Ksp. Let’s say we place copper one bromide, which has a Ksp of 6.3 x 10-9, in aqueous solution, which will then generate copper ions
25:10
and bromide ions to some extent. What will be the molar solubility of this substance, meaning how many moles per liter of the formula unit will dissolve? To find this, we can write the solubility product expression, and use this to construct a simple ICE chart. In this, we will not list any values for copper bromide, as this is a solid. The ions themselves will start at zero, the change will be X because of the 1 to 1 to 1 ratio, and this means that the equilibrium
25:40
concentrations of the ions will also be equal to X. Therefore the Ksp will be equal to X2, and X will be 7.9 x 10-5 moles per liter. This is the molar solubility of copper one bromide, which in this case is simply the concentration of copper ions and bromide ions in solution at equilibrium. This can be a little trickier if the Ksp equation involves exponents, but the process is essentially the same. Let’s look at calcium hydroxide, which has a Ksp of 8.0 x 10-6. Say we
26:16
place this in aqueous solution, which will produce some calcium ions and hydroxide ions. Again, let’s write the Ksp expression, noticing that the hydroxide concentration must be squared, as there will be two hydroxide ions for every formula unit of calcium hydroxide that dissolves. Then when we make the ICE table, we will notice that again the initial concentration of each ion will be zero, but the change will be X and 2X respectively, due to the stoichiometry of the equation,
26:45
and the equilibrium concentrations will also be X and 2X. Putting these into the Ksp expression, we get that the Ksp will be equal to 4X3. Now solving for X, we divide by four and then take the cube root, leaving us with 1.3 x 10-2 for X, which will therefore be the molar solubility, which we can interpret as the solubility of the solid expressed in moles per liter, or the moles of the solid that will dissolve per liter of water. So by introducing the solubility
27:19
product, we have enhanced our comprehension of solubility as a quantifiable phenomenon. With our new understanding of solubility and equlibria, we can learn about how we can control precipitation using something called the common ion effect. The common ion effect is an application of Le Chatelier’s principle. We know that under this principle, if we were to add an additional amount of one of the components in an equilibrium, the equilibrium would shift towards
27:48
the other side, using up some of that additional component and creating more of the substances on the other side. Let’s now look at an equilibrium between, say, silver iodide and the aqueous silver and iodide ions. If we are interested in forming a silver iodide precipitate, one way we can do this is to add some potassium iodide. Potassium iodide shares a common ion with silver iodide,
28:14
that being the iodide ion. If there is more iodide ion in solution, that will affect the equilibrium, and it will shift left to use up the excess iodide ions. If the equilibrium shifts left, that means more silver iodide precipitate will form, so this is a method of promoting precipitation. Let’s look at this more quantitatively. Cadmium sulfide has a Ksp value of 1 x 10-28. If we place some
28:44
cadmium sulfide in a 0.01 molar cadmium bromide solution, what will be the molar solubility of the cadmium sulfide? So first we must recognize that the common ion in this situation is cadmium, as it is present in both compounds. Let’s then set up an icebox for the cadmium sulfide equilibrium. We know the initial concentration of cadmium ion is 0.01, from the cadmium bromide, and there is
29:12
no sulfide at first. Then, a tiny amount of the cadmium sulfide will also dissolve, so we must add a change of X to each initial concentration, to get our equilibrium concentrations, which will be 0.01 + X, and X. We can then plug these terms into the Ksp expression, like so. Now, since the Ksp is so incredibly tiny when compared with the initial cadmium ion concentration, we can consider X,
29:42
or the additional concentration of cadmium that will come from cadmium sulfide, to be negligible. In other words, the equilibrium concentration of cadmium will certainly round down to 0.01, so let’s just call it 0.01, which simplifies the expression considerably. Solving for X, since the sulfide concentration will be equal to the molar solubility of cadmium sulfide, we get a molar solubility of 1.0 x 10-26. Without the cadmium bromide, the molar solubility would
30:17
simply have been the square root of the solubility product, which would have been significantly larger than the value we just calculated, so the presence of the cadmium ions in solution has reduced the molar solubility and will thus promote precipitation. If the salt involves something like hydroxide ions, manipulating pH would be another way of controlling precipitation, as changing the pH changes the amount of hydronium and hydroxide ions in solution,
30:47
thus acting as a stress on the concentration of the ions in the solid. So raising the pH would generate more hydroxide ions and promote the precipitation of an ionic solid containing hydroxide, in certain scenarios. And that wraps up our Unit 7 review. I’ll see you over in Unit 8, where we will discuss acids and bases.
Here to help you learn
Sign in to keep a private conversation with your course tutor.
Ask me anything about the material…
Sign in to chat