EQUILIBRIUM CONSTANT @ LE CHATELIER' PRINCIPAL

EQUILIBRIUM CONSTANTS and LE CHATELIER'S PRINCIPLE This page looks at the relationship between equilibrium constants and Le Chatelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as you change the conditions of a reaction, although the equilibrium constant may remain the same. Be warned that this page assumes a good understanding of Le Chatelier's Principle and how to write expressions for equilibrium constants.
Changing concentrations The facts Equilibrium constants aren't changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium is changed if you change the concentration of something present in the mixture. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. Suppose you have an equilibrium established between four substances A, B, C and D. According to Le Chatelier's Principle, if you decrease the concentration of C, for example, the position of equilibrium will move to the right to increase the concentration again.
	Note:  The reason for choosing an equation with "2B" will become clearer when I deal with the effect of pressure further down the page.
Explanation in terms of the constancy of the equilibrium constant The equilibrium constant, Kc for this reaction looks like this: If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased? This is actually the wrong question to ask! We need to look at it the other way round. Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does? If you decrease the concentration of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before. The position of equilibrium moves - not because Le Chatelier says it must - but because of the need to keep a constant value for the equilibrium constant. If you decrease the concentration of C: Changing pressure This only applies to systems involving at least one gas. The facts Equilibrium constants aren't changed if you change the pressure of the system. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium may be changed if you change the pressure. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favouring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium. Explanation Where there are different numbers of molecules on each side of the equation Let's look at the same equilibrium we've used before. This one would be affected by pressure because there are 3 molecules on the left but only 2 on the right. An increase in pressure would move the position of equilibrium to the right. Because this is an all-gas equilibriium, it is much easier to use Kp: Once again, it is easy to suppose that, because the position of equilibrium will move to the right if you increase the pressure, Kp will increase as well. Not so! To understand why, you need to modify the Kp expression. Remember the relationship between partial pressure, mole fraction and total pressure?
	Note:  If you aren't happy with this, read the beginning of the page about Kp before you go on. Use the BACK button on your browser to return to this page.
Replacing all the partial pressure terms by mole fractions and total pressure gives you this: If you sort this out, most of the "P"s cancel out - but one is left at the bottom of the expression. Now, remember that Kp has got to stay constant because the temperature is unchanged. How can that happen if you increase P? To compensate, you would have to increase the terms on the top, xC and xD, and decrease the terms on the bottom, xA and xB. Increasing the terms on the top means that you have increased the mole fractions of the molecules on the right-hand side. Decreasing the terms on the bottom means that you have decreased the mole fractions of the molecules on the left. That is another way of saying that the position of equilibrium has moved to the right - exactly what Le Chatelier's Principle predicts. The position of equilibrium moves so that the value of Kp is kept constant. Where there are the same numbers of molecules on each side of the equation In this case, the position of equilibrium isn't affected by a change of pressure. Why not? Let's go through the same process as before: Substituting mole fractions and total pressure: . . . and cancelling out as far as possible: There isn't a single "P" left in the expression. Changing the pressure can't make any difference to the Kp expression. The position of equilibrium doesn't need to move to keep Kp constant. Changing temperature The facts Equilibrium constants are changed if you change the temperature of the system. Kc or Kp are constant at constant temperature, but they vary as the temperature changes. Look at the equilibrium involving hydrogen, iodine and hydrogen iodide: The Kp expression is: Two values for Kp are: temperature Kp 500 K 160 700 K 54 You can see that as the temperature increases, the value of Kp falls.
	Note:  You might possibly be wondering what the units of Kp are. This particular example was chosen because in this case, Kp doesn't have any units. It is just a number. The units for equilibrium constants vary from case to case. It is much easier to understand this from a book than from a lot of maths on screen. You will find this explained in my chemistry calculations book.
This is typical of what happens with any equilibrium where the forward reaction is exothermic. Increasing the temperature decreases the value of the equilibrium constant. Where the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant.
	Note:  Any explanation for this needs knowledge beyond the scope of any UK A level (or equivalent) syllabus.
The position of equilibrium also changes if you change the temperature. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat. In the equilibrium we've just looked at, that will be the back reaction because the forward reaction is exothermic. So, according to Le Chatelier's Principle the position of equilibrium will move to the left. Less hydrogen iodide will be formed, and the equilibrium mixture will contain more unreacted hydrogen and iodine. That is entirely consistent with a fall in the value of the equilibrium constant. Adding a catalyst The facts Equilibrium constants aren't changed if you add (or change) a catalyst. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium is not changed if you add (or change) a catalyst. Explanation A catalyst speeds up both the forward and back reactions by exactly the same amount. Dynamic equilibrium is established when the rates of the forward and back reactions become equal. If a catalyst speeds up both reactions to the same extent, then they will remain equal without any need for a shift in position of equilibrium.

EQUILIBRIUM CONSTANT

EQUILIBRIUM CONSTANTS: Kp This page explains equilibrium constants expressed in terms of partial pressures of gases, Kp. It covers an explanation of the terms mole fraction and partial pressure, and looks at Kp for both homogeneous and heterogeneous reactions involving gases. The page assumes that you are already familiar with the concept of an equilibrium constant, and that you know about Kc - an equilibrium constant expressed in terms of concentrations
	Important:  If you have come directly to this page via a search engine (including the Google site search on the Main Menu page), you should first read the page on equilibrium constants - Kc before you go on - unless you are already fully confident about how to write expressions for Kc. You will find a link back to this page at the bottom of the Kc page.
Defining some terms Before we can go any further, there are two terms relating to mixtures of gases that you need to be familiar with. Mole fraction If you have a mixture of gases (A, B, C, etc), then the mole fraction of gas A is worked out by dividing the number of moles of A by the total number of moles of gas. The mole fraction of gas A is often given the symbol xA. The mole fraction of gas B would be xB - and so on. Pretty obvious really! For example, in a mixture of 1 mole of nitrogen and 3 moles of hydrogen, there are a total of 4 moles of gas. The mole fraction of nitrogen is 1/4 (0.25) and of hydrogen is 3/4 (0.75). Partial pressure The partial pressure of one of the gases in a mixture is the pressure which it would exert if it alone occupied the whole container. The partial pressure of gas A is often given the symbol PA. The partial pressure of gas B would be PB - and so on. There are two important relationships involving partial pressures. The first is again fairly obvious. The total pressure of a mixture of gases is equal to the sum of the partial pressures. It is easy to see this visually: Gas A is creating a pressure (its partial pressure) when its molecules hit the walls of its container. Gas B does the same. When you mix them up, they just go on doing what they were doing before. The total pressure is due to both molecules hitting the walls - in other words, the sum of the partial pressures. The more important relationship is the second one: Learn it! That means that if you had a mixture made up of 20 moles of nitrogen, 60 moles of hydrogen and 20 moles of ammonia (a total of 100 moles of gases) at 200 atmospheres pressure, the partial pressures would be calculated like this: gas mole fraction partial pressure nitrogen 20/100 = 0.2 0.2 x 200 = 40 atm hydrogen 60/100 = 0.6 0.6 x 200 = 120 atm ammonia 20/100 = 0.2 0.2 x 200 = 40 atm Partial pressures can be quoted in any normal pressure units. The common ones are atmospheres or N m-2 (newtons per square metre). Kp in homogeneous gaseous equilibria A homogeneous equilibrium is one in which everything in the equilibrium mixture is present in the same phase. In this case, to use Kp, everything must be a gas. A good example of a gaseous homogeneous equilibrium is the conversion of sulphur dioxide to sulphur trioxide at the heart of the Contact Process: Writing an expression for Kp We are going to start by looking at a general case with the equation: If you allow this reaction to reach equilibrium and then measure (or work out) the equilibrium partial pressures of everything, you can combine these into the equilibrium constant, Kp. Just like Kc, Kp always has the same value (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with. Kp has exactly the same format as Kc, except that partial pressures are used instead of concentrations. The gases on the right-hand side of the chemical equation are at the top of the expression, and those on the left at the bottom.
	Beware!  People are sometimes tempted to write brackets around the individual partial pressure terms. Don't do it! Even if you intend to write normal round brackets, it is too easy in an exam to write them as square brackets instead. This makes it look as if you are confusing Kp with Kc. Examiners don't like it, and you could be penalised.
The Contact Process equilibrium You will remember that the equation for this is: Kp is given by: The Haber Process equilibrium The equation for this is: . . . and the Kp expression is: Kc in heterogeneous equilibria A typical example of a heterogeneous equilibrium will involve gases in contact with solids. Writing an expression for Kp for a heterogeneous equilibrium Exactly as happens with Kc, you don't include any term for a solid in the equilibrium expression. The next two examples have already appeared on the Kc page. The equilibrium produced on heating carbon with steam Everything is exactly the same as before in the expression for Kp, except that you leave out the solid carbon. The equilibrium produced on heating calcium carbonate This equilibrium is only established if the calcium carbonate is heated in a closed system, preventing the carbon dioxide from escaping. The only thing in this equilibrium which isn't a solid is the carbon dioxide. That is all that is left in the equilibrium constant expression. Calculations involving Kp On the Kc page, I've already discussed the fact that the internet isn't a good medium for learning how to do calculations. If you want lots of worked examples and problems to do yourself centred around Kp, you might be interested in my book on chemistry calculations.
	Note:  If you are interested in my chemistry calculations book you might like to follow this link.
Where would you like to go now? To the equilibrium menu . . . To the Physical Chemistry menu . . . To Main Menu . . .

EQUILIBRIUM CONSTANT

EQUILIBRIUM CONSTANTS: Kc This page explains what is meant by an equilibrium constant, introducing equilibrium constants expressed in terms of concentrations, Kc. It assumes that you are familiar with the concept of a dynamic equilibrium, and know what is meant by the terms "homogeneous" and "heterogeneous" as applied to chemical reactions.
	Important:  If you aren't sure about dynamic equilibria it is important that you follow this link before you go on. If you aren't sure what homogeneous and heterogeneous mean, you would find it useful to follow this link and read the beginning of the page that you will find (actually on catalysis). Use the BACK button on your browser to return to this page.
We need to look at two different types of equilibria (homogeneous and heterogeneous) separately, because the equilibrium constants are defined differently. A homogeneous equilibrium has everything present in the same phase. The usual examples include reactions where everything is a gas, or everything is present in the same solution. A heterogeneous equilibrium has things present in more than one phase. The usual examples include reactions involving solids and gases, or solids and liquids. Kc in homogeneous equilibria This is the more straightforward case. It applies where everything in the equilibrium mixture is present as a gas, or everything is present in the same solution. A good example of a gaseous homogeneous equilibrium is the conversion of sulphur dioxide to sulphur trioxide at the heart of the Contact Process: A commonly used liquid example is the esterification reaction between an organic acid and an alcohol - for example: Writing an expression for Kc We are going to look at a general case with the equation: No state symbols have been given, but they will be all (g), or all (l), or all (aq) if the reaction was between substances in solution in water. If you allow this reaction to reach equilibrium and then measure the equilibrium concentrations of everything, you can combine these concentrations into an expression known as an equilibrium constant. The equilibrium constant always has the same value (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with. It is also unaffected by a change in pressure or whether or not you are using a catalyst. Compare this with the chemical equation for the equilibrium. The convention is that the substances on the right-hand side of the equation are written at the top of the Kc expression, and those on the left-hand side at the bottom. The indices (the powers that you have to raise the concentrations to - for example, squared or cubed or whatever) are just the numbers that appear in the equation.
	Note:  If you have come across orders of reaction, don't confuse this with the powers that appear in the rate equation for a reaction. Those powers (the order of the reaction with respect to each of the reactants) are experimentally determined. They don't have any direct connection with the numbers that appear in the equation You may come across attempts to derive the expression for Kc by writing rate equations for the forward and back reactions. Except in a very limited number of very simple examples, this can't be done! These attempts make the fundamental mistake of obtaining the rate equation from the chemical equation. That's WRONG! Deriving an expression for Kc is impossible at this level of chemistry. It isn't relevant to this page, but if you want to find out more about orders of reaction, you might like to follow this link at some time in the future.
Some specific examples The esterification reaction equilibrium A typical equation might be: There is only one molecule of everything shown in the equation. That means that all the powers in the equilibrium constant expression are "1". You don't need to write those into the Kc expression. As long as you keep the temperature the same, whatever proportions of acid and alcohol you mix together, once equilibrium is reached, Kc always has the same value. At room temperature, this value is approximately 4 for this reaction. The equilibrium in the hydrolysis of esters This is the reverse of the last reaction: The Kc expression is: If you compare this with the previous example, you will see that all that has happened is that the expression has turned upside-down. Its value at room temperature will be approximately 1/4 (0.25). It is really important to write down the equilibrium reaction whenever you talk about an equilibrium constant. That is the only way that you can be sure that you have got the expression the right way up - with the right-hand substances on the top and the left-hand ones at the bottom. The Contact Process equilibrium You will remember that the equation for this is: This time the Kc expression will include some visible powers: Although everything is present as a gas, you still measure concentrations in mol dm-3. There is another equilibrium constant called Kp which is more frequently used for gases. You will find a link to that at the bottom of the page. The Haber Process equilibrium The equation for this is: . . . and the Kc expression is: Kc in heterogeneous equilibria Typical examples of a heterogeneous equilibrium include: The equilibrium established if steam is in contact with red hot carbon. Here we have gases in contact with a solid. If you shake copper with silver nitrate solution, you get this equilibrium involving solids and aqueous ions: Writing an expression for Kc for a heterogeneous equilibrium The important difference this time is that you don't include any term for a solid in the equilibrium expression. Taking another look at the two examples above, and adding a third one: The equilibrium produced on heating carbon with steam Everything is exactly the same as before in the equilibrium constant expression, except that you leave out the solid carbon. The equilibrium produced between copper and silver ions Both the copper on the left-hand side and the silver on the right are solids. Both are left out of the equilibrium constant expression. The equilibrium produced on heating calcium carbonate This equilibrium is only established if the calcium carbonate is heated in a closed system, preventing the carbon dioxide from escaping. The only thing in this equilibrium which isn't a solid is the carbon dioxide. That is all that is left in the equilibrium constant expression. Calculations involving Kc There are all sorts of calculations you might be expected to do which are centred around equilibrium constants. You might be expected to calculate a value for Kc including its units (which vary from case to case). Alternatively you might have to calculate equilibrium concentrations from a given value of Kc and given starting concentrations. This is simply too huge a topic to be able to deal with satisfactorily on the internet. It isn't the best medium for learning how to do chemistry calculations. It is much easier to do this from a carefully structured book giving you lots of worked examples and lots of problems to try yourself. If you have found this site useful, you might like to have a look at my book on chemistry calculations. It covers equilibrium constant calculations starting with the most trivial cases, and gradually getting harder - up to the moderately difficult examples which may be asked in a UK A' level examination.
	Note:  If you are interested in my chemistry calculations book you might like to follow this link.
Where would you like to go now? To look at Kp . . . To the equilibrium menu . . . To the Physical Chemistry menu . . . To Main Menu . . .