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Kc is the equilibrium constant and is the ratio of the activity of the reactants (numerator) to the activity of the product (denominator). The activity of each component is raised to the power of its corresponding chemical stoichiometric coefficient. Since the activity of each chemical is unitless, the equilibrium constant will also have no units.
Example:
For equilibrium of chemicals in the gas phase, each activity will be measured by its partial pressure (units of pressure) multiplied by its fugacity (units of 1/pressure), so the activity of each gas participating in the equilibrium will be unitless, and the corresponding ratio (Kc) will be unitless.
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What are Kp aND KC?
Kc is the equilibrium constant of a chemical reaction; Kp is the equilibrium constant of a chemical reaction at constant pressure.
How can the magnitude of an equilibrium constant indicate the extent of a reaction?
The reaction proceeds to a very large extent if Kc >> 1. On the other hand, the reaction hardly proceeds if Kc << 1.
What does K subscript c mean?
The constant Kc appears in the equation ~ Kp= Kc(RT)Delta n and Kc = Kp(RT)Delta -nit is derived from the ideal gas law equation PV=nRT,where P is isolated so that P=(n/V)RT, and n/V is converted to a C for concentration, (#mols/Liters being a concentration). Therefore, the constant Kc is merely the constant used at a specific concentration (which is not the concentration at equilibrium), but only when pressure changes are also involved.
What is the difference between Kc and Kw for water?
Kc is the constant for concentration and Kw is the constant for water. Kc[h20] = [OH-][H+] which becomes Kw= [OH-][H+]
What does the equilibrium constant depends on change in?
EQUILIBRIUM CONSTANTS and LE CHATELIER'S PRINCIPLEThis 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.Important: If you aren't happy about the basics of equilibrium, explore the equilibrium menu before you waste your time on this page.This page should only be read when you are confident about everything else to do with equilibria.Changing concentrationsThe factsEquilibrium 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 constantThe 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 pressureThis only applies to systems involving at least one gas.The factsEquilibrium 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.ExplanationWhere there are different numbers of molecules on each side of the equationLet'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 equationIn 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 temperatureThe factsEquilibrium 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:temperatureKp500 K160700 K54You 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 catalystThe factsEquilibrium 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.ExplanationA 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.Note: If you know about the Arrhenius equation, it isn't too difficult to use it to show that the ratio of the rate constants for the forward and back reactions isn't affected by adding a catalyst. Although the activation energies for the two reactions change when you add a catalyst, they both change by the same amount.I'm not going to do this bit of algebra, because it would never be asked at this level (UK A level or equivalent).Exploring some of this using a simple computer programThe link below will take you to a page where you can explore the effect of changing conditions on the reaction:The page comes from Davidson College in America. It needs you to have Java enabled in your browser.You are told that the reaction is endothermic, and can change things like the temperature, the volume of the mixture, and the amounts of all the reactants to see what would happen.It would be best if you worked out what you expected to happen before you change anything. You change things by moving the grey sliders.You will notice that there is no direct way of changing the pressure. Instead, you have to change the volume. Obviously, if you decrease the volume, keeping the amounts of everything constant, that increases the pressure.If you do this to change the pressure, concentrate on the red bars showing what happens to the number of moles of substances present. The blue bars are more confusing. These represent concentrations, and these will change not only because of the change in quantities present, but also because of the change in volume. That's confusing!Effect of adding an inert gasAn inert gas (or noble gas) such as heliumis one that does not react with other elements or compounds. Adding an inert gas into a gas-phase equilibrium at constant volume does not result in a shift.[4]This is because the addition of a non-reactive gas does not change the partial pressures of the other gases in the container. While it is true that the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium. If, however, the volume is allowed to increase in the process, the partial pressures of all gases would be decreased resulting in a shift towards the side with the greater number of moles of gas. There is a short form to remember this: LBMF (little boy married fiona); L stands for less pressure, B - backward reaction, M - more pressure, and F - forward reaction.Effect of a catalystA catalyst speeds up the rate of a reaction by providing additional mechanism(s). Adding a catalyst allows for alternative pathways to be made, where the particles can be absorbed onto the catalyst temporarily before being re-bonded into a new arrangement. The intended effect in adding a catalyst is to lower the activation energy, which frequently increases the rate of reaction. However, the activation energy is lowered by the same amount for the forward and reverse reactions. There is the same increase in reaction rates for both reactions. As a result, a catalyst does not affect the position of the equilibrium. It only affects the time or energy that is required to achieve equilibrium.4) Effect of Catalyst: Catalysts increase the rates of both the forward and reverse reactions equally. Thus, they reduce the time to reach the equilibrium. They have no effect on either the yield of the reaction or the equilibrium constants.
What are Kp aND KC?
Kc is the equilibrium constant of a chemical reaction; Kp is the equilibrium constant of a chemical reaction at constant pressure.
What is the significance of Kc in equilibrium expression?
Kc is the equilibrium constant.
What is the effect of pressure and concentration on Kc and Kp values?
Kc is the equilibrium constant of a chemical reaction related to concentrations. Kp is the equilibrium constant of a chemical reaction related to pressures. Generally, in normal conditions the effect of temperature is not so important.
How can the magnitude of an equilibrium constant indicate the extent of a reaction?
The reaction proceeds to a very large extent if Kc >> 1. On the other hand, the reaction hardly proceeds if Kc << 1.
What are the units for the equilibrium constant k?
I'm taking an awesome chemistry final tomorrow. So, I'm not a massive failure at this: k=mol/liters Kc can only determine by experiment , not by evaluations of equations. so when writting the eq of Kc= [] products /[reactants], do not use units for [], as Kc has no units. Kc, only affected by temperature...
What happens to the value of kc when Catalyst is added to a chemical system at equilibrium?
It will remain Same Or Unchanged
What does K subscript c mean?
The constant Kc appears in the equation ~ Kp= Kc(RT)Delta n and Kc = Kp(RT)Delta -nit is derived from the ideal gas law equation PV=nRT,where P is isolated so that P=(n/V)RT, and n/V is converted to a C for concentration, (#mols/Liters being a concentration). Therefore, the constant Kc is merely the constant used at a specific concentration (which is not the concentration at equilibrium), but only when pressure changes are also involved.
What is the difference between Kc and Kw for water?
Kc is the constant for concentration and Kw is the constant for water. Kc[h20] = [OH-][H+] which becomes Kw= [OH-][H+]
What if kc expression is greater than 1 in chemistry?
If the Kc expression is greater than 1 in chemistry, it means that the concentration of products in the equilibrium mixture is higher than the concentration of reactants. This suggests that the reaction favors the formation of products at equilibrium.
What do you mean by equilibrium constant?
It is the ratio of the concentrations of products to the concentrations of reactants.
What tools were used to make measuring units?
kdvn wfkjvnwejkfc kc fw2rjng;klwe;fj2r
What nicknames does KC Concepcion go by?
KC Concepcion goes by KC, Kace, and Kooks.
