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Kc is an equilibrium constant that links the concentration of reactants and the concentration of products in a reversible reaction at equilibrium.
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Read More »Let’s say that you have a solution made up of two reactants in a reversible reaction. If you leave them for long enough, they’ll eventually reach a state of dynamic equilibrium. This is characterised by two key things: The rate of the forward reaction and the rate of the backward reaction are equal. The concentrations of products and reactions remain the same. But what if you want to know the composition of this equilibrium mixture? If you try to measure the amounts of products or reactants in the solution, it’s likely that you’ll end up disturbing the system. Instead, we can use the equilibrium constant. In this article, we’re going to focus specifically on the equilibrium constant Kc. The equilibrium constant is a value that links the amounts of reactants and products in a reversible reaction at equilibrium. This article is about the equilibrium constant, Kc . . We will begin by explaining what Kc is and go through the formula used to calculate it. We’ll then explore how we calculate the units for Kc and work through two examples together. After that, we'll look at what you can infer from the magnitude of Kc. Finally, we’ll touch on what factors affect the value of Kc. What is the equilibrium constant Kc? As we mentioned above, the equilibrium constant is a value that links the amounts of reactants and products in a mixture at equilibrium. There are a few different types of equilibrium constant, but today we’ll focus on Kc. Kc is an equilibrium constant that links the concentration of reactants and the concentration of products in a reversible reaction at equilibrium. There are two things to note when it comes to Kc: The higher the value of Kc, the higher the proportion of product compared to reactant at equilibrium. The value of Kc for a particular reaction at a certain temperature is always the same, no matter how much of the products or reactants you start with. What is the equation for the equilibrium constant Kc? Let’s take a general equilibrium reaction, shown below. Kc measures concentration. This means that our products and reactants must be liquid, aqueous, or gaseous. To start with, we’ll look at homogeneous dynamic equilibria - these are systems in which all the reactants and products are in the same state. Later we’ll look at heterogeneous equilibria. In this reaction, reactants A and B react to form products C and D in the molar ratio a:b:c:d. Of course, because this is a reversible reaction, you could look at it from the other way - C and D react to form A and B. However, we’ll only look at it from one direction to avoid complicating things further. What is the equation for Kc? Well, it looks like this:
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Read More »The question tells us that at equilibrium, there are 0.4 moles of HCl present. Write this value into the table. We can now work out the change in moles of HCl. At the start of the reaction, there wasn’t any HCl at all. At equilibrium, there are 0.400 moles. This is a change of +0.400. If we take a look at the equation for the equilibrium reaction, we can see that for every two moles of HCl formed, one mole of H 2 and one mole of Cl 2 is used up. The molar ratio is therefore 1:1:2. To find out the number of moles of H 2 and Cl 2 used up in the reaction, divide the number of moles of HCl formed - the change in moles - by 2. So 0.200 moles of H 2 and 0.200 moles of Cl 2 are used up in the reaction, to form 0.400 moles of HCl. The change in moles for these two species is therefore -0.200. Once we know the change in number of moles of each species, we can work out the number of moles at equilibrium. To do this, add the change in moles to the number of moles at the start of the reaction. Write these into your table. Remember that Kc uses equilibrium concentration, not number of moles. But because we know the volume of the container, we can easily work this out. Remember to turn your volume into . In this case, . Concentration = number of moles volume Your table should now be looking like this: Now we can look at Kc. Remember that for the reaction . For our equation , Kc looks like this: Notice that in the equation, the molar ratio of H 2 :Cl 2 :HCl is 1:1:2. In Kc, we must therefore raise the concentration of HCl to the power of 2. Next, we can put our values for concentration at equilibrium into the equation for Kc: The question gives all values to 3 significant figures, and so we must too. The final step is to find the units of Kc. To do this, put the units of each of the concentrations into the equation for Kc and cancel them down. In this case, they cancel completely to give 1. Here, Kc has no units: So our final answer is 1.33. Here’s another question. 5.0 moles of O 2 and 5.0 moles of SO 2 reach dynamic equilibrium in a container of volume 12 dm3. The equilibrium contains 3.0 moles of SO 3 . Find Kc and give its units. The following equation may help you: Let’s write out our table, as before: At equilibrium, we have 3 moles of SO 3 . The change of moles is therefore +3. Because our molar ratio is 1:2:2, the change in moles for O 2 must be -0.15 and the change in moles for SO 2 must be -0.3. We can now work out the number of moles of each species at equilibrium and their concentrations, using the volume given of 12 dm3: Your table should look like this: The equation for Kc is as follows: Subbing in our concentrations gives: To find the units, we need to cancel the units of the concentrations down: Our overall answer is therefore 7.7 mol-1 dm3. Struggling to get to grips with calculating Kc? Here’s a handy flowchart that should simplify the process for you. A flowchart you can use to calculate Kc. Anna Brewer, StudySmarter Originals Working backwards from Kc Sometimes, you may be given Kc for a reaction and have to work out the number of moles of each species at equilibrium. This is a little trickier and involves solving a quadratic equation. Let’s work through an example together. 1 mole of ethyl ethanoate and 5 moles of water react together to form a dynamic equilibrium in a container with a volume of . The Kc for this reaction is 10.0. Find the number of moles of each substance at equilibrium, using the following equation to help you: Let’s start by writing out the values that we do know in a table. We know that at the start, we have 1 mole of ethyl ethanoate and 5 moles of water. To form an equilibrium, some of the ethyl ethanoate and water will react to form ethanol and ethanoic acid. We also know that the molar ratio is 1:1:1:1. For each mole of ethyl ethanoate that is used up, one mole of water will also be used up, forming one mole each of ethanol and ethanoic acid. However, we don’t know how much of the ethyl ethanoate and water will react. We can show this unknown value using the symbol x. We started with 1 mole of ethyl ethanoate. If x moles of this react, then our equilibrium mixture will contain 1 - x moles of ethyl ethanoate. Likewise, we started with 5 moles of water. Because the molar ratio is 1:1:1:1, x moles of water will also react, and so the number of moles of water at equilibrium is 5 - x. How much ethanol and ethanoic acid do we have at equilibrium? We started with 0 moles of each, and know from the molar ratio that we will produce x moles of each. This means that at equilibrium, we have exactly x moles of ethanol and x moles of ethanoic acid. If you make a table showing all the values, it should look something like this: To find the concentration of each species at equilibrium, we divide the number of moles of each species at equilibrium by the volume of the container. In this case, the volume is 1 dm3. Anything divided by 1 gives itself, so here the equilibrium concentration is the same as the equilibrium number of moles. Now let’s write an equation for Kc. We can sub in our values for concentration. In the question, we were also given a value for Kc, which we can sub in too. This means that the only unknown is x: Multiply both sides of the equation by (1-x) (5-x): Expand the brackets to make a quadratic equation in terms of x and rearrange to make it equal 0: You can now solve this using your calculator. You should get two values for x: 5.69 and 0.976. How do you know which one is correct? Well, remember that x equals the number of moles of ethyl ethanoate and water that reacted to form a dynamic equilibrium. We only started with 1 mole of ethyl ethanoate. If 5.69 moles of ethyl ethanoate reacted, then we would be left with -4.69 moles, which isn’t possible - you can’t have a negative number of moles! Therefore, x must equal 0.976. To finish this question, we can now find the number of moles of each species at equilibrium:
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Read More »You might have noticed that we have only calculated Kc for homogeneous systems. These are systems where all the products and reactants are in the same state - for example, all liquids or all gases. However, we can calculate Kc for heterogeneous mixtures too if some of the species are solids. In these cases, the equation for Kc simply ignores the solids. Take the following example: For this reaction, . We ignore the concentrations of copper and silver because they are solids. Magnitude of the equilibrium constant Kc From the magnitude of Kc, we can infer some important things about the reaction at that specific temperature: If Kc is less than 1 , then the denominator of the equation for Kc must be larger than the numerator. Therefore, we have a higher concentration of reactants than products at equilibrium. This means that the position of the equilibrium lies to the left and the backward reaction predominates. , then the denominator of the equation for Kc must be larger than the numerator. Therefore, we have a higher concentration of reactants than products at equilibrium. This means that the position of the equilibrium and the backward reaction predominates. If Kc equals 1 , then the numerator and the denominator of the equation for Kc must be the same. Therefore, we have equal concentrations of reactants and products at equilibrium. This means that the position of the equilibrium lies in the middle . , then the numerator and the denominator of the equation for Kc must be the same. Therefore, we have equal concentrations of reactants and products at equilibrium. This means that the position of the equilibrium . If Kc is greater than 1, then the numerator of the equation for Kc must be larger than the denominator. Therefore, we have a higher concentration of products than reactants at equilibrium. This means that the position of the equilibrium lies to the right and the forward reaction predominates. Factors affecting the equilibrium constant Kc Finally, let’s take a look at factors that affect Kc. It’s actually quite easy to remember - only temperature affects Kc. Pressure, concentration and the presence of a catalyst have no effect on Kc whatsoever. Take this example reaction:
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