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Ch.15 - Chemical Equilibrium

Chapter 15, Problem 86

A reaction vessel at 27 °C contains a mixture of SO2 (P = 3.00 atm) and O2 (P = 1.00 atm). When a catalyst is added, this reaction takes place: 2 SO2( g) + O2( g) ⇌ 2 SO3( g). At equilibrium, the total pressure is 3.75 atm. Find the value of Kc.

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Welcome back everyone to another video, a reaction vessel at 27 °C contains a mixture of so two at a pressure of three atmospheres and 02 at a pressure of one atmosphere. When a catalyst is added, this reaction takes place two. So two plus 02 is an equilibrium with two. So three at equilibrium, the total pressure is 3.75 atmospheres find the value of KC which is the equilibrium constant in terms of concentrations. And we're given four answer choices. A 0.053 B 1.0 C 0.72 and D 1.3. So we're going to write down the equilibrium equation. And essentially that would be our equation given in the problem to so two plus 02 in equilibrium web two episode three. The best way is to actually introduce an I ce table initial change and equilibrium, we know that according to the problem, the initial amount or the initial partial pressure of so two would be three atmospheres. Now we also know that for ot oxygen, we have one atmosphere and there is no mention of the product. So we can essentially understand that there is no product formed yet because we are just mixing three and one atmosphere respectively for each reactant. Now, for the change, we are going to assume that we are consuming X atmospheres of 02, we need to include a negative sign because that's the reactant which also tells us that we are going to consume two X for so two. So we're including negative two X for the product, we're producing two acts according to the stom Mery. And at equilibrium, we can get the following amounts or pressures. 3.00 minus two, X, 1.00 minus X and two X. For the product, we know that the total pressure at equilibrium is 3.75. So if we add all of them together, we should end up with 3.75. So let's do that. We're going to take 3.00 minus two X. We're going to add a 1.00 minus X and we're going to add two X. This should give us 3.75 atmospheres and we can easily sulfur acts from here, right? So essentially, if we simplify the equation, we get four minus X is equal to 3.75. And now if we simplify the equation X is equal to 0.25 atmospheres, now we can easily find the equilibrium pressures for each species. So the partial pressure of so two at equilibrium would be three point 00 minus two X. That's three minus 0.5. We got 2.50 for the partial pressure of 02. That's one minus a one minus 0.25 gives us 0.75 atmospheres. And now the partial pressure of so three is equal to two X. If we multiply two by 0.25 we get 0.5. Well done, we have a lot of this problem solved, but we still need to understand what we're trying to achieve here. So we're trying to calculate the KC value. And before we do that, we can easily identify the KP value. Let's remember the relationship between the two. So here we're essentially going to remember that KP is equal to KC multiplied by RT raised to the power of delta N. Now, if we think about this equation, we are going to rearrange it and we're going to say that KC is equal to KP divided by RT, raise the power of delta. And, and from here, we can continue calculating the required parameters. First of all, what about KP? Well, according to the equation, we need to take the partial pressure of so three, which is our product squared because we have a coefficient of two and we need to divide by the partial pressure of so two squared multiplied by the partial pressure of 02. If we plug in the numbers that we got previously for so three, we take 0.50 squared. Now, for the bottom terms we're going to use. So two squared, that's 2.50 squared. And we're going to multiply by 020.75. Let's see what we get here. So the KP value should be 0.05333. Let's use four significant figures. And now let's apply the equation for KC. On top, we use our KP value. So let's include 0.05333. And now on the bottom, let's include RT, that'd be 0.08206. All right, that's the value we use. And let's multiply by the temperature. What is our temperature? Well, essentially, if we have 27 °C, we need to convert that into Kelvin by adding 273.15 Kelvin. And now delta M is the change in the moles for the reaction. So, on the product side, we have two, on the reactant side, we have two plus one, right two from so 21 from 02. And its essential tells us that delta and is negative one, we are going to use a heat power of negative one. And this gives us the KC value of 1.3. So now if we have our KC value, we can go back to the answer choices and let's identify the correct option, which is option D 1.3 is the value of KC for this problem. Thank you for watching.
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