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ემზადებით გამოცდისთვის? მოემზადეთ ამ 4 გაკვეთილის დახმარებით შემდეგ თემაზე: Kinetics
იხილეთ 4 გაკვეთილი
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- A catalyst is a substance that increases the rate of a reaction, but it itself is not consumed in the overall reaction. So let's look at the decomposition of hydrogen peroxide, so H2O2 is hydrogen peroxide, and when it decomposes you get water and also oxygen. This reaction occurs at room temperature. However, it's very slow, so this is a slow reaction. So to speed it up you need to add a catalyst and if you're doing a demonstration like the famous elephant's toothpaste demonstration in general chemistry, you need to add a source of iodide ions. So this is one of the catalysts that you could use. You could use potassium iodide or sodium iodide. So you add a source of iodide anions, and that makes this reaction fast. So your iodide anion is your catalyst; it increases the rate of a reaction. Let's take a look at the mechanism for the reaction when we add our iodide anion as our catalyst. So in the first step of the mechanism, you can see we have H2O2 and our iodide catalyst, and this forms the hypoiodite ions. So this is our intermediate, so the hypoiodite anion is our intermediate and we also are given the information that this first step of the mechanism is the slow step. And the second step of the mechanism, alright we have another molecule of hydrogen peroxide reacts with our intermediate, our hypoiodite ion and we get our oxygen, and this step is fast. Remember, for a mechanism, a possible mechanism must have elementary steps that add up to the overall reaction. So if we add our two steps together, we should get our overall reaction. So we're gonna add all of our reactants together, so that would be H2O2 plus I- plus another H2O2 plus IO- and that should give us our products. So our products, circle all of them over here, we have H2O plus IO- plus H2O plus O2 plus I-. So we have a lot going on there. Let's see what we can cancel out. So what do we have on both sides? Well we can cancel out the iodide, that's on the left and that's on the right. That's our catalyst. It increases the rate of the reaction but it's not consumed. You can see we're using it in the first step, but the iodide anion is regenerated in the second step. So overall, our catalyst is not consumed. And then we have our intermediate. The hypoiodite ion is on the left side and on the right side, so we can cancel that out. Our intermediate is created in the first step, but then it's consumed in the second step. So what are we left with here? We'll be left with two H2O2, so we have two H2O2 for our reactants, and then on the right, we would have two H2Os, so two waters and also oxygen, so plus O2. So we get back, we get back our original reaction, our overall reaction. Also, a possible mechanism must be consistent with the experimental rate law for the overall reaction. And we've seen how to do that in an earlier video. To write your rate law, you need to first recognize the rate determining step in your mechanism. And the rate determining step is the slow step in a mechanism. So step one is our rate determining step. And we can write the rate law for the reaction from the rate determining step, which we know is an elementary reaction. Alright, this is an elementary reaction, it's bimolecular so let's go ahead and write down our rate law. Alright, so the rate of the reaction should be equal to our rate constant, k, so whatever the rate constant happens to be for this, times the concentration of H2O2. So we have times the concentration of H2O2 and since we have a coefficient of one here, remember for an elementary reaction we can turn that coefficient into an exponent. So we have to the first power, and then we also have the concentration of iodide anions, so the concentration of I- and once again our coefficient is a one here, so since this is an elementary reaction, we take our coefficient and turn it into our exponent. And this is the rate law that we would predict using the rate determining step, and this is in agreement with the experimental rate law for our overall reaction. So this is a possible mechanism for our overall reaction, the decomposition of hydrogen peroxide. So how does our catalyst actually increase the rate of our reaction? Well let's look at an energy profile for our uncatalyzed reaction, so the conversion of hydrogen peroxide into water and oxygen. And here's the uncatalyzed version. We start with a certain energy for our reactants and we know at the top here, that represents the energy of the transition state. And the difference between those two would be our activation energy, alright. So that represents our activation energy for our uncatalyzed reaction. When we add the source of iodide ion, when we add our catalyst, this actually provides a different mechanism, a lower energy mechanism, and we know that mechanism occurred in two steps, so let me go ahead and sketch this. The energy of our reactants is the same, right, It's the same level, but we're going to decrease the activation energy, so let me go ahead and draw this in here. So it might look something like this, I'm sure I'm not drawing it perfectly. So let's say that's what our energy profile looks like with the addition of our catalyst. So this, this would be the transition state for the first step of our mechanism, and you can see the activation energy has decreased. So here we have a lower activation energy. So remember from an earlier video if you decrease the activation energy you increase the rate of your reaction, which you can see with the Arrhenius equation. So that's what the catalyst does, and then we reach this point right here, this valley if you will would represent the energy of the intermediate, the hypoiodite ion, and then, we have a second activation energy for the second step of our mechanism. So this would be the first step of our mechanism, so I'm gonna write Ea1, and this would be the activation energy for the second step of our mechanism which is Ea2. So the activation energy for the first step is higher because that's our rate determining step. So again, the catalyst does not affect the energy of your reactants or of your products. It's still the same energy for both of those. What the catalyst does is decrease the necessary activation energy which increases the rate of your reaction.