Video: Applying Le Chatelier’s Principle to Hydrogen Iodide Formation

Gaseous hydrogen and iodine react reversibly to produce hydrogen iodide gas, as shown in the equation: H₂ + I₂ ⇌ 2HI. The enthalpy change in this reaction is −9.48 kJ/mol. Which of the following statements correctly describes the effect of temperature or pressure on the position of equilibrium? [A] Increasing pressure strongly favors the backward reaction. [B] Cooling and removing iodine both increase the equilibrium yield of hydrogen iodide. [C] Decreasing pressure has no effect on the equilibrium yield of hydrogen iodide. [D] Heating shifts the position of the equilibrium to the right. [E] Removing hydrogen iodide strongly favors the backward reaction.

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Video Transcript

Gaseous hydrogen and iodine react reversibly to produce hydrogen iodide gas, as shown in the equation: H₂ plus I₂ are in equilibrium with 2HI. The enthalpy change in this reaction is minus 9.48 kilojoules per mole. Which of the following statements correctly describes the effect of temperature or pressure on the position of equilibrium? a) Increasing pressure strongly favors the backward reaction. b) Cooling and removing iodine both increase the equilibrium yield of hydrogen iodide. c) Decreasing pressure has no effect on the equilibrium yield of hydrogen iodide. d) Heating shifts the position of the equilibrium to the right. Or e) Removing hydrogen iodide strongly favors the backward reaction.

Let’s have a look at some of the key information that we’re given in this question. The first thing to note is that the question specifically tells us that both the reactants and the product are gaseous. So let’s label this on our equation. We’re also told in this first sentence that this reaction is reversible. And this is why we have an equilibrium arrow in our equation. We’ve now got all the important information from this first sentence. So we can get rid of it to give ourselves a bit more space. There.

Now, let’s pick out the important information from the next part of the question. We are told that the enthalpy change for this reaction is negative 9.48 kilojoules per mole. By this reaction, what they mean is the forward reaction, the one described earlier. So let’s label the forward and backward reactions just to be safe. When the enthalpy change of a reaction is a negative number, that means that the reaction is exothermic. So we can label the forward reaction as exothermic. Again, let’s give ourselves a little bit more room. So if our forward reaction is exothermic, then our backward reaction must be the opposite, endothermic.

So now that we’ve summarized all the information we’ve been given, we need to consider how all of this information can help us answer the question. Remember that we’re being asked about the effect of temperature or pressure on the position of equilibrium. When considering what happens to the position of equilibrium in a reversible reaction, we need to refer to a fundamental principle.

Hopefully, you’ve remembered that the principle we’re looking for is Le Chatelier’s principle. This principle states that when a change is made to a system which is at equilibrium, the position of equilibrium moves to counteract that change. You can think of this like a seesaw, where you have the reactants on one end of the seesaw and the product on the other. Le Chatelier’s principle means that the reaction always likes this seesaw to be level with the ground, as in balanced.

Now, let’s go through each possible answer and see what effect it has on the position of equilibrium or the balance of our seesaw. So the first possible answer “Increasing pressure strongly favors the backward reaction.” To work out what effect changing the pressure has on our equilibrium, we need to count how many gas molecules we have on each side of the reaction. This is where adding the state symbols to our reaction equation has been really helpful. We can see straight away that, on the product side, we have two molecules of hydrogen iodide gas. On the reactant side, we have one molecule of hydrogen gas and one of iodine gas, which is two in total.

When you increase the pressure on a reversible reaction, Le Chatelier’s principle says that the system will try to change the equilibrium to counteract that increase in pressure. So if we were to have a reaction which had one molecule of gas on reactants’ side and two molecules of gas on the product’s side and we increase the pressure, the equilibrium position is going to change to try to decrease the pressure again.

The best way to decrease the pressure is to reduce the number of gas molecules in the system. So the equilibrium will shift to favor whichever action turns the two gas molecules into one gas molecule. It doesn’t matter whether this is the forward or backward reaction. All that matters is the number of gas molecules needs to decrease to counteract that increase in pressure.

Another way to think of this is about gas molecules in a container. In this reaction, we have two gas molecules in the container on the left and four in the container on the right. If we then increase the pressure, perhaps by making the container smaller, we can see that we have the same number of molecules of gas. But the space that they reside in is much smaller. The container on the right, which has four molecules of gas, is going to have a high pressure.

Le Chatelier’s principle says that the system wants to change to try to decrease that pressure. And the best way to do that is to favor the backward reaction so that, then, we only have two molecules of gas. Because there are only two molecules in this container on the left, they have a bit more space to work around in. And that means the pressure is lower.

So going back to the original question, a), “Increasing pressure strongly favors the backward reaction,” we can now see that this is not correct. Both sides of our particular reaction have the same number of gas molecules. This means that increasing the pressure isn’t going to favor either direction because, no matter which way the equilibrium moves, it’s still gonna have two molecules of gas. So we can rule this answer out.

Next, we have “Cooling and removing iodine both increase the equilibrium yield of hydrogen iodide.” Let’s have a look at the effect of temperature on the equilibrium position. When considering the effect of temperature, we need to look to see which reaction is exothermic and which is endothermic. In our example, the forward reaction is exothermic. And the backward reaction is endothermic. So if we increase the temperature, Le Chatelier’s principle says that the position of equilibrium wants to move in order to counteract that change. So the system will try to move the equilibrium position in order to decrease the temperature again.

So which direction of the reaction would be favored if we want to try to decrease the temperature? The answer is that the position of equilibrium will move to favor the endothermic reaction. Remember that endothermic reactions take in heat whereas exothermic reactions produce heat. Conversely, if we decrease the temperature by cooling, the equilibrium position will move to try to increase the temperature back up, which means favoring the exothermic reaction since this is the one that produces heat.

So in our question, it suggests cooling this reaction. Cooling the reaction will favor the exothermic forward reaction, meaning we will produce more hydrogen iodide. So cooling the reaction will increase the equilibrium yield of hydrogen iodide.

But what about the second part, removing iodine? If we decrease the amount of iodine in this reaction, it becomes unbalanced. And the system will try to move the position of equilibrium to counteract this. We want to alter the position of equilibrium so that we produce more iodine. In order to produce more iodine, we have to use up our product, our hydrogen iodide. So removing iodine decreases the yield of hydrogen iodide. So whilst part of answer b) is true, part of it is also not true. So this is not the correct answer.

The third answer is “Decreasing pressure has no effect on the equilibrium yield of hydrogen iodide.” Remember that, to work out the effect of pressure on a reaction at equilibrium, we need to count the number of gas molecules on both sides of the equation. In our case, we have two gas molecules on both sides, which means that changing the pressure has no effect on the equilibrium yield of either the products or the reactants. This is because, regardless of whether the forward or backward reaction is favored, we do not change the pressure since that’s the same number of gas molecules each side. So this answer is accurate.

But just to be safe, let’s check the last two as well. d) says that “Heating shifts the position of the equilibrium to the right.” Shifting the equilibrium position to the right means that we’re favoring the forward reaction. When considering the effect of temperature on equilibrium, remember that increasing the temperature favors the endothermic reaction because endothermic reactions take in heat. And decreasing the temperature favors the exothermic reaction. Here, we’re told that heating shifts the position in favor of the forward reaction. But in our case, the forward reaction is exothermic. So really, if we want to favor the forward reaction, the exothermic reaction, we need to decrease the temperature, not increase it. So this is incorrect.

And finally, e), “Removing hydrogen iodide strongly favors the backward reaction.” So if we look once again at our balance, if we remove some of the hydrogen iodide, the reaction will shift to try to counteract this change and produce more hydrogen iodide. It does this by favoring the reaction which converts the H₂ and I₂ into hydrogen iodide. This is actually the forward reaction. So removing hydrogen iodide favors the forward reaction, not the backward reaction. So answer e) is incorrect.

So the correct answer to this question is that decreasing the pressure has no effect on the equilibrium yield of hydrogen iodide. And this is because there is the same number of gas molecules on both sides. So changing the pressure has no effect on the equilibrium position.

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