Video Transcript
The Haber process involves the reversible reaction of nitrogen and hydrogen. Give a balanced chemical equation for this reaction.
In a closed system, like that used in the Haber process, a reversible reaction will produce an equilibrium. This means that the chemical equation should include an equilibrium arrow. Our reactants are nitrogen and hydrogen gas, which consist of diatomic molecules, N2 and H2. So we have the start of our equation, N2 plus H2. So what’s the product? The Haber process was invented to produce ammonia in large quantities. So the product of this equilibrium reaction is NH3. Now that we have the reactants and the products sorted, we can move on to balancing.
To start with, we have two nitrogen atoms on the reactants’ side and one on the products’ side, while for hydrogen we have two on the left and three on the right. We could pick either element to balance first, but it’s easiest if we start with the nitrogen. This brings us to two nitrogens on each side and six hydrogens on the products’ side. Tripling up the amount of hydrogen on the reactants’ side gives us six hydrogens on both sides, and the reaction is now balanced. If you wanted, you could’ve fixed the amount of ammonia, halved the amount of nitrogen, and used one and a half hydrogen molecules. Either way is correct, but the first way has round numbers, so it’s easy to read.
Why is this reaction described as reversible?
Let’s have a look back at the equation we figured out in part (a). The equilibrium arrow in the equation indicates that both the forward and the reverse reactions are happening simultaneously. So in the forward reaction, one nitrogen molecule is reacting with three hydrogen molecules to produce two ammonia molecules while at the same time the reverse reaction is happening, where ammonia molecules are decomposing into nitrogen molecules and hydrogen molecules.
The question is asking why this reaction is described as reversible. The answer is that some of the product, ammonia, converts back to the original reactants, nitrogen and hydrogen, under the reaction conditions. Since the reverse reaction occurs under the same conditions as the forward reaction, the reaction is reversible. One of the key parts of this answer is that it has to be under the reaction conditions. Take, for example, the irreversible reaction of magnesium and oxygen when you’re burning magnesium in air. It would be possible to change the reaction conditions and electrolyze magnesium oxide back to its constituent elements. But that wouldn’t make the reaction reversible because it wouldn’t be under the same reaction conditions.
If the reaction is performed at a higher gas pressure, what happens to the reaction rate and percentage yield?
The term reaction rate can apply to both the forward and the reverse reaction, while the percentage yield applies exclusively to the forward reaction. Let’s look at our equation and decompose it into the forward and reverse reactions again. Here’s the forward and the reverse reaction expressed separately. Since we’re dealing with a question regarding gas pressure, it’s important we include the state symbols. Under the conditions of the Haber process, ammonia is a gas, along with nitrogen and hydrogen.
Since this question is asking us to predict what happens when the conditions for an equilibrium are changed, we’re going to need to use Le Chatelier’s principle. Le Chatelier’s principle states that for a dynamic equilibrium, if the conditions change, the position of equilibrium will move to counteract the change. Broadly speaking, the position of equilibrium reflects the relative concentrations of the products and the reactants. What often happens with an equilibrium is that either the forward or reverse reaction is favored by a change. And this triggers a change in the concentration of the products and reactants.
So the question is, “what will happen to the reaction rate and the percentage yield when we compress the gas?” When we compress the gas, the gas particles move closer together. This means that there’re more collisions, and the rate of the reactions will increase. This applies to both the forward and the reverse reactions. So for the Haber process, a higher gas pressure will lead to an increase in the reaction rate.
Now, what about the percentage yield? The percentage yield of a reaction is calculated based on the amount of product produced. And it’s equal to the actual amount of product divided by the theoretical maximum amount, all multiplied by 100 percent. The actual amount we get from the equilibrium in the Haber process is dictated by the position of the equilibrium. We can go back to Le Chatelier’s principle to see what effect increasing the gas pressure has on the position of equilibrium.
To make progress, we can fall back on the ideal gas law, which shows us a relationship between the amount of substance in a gas and its pressure. If the volume and the temperature of a gas are kept constant, the pressure and the amount of gas are proportional to one another. This means that the fewer molecules there are, the lower the pressure. In the forward reaction of the Haber process, four molecules transform into two — therefore, the pressure goes down — while in the reverse reaction two molecules become four. So the pressure goes up.
Le Chatelier’s principle says that the position of the equilibrium will move to counteract the change. Therefore, increasing the gas pressure will favor the forward reaction. A faster forward reaction will increase the yield. So we’ve demonstrated that an increase in the gas pressure for the Haber process will increase the reaction rate and the yield. So the answer to the question, “if the reaction is performed at a higher gas pressure, what happens to the reaction rate and percentage yield?” is that the reaction rate and percentage yield both increase.
If the reaction is performed at a lower temperature, what happens to the reaction rate and percentage yield?
I’ve kept some of the information from the previous question because it’s pertinent to this one as well. Here, the change to our equilibrium conditions is the lower temperature. But we’re looking again at the reaction rate and the percentage yield. The first part, the effect of lower temperature on the reaction rate, is straightforward. A reaction rate is roughly proportional to the temperature. This is because the more energy gas molecules have, the more likely they are to react. So the lower the temperature and the less energy these molecules have, the lower the reaction rate. Therefore, lowering the temperature will decrease the reaction rate.
As for the percentage yield, we can rely on the fact that lowering the reaction temperature will favor whichever reaction releases heat. This will cause the position of equilibrium to shift and affect the percentage yield. For this question, you’ll need to remember that the enthalpy change for the forward reaction is negative, minus 92 kilojoules per mole. This means that 92 kilojoules of energy is released into the surroundings for each mole of nitrogen consumed, each three moles of hydrogen consumed, or each two moles of ammonia generated. You don’t necessarily need to remember the value. The main thing is that the value is negative. Energy is being released into the surroundings.
So the forward reaction, the reaction of nitrogen and hydrogen, is exothermic, while the reverse reaction is endothermic. It involves a drawing in of energy from the surroundings. So the forward reaction releases heat, raising the temperature, while the reverse reaction absorbs heat, reducing the temperature. So lowering the temperature favors the forward reaction. So the position of the equilibrium shifts to the products’ side, to the right, increasing the percentage yield of ammonia.
So the answer to the question “if the reaction is performed at a lower temperature, what happens to the reaction rate and the percentage yield?” is the reaction rate decreases and the percentage yield increases. So it takes longer for us to reach the position of equilibrium, starting with pure nitrogen and hydrogen. But we produce more ammonia overall.
