Methanol (CH₃OH) can be made by the reaction of CO with H₂: CO(𝑔) + 2 H₂(𝑔) ⇌ CH₃OH(𝑔) (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature?

Video transcript

hi everyone for this problem. It reads the reaction of nitric oxide with oxygen gas produces nitrogen dioxide to maximize the equilibrium yield of nitrogen dioxide. Should the temperature be increased or decreased? And were given a hint to use thermal chemical data to calculate the standard entropy change. So the answer, the question that we're trying to answer here is about the temperature and whether or not it should be increased or decreased. Something that we need to think about here is how does temperature affect our equilibrium? So our reaction can be either endo thermic or exons thermic. If our reaction is endo thermic, that means we can treat heat as a reactant. Okay. And we know that when we have an endo thermic reaction, it's change in entropy is going to be positive. Okay, on the other hand, we can have an exhaust thermic reaction and in that case we treat heat as a product and the entropy change for exa thermic reactions is negative. Okay, so the first thing that we're going to need to do is determine whether or not our reaction is eggs are thermic or endo thermic because we weren't given that information but we can use thermo chemical data to calculate that. Okay, so our standard entropy change is going to be our standard entropy change of our product minus our standard entropy change of our reactant. Okay, so this is data we're going to look up and we're going to do it Um by multiplying by the standard heats of formation for what's in our for what's in our reaction? Okay. So for our product we have the nitrogen dioxide and we have two moles of that. So we're going to take two moles of nitrogen dioxide and multiply it by its standard heat of formation, which is 33.2. Okay. And this is going to be minus our standard entropy. Our standard heats of formation of our reactant. Okay, so we have to react ints we have two moles of nitric oxide. And the standard heat of formation for that is 91.3. Okay. And oxygen gas is element in its elemental state. And so it's standard formation is going to be zero. Okay, so now once we do this calculation, we get a standard entropy change to be negative 116.2 kg jewels. Which tells us we have an exhaust thermic reaction. So now we know because we have an extra thermic reaction, we're going to treat heat as a product. So let's rewrite out our reaction where we're treating the heat as a product. Okay, so we have our nitric oxide plus our oxygen gas that equilibrium to yield nitrogen dioxide plus heat. Alright, so we want to know whether or not the temperature should be increased or decreased. So let's look at what happens in both situations to see which one is going to make the most sense. Okay, because we want to maximize the equilibrium yield of nitrogen dioxide. Okay, So if we decrease the temperature that means we're removing heat and what chandeliers principle tells us is that when a system at equilibrium has a stress put on it, it's going to try to offset that stress. So if we're removing heat as a as a product here, the system is going to try to replace that heat that was just moved. So our reaction is going to shift, right, okay, and by shifting right, this is going to increase our concentration of nitrogen dioxide. Okay, So this is going to increase our concentration of nitrogen dioxide. On the other hand, if we increase the temperature, this means we're adding heat and our system is going to shift to the left to offset what to offset that heat that was just added. So by shifting, that should say left, not right. So when we shift to the left, that means our concentration of nitrogen dioxide is going to decrease. Okay, So what's going to maximize the equilibrium yield of nitrogen dioxide is if we have an increase of nitrogen dioxide and so our temperature should be decreased in order to do that. Okay, so the final answer for this problem is the temperature should be decreased. That's the end of this problem. I hope this was helpful