Methanol (CH 3 OH) can be made by the reaction of CO with H 2 : CO(𝑔) + 2 H 2 (𝑔) ⇌ CH 3 OH(𝑔) (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Hi everyone for this problem. It reads phosphorus Penta chloride dissociates to produce gashes, phosphorous tri chloride and chlorine gas to further promote the dissociation of phosphorus Penta chloride. Would you increase or decrease the pressure of the system? Okay, so here we know we have a system that's at equilibrium and we want to know whether or not increasing or decreasing the pressure of the system, if that's going to promote the dissociation of phosphorus Penta chloride. So, for this problem, we're dealing with lush outliers principle, which tells us that when a system is at equilibrium and a stressor is put on that system, our system is going to try to readjust to minimize the effect of that stressor. And this problem specifically our stressor is pressure and we want to know how this pressure is going to affect phosphorus Penta chloride. Okay, so let's go ahead and write out our equation or reaction here so that we can see what the effect is going to be. So, when we're dealing with gasses, we need to pay attention to the number of moles in the product and then the reactant because that has an effect on where our equilibrium is going to shift. Okay, so if we have an increase in pressure, that means our system is going to shift towards less moles of gas. But if we have a decrease in pressure, our system is going to want to shift towards the side with more moles of gas. So this is why it's important to look at that. So let's do that on our reactant side. We have one mole and on our product side we have two moles. Okay, so we want to know specifically about phosphorus Penta chloride. Okay, we want to know would increasing or decreasing the pressure of the system? What which one is going to further promote the dissociation of phosphorus Penta chloride. So dissociation involve bond breaking, which is endo thermic. And so if we're breaking bonds, we're going to want to decrease the pressure of the system to promote the dissociation of phosphorus Penta chloride. Okay, so our answer to this question is decrease the pressure of the system by doing that. That is going to shift towards the more moles of gas. And as you can see by shifting right where um decreasing our concentration of reactant and increasing our concentration of product. So this is the answer to this problem. And that is the end of this problem. I hope this was helpful

Ch.15 - Chemical Equilibrium

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Brown 15th Edition

Ch.15 - Chemical Equilibrium

Problem 66c

Chapter 15, Problem 66c

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

Verified step by step guidance

Understand the concept of Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

Identify the number of moles of gas on both sides of the reaction: 1 mole of CO and 2 moles of H₂ on the left (total 3 moles) and 1 mole of CH₃OH on the right.

Recognize that increasing the pressure of a system in equilibrium favors the side of the reaction with fewer moles of gas. In this reaction, the right side with CH₃OH has fewer moles of gas compared to the left side.

Conclude that to maximize the equilibrium yield of methanol, a high pressure should be used because it will shift the equilibrium towards the formation of methanol, which has fewer moles of gas.

Consider other factors that might affect the yield, such as temperature and catalysts, but for pressure specifically, a high pressure is favorable for increasing the yield of methanol in this reaction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. In the context of gas reactions, increasing pressure favors the side of the reaction with fewer moles of gas, while decreasing pressure favors the side with more moles.

Mole Ratio in Reactions

In the given reaction, the mole ratio of reactants to products is crucial for understanding how pressure affects equilibrium. The reaction CO(g) + 2 H2(g) ⇌ CH3OH(g) has three moles of gas on the reactant side and one mole on the product side, indicating that increasing pressure will favor the formation of methanol.

Equilibrium Constant (K)

The equilibrium constant (K) quantifies the ratio of concentrations of products to reactants at equilibrium. Changes in pressure can influence the concentrations of gaseous reactants and products, thereby affecting the position of equilibrium and the yield of methanol. A higher pressure shifts the equilibrium towards the side with fewer gas moles, potentially increasing K for the product.