The reaction 2 NaHCO 3 (s) ⇌ Na 2 CO 3 (s) + H 2 O (g) + CO 2 (g) is endothermic. What would you do in order to maximize the yield of Na 2 CO 3 (s)?

So here it says the reaction of 2 moles of sodium bicarbonate yields 1 mole of sodium carbonate, 1 mole of water gas, and 1 mole of carbon dioxide And we're told that it is endothermic. Endothermic means that we're absorbing heat, so heat would be a reactant. Now what would you do in order to maximize the yield of sodium carbonate? So we're trying to make more of this product here, which means we want the reaction to shift towards the product side. So we we want this side to be increasing. So if we take a look, first option, lower the temperature and increase the volume of the container. Alright, so if we lower the temperature, Le Chatelier's principle says that we'll shift to wherever the heat is. The heat is a reactant so it'd be shifting towards the reactant side which would mean the product that I want to increase would actually be decreasing because we're shifting away from it. Increase the volume of the container. According to Le Chatelier's principle, we'd shift to the side with more moles of gas. We have more moles of gas on the product side so that would make it go towards the product side. So one thing is making it go to the left. One thing is making it go to the right. So overall, these kind of cancel each other out. We want the changes to all agree with one another. We want them all to move it towards the right side to help increase the amount of sodium carbonate. Alright. Raise the temperature. So according to Le Chatelier's principle, you would shift away from heat so we'd shift towards the product side. But then it says we're adding CO2 and H2O. According to Le Chatelier's principle, we would shift away from these products since we're adding them. So it would go this way. Lower the temperature again. We said that it would shift to the reactant side towards the heat. Decreasing the volume of the container, which is the same thing as increasing pressure, we would shift away from the side with more moles of gas. Okay. So we're going to shift to the side with less moles of gas here which would be the reactants. But here, this is not what we want. We want both arrows to be pointing towards the products. So adding CO2 and H2O. So again that would make it shift this way. Increasing the volume, we said, would go this way. So that leaves us with option e. Let's see if that makes the most sense. We're increasing the volume. So according to principle, we would shift to the side with more moles of gas which is the right side, the product side, and raise the temperature. Raising the temperature, we shift away from wherever the heat is. Heat is a reactant, so we'd shift to the product side. Both conditions are saying that we're shifting towards the product side to help increase the amount of product. One of which is the one that we want to increase. So option E would have to be the best choice. Both conditions help to maximize the yield of our product of sodium carbonate. So just keep in mind all the fundamental concepts we went over in terms of Le Chatelier's principle and apply them to any types of questions where they're asking is this increasing or is that decreasing and which direction are we going to shift towards. Remember, this works in conjunction as well, with reaction quotient, which can also be used to help us determine which way will our chemical reaction shift to reach equilibrium.

The reaction 2 NaHCO 3 (s) ⇌ Na 2 CO 3 (s) + H 2 O (g) + CO 2 (g) is endothermic. What would you do in order to maximize the yield of Na 2 CO 3 (s)?

Raise the temperature and add CO 2 (g) and H 2 O(g).

Add CO 2 (g) and H 2 O(g) and increase the volume of the container.

Table of contents

Skip topic navigation

1. Chemical Measurements1h 50m
2. Tools of the Trade1h 17m
3. Experimental Error1h 52m
4 & 5. Statistics, Quality Assurance and Calibration Methods1h 57m
6. Chemical Equilibrium3h 41m
7. Activity and the Systematic Treatment of Equilibrium1h 0m
8. Monoprotic Acid-Base Equilibria1h 53m
9. Polyprotic Acid-Base Equilibria2h 17m
10. Acid-Base Titrations2h 37m
11. EDTA Titrations1h 34m
12. Advanced Topics in Equilibrium1h 16m
13. Fundamentals of Electrochemistry2h 19m
14. Electrodes and Potentiometry41m
15. Redox Titrations1h 14m
16. Electroanalytical Techniques57m
17. Fundamentals of Spectrophotometry50m

6. Chemical Equilibrium

Le Chatelier's Principle

Analytical Chemistry

6. Chemical Equilibrium

Le Chatelier's Principle

Struggling with Analytical Chemistry?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

The reaction 2 NaHCO₃(s) ⇌ Na₂CO₃(s) + H₂O (g) + CO₂(g) is endothermic. What would you do in order to maximize the yield of Na₂CO₃(s)?

Lower the temperature and increase the volume of the container.

Raise the temperature and add CO₂(g) and H₂O(g).

Lower the temperature and decrease the volume of the container.

Add CO₂(g) and H₂O(g) and increase the volume of the container.

Increase the volume of the container and raise the temperature.

Verified step by step guidance

Identify the nature of the reaction: The given reaction is endothermic, meaning it absorbs heat. According to Le Chatelier's principle, increasing the temperature will favor the formation of products in an endothermic reaction.

Analyze the reaction equation: 2 NaHCO3 (s) ⇌ Na2CO3 (s) + H2O (g) + CO2 (g). Notice that the reaction produces gases (H2O and CO2) on the product side.

Consider the effect of volume change: Increasing the volume of the container will decrease the pressure. According to Le Chatelier's principle, the system will shift towards the side with more moles of gas to counteract this change, which is the product side in this case.

Combine the effects of temperature and volume: Raising the temperature and increasing the volume both shift the equilibrium towards the products, maximizing the yield of Na2CO3 (s).

Conclude the strategy: To maximize the yield of Na2CO3 (s), increase the volume of the container and raise the temperature, as both actions favor the formation of products in this endothermic reaction.