At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO 3 : 2 KClO 3 (s) → 2 KCl(s) + 3 O 2 (g) ΔH = -89.4 kJ (c) The decomposition of KClO 3 proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of KClO 3 from KCl and O 2 , is likely to be feasible under ordinary conditions? Explain your answer.

Hello. Everyone in this video we're given to specific reactions is saying that the C. 02 production is a lot more favorable than the C. O. Production. And we're trying to explain why. So let's take a deeper look into this. So we can kind of determine this by looking at the bonds that are being broken or the bonds being created. So let's see here, we can first take a look at the bonds that are being broken. So first we can see that the 02 is going to turn into just the '00 so the bonds between each oxygen atom is just going to break. But this happens to both the C. 02 production as well as the C. O. Production. They both have 0. 2 going to go by itself and in this case then the oxygen is going to not really help us in this situation. So we move on to the next thing that we have to look at, we're just going to be the bonds that are being made. So from the ceo production or the ceo compound, we can see here that we have a triple bond. And for our C. 02 molecule we have double bonds and the bonds is just the carbon attaching to the oxygen. This has a triple bond. This has a double bond. If you were alert if you recall from general chemistry in the beginning is that triple bonds are a lot stronger and because they're a lot stronger it requires more energy to break or form. So stronger requires more energy. All right. So in this case if we think of it as doing homework, if we have something that's hard to do it's easier to do of course you want to do the one that's easier. So in this chemistry um going back to chemistry words is that C. O. So this right here is a lot less favorable to do because it's hard to make because it is stronger and it requires more energy to do and because it requires more energy and also releases less energy it makes this whole process exhaust thermic. So the answer to this question scroll down just a little bit. So the answer is that Co two is more favorable because it's more Excel and this is going to be our final answer for this problem.

Ch.5 - Thermochemistry

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Brown 14th Edition

Ch.5 - Thermochemistry

Problem 46c

Chapter 5, Problem 46c

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ (c) The decomposition of KClO₃ proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of KClO₃ from KCl and O₂, is likely to be feasible under ordinary conditions? Explain your answer.

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Identify the reaction type: The given reaction is a decomposition reaction where potassium chlorate (KClO3) breaks down into potassium chloride (KCl) and oxygen gas (O2) upon heating.

Analyze the reaction's enthalpy change (ΔH): The reaction has a negative ΔH value of -89.4 kJ, indicating that it is exothermic. This means that energy is released when KClO3 decomposes.

Consider the reverse reaction: The reverse reaction would be the synthesis of KClO3 from KCl and O2. Since the forward reaction is exothermic, the reverse reaction would be endothermic, requiring an input of energy (89.4 kJ).

Evaluate the feasibility under ordinary conditions: Endothermic reactions, such as the reverse reaction here, are generally not spontaneous at room temperature without an external source of energy.

Conclude based on spontaneity: Given that the reverse reaction is endothermic and would require continuous energy input to proceed, it is unlikely to be feasible under ordinary conditions without specific, controlled circumstances.

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Key Concepts

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

Thermodynamics and Enthalpy

Thermodynamics is the study of energy transformations, and enthalpy (ΔH) is a key concept that measures the heat content of a system. In the given reaction, the negative ΔH indicates that the decomposition of KClO3 is exothermic, releasing energy. This suggests that the reverse reaction, which is endothermic, would require an input of energy to proceed, making it less favorable under ordinary conditions.

Spontaneity of Reactions

A reaction's spontaneity is determined by its Gibbs free energy change (ΔG), which considers both enthalpy and entropy. The decomposition of KClO3 is spontaneous at elevated temperatures, indicating that the products are favored. Conversely, the formation of KClO3 from KCl and O2 would likely have a positive ΔG under normal conditions, suggesting that it is not spontaneous and thus less feasible.

Reaction Conditions and Kinetics

The feasibility of a chemical reaction also depends on the conditions under which it occurs, including temperature, pressure, and concentration. The reverse reaction of forming KClO3 from KCl and O2 may require specific conditions, such as high temperatures or catalysts, to overcome the activation energy barrier. Without these conditions, the reaction is unlikely to proceed efficiently in a typical laboratory setting.

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Textbook Question

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates Ag⁺(aq) + Cl^-(aq) → AgCl(s) H = -65.5 kJ (a) Calculate H for the production of 0.450 mol of AgCl by this reaction. (b) Calculate H for the production of 9.00 g of AgCl. (c) Calculate H when 9.25⨉10^-4 mol of AgCl dissolves in water.

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Textbook Question

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO₃: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (a) 1.36 mol of O₂

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH = -726.5 kJ (a) What is the enthalpy change for the reverse reaction?

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Textbook Question

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH = -726.5 kJ (b) Balance the forward reaction with whole-number coefficients. What is ΔH for the reaction represented by this equation?

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Textbook Question

Consider the combustion of liquid methanol, CH₃OH(l): CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l) ΔH = -726.5 kJ (d) If the reaction were written to produce H₂O(g) instead of H₂O(l), would you expect the magnitude of ΔH to increase, decrease, or stay the same? Explain.