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Ch.5 - Thermochemistry

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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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.
Related Practice
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

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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 (b) 10.4 g of KCl.

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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 (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 decomposition of liquid benzene, C6H6(l), to gaseous acetylene, C2H2(g): C6H6(l) → 3 C2H2(g) ΔH = +630 kJ (a) What is the enthalpy change for the reverse reaction?

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