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Write the balanced chemical equation for the combustion of methane, $\mathrm{CH}_{4}(g),$ to give carbon dioxide and water vapor. Explain why it is difficult to predict whether $\Delta S$ is positive or negative for this chemical reaction.
difficult to predict the change in entropy of the reaction.
Chemistry 102
Chapter 16
Thermodynamics
Carleton College
University of Central Florida
University of Toronto
Lectures
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In thermodynamics, the zer…
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Write the balanced chemica…
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Gaseous methane $\left(\ma…
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Write balanced chemical eq…
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Find $\Delta S_{\text { rx…
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Write the chemical equatio…
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Mcthanol burns in oxygen t…
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Methanol $\left(\mathrm{CH…
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Burning methane in oxygen …
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The combustion of methane …
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were asked to write a conduction combustion reaction for nothing. First, a couple of combustion reaction basics. With these types of reactions, you're always gonna have pretty much the same to reactant and the same to products. So our reactors are going to be the hydrocarbon, and in this case it's methane, and we have oxygen as another beer to go to, and the products are going to be carbon dioxide and water. Now let's look at our reaction for methane. So based on the information, I just told you, we have this basic skeleton for reaction with nothing. The hydrocarbon plus oxygen in the reactant side, and carbon dioxide and water in the product side. And all of these are in gash estate, so these blank spaces air for the coefficient so we can go ahead and look to see if our reactions balanced. Let's look at the carbons first. There's a one on the React inside and one of the product sides of that check cell, and that was looking at auctions. There's to under react inside, but three on the product side, so we're gonna go ahead and double this one. So now we have four on the rack inside. And in order to get four on the product side, we're actually gonna need to double this each tool. Now, let's check the hydrogen for Hodgins here on the reactant side and now four heart surgeons on the product sight as well. So this is our balanced reaction. So both of the reactant and the product side of this reaction have three moles of gas, as you can see one from the methane to from the oxygen. And here we have one from the carbon dioxide and two from the water raper. So because there's three moles of gas on each side, it is hard to predict wth e entropy of the system. And therefore we can't say if the reactor's have more inter Pete or the products have more entropy, since they both have three moles of gas
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