The chemical equation above represents the combustion of glucose, and the table provides the approximate standard absolute entropies, SÂ°, for some substances. Based on the information given, which of these equations can be used to calculate an approximation of SÂ° for H2O(g) ?

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09-12-2022
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The chemical equation above represents the combustion of glucose, and the table provides the approximate standard absolute entropies, SÂ°, for some substances. Based on the information given, which of these equations can be used to calculate an approximation of SÂ° for H2O(g) ?

Answer :

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RupaBera RupaBera · Answer 1 · 2022-12-14T15:24:12-05:00

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topeadeniran2 topeadeniran2 · Answer 2 · 2022-12-19T09:26:49-05:00

One possible equation that can be used to calculate an approximation of SÂ° for H2O(g) is:

SÂ°(H2O(g)) = SÂ°(C6H12O6(s)) + 6SÂ°(O2(g)) - 6SÂ°(CO2(g)) - 6SÂ°(H2O(l))

What is the possible equation?

This equation is based on the principle that the entropy change of a reaction is equal to the sum of the entropies of the products minus the sum of the entropies of the reactants. Therefore, by rearranging the given chemical equation, we can isolate the entropy of H2O(g) on one side and express it in terms of the entropies of the other substances involved in the combustion of glucose.

To use this equation, we need to know the values of SÂ° for C6H12O6(s), O2(g), CO2(g), and H2O(l). For C6H12O6(s), we can assume that it has a similar entropy to other solid organic compounds with similar molecular weights and structures. For example, sucrose (C12H22O11) has an entropy of 228.2 J/mol K, and starch (C6H10O5)n has an entropy of 150 J/mol K per repeating unit. A reasonable estimate for C6H12O6(s) would be somewhere between these values, say 180 J/mol K.

For H2O(l), we can use the fact that the entropy of a substance increases with its phase change from solid to liquid to gas. The table gives the entropy of H2O(s) as 69.9 J/mol K, and we can use the entropy of fusion of water, which is 22.0 J/mol K, to estimate the entropy of H2O(l) as 69.9 + 22.0 = 91.9 J/mol K.

Using these estimates, we can plug in the values into the equation and solve for SÂ°(H2O(g)):

SÂ°(H2O(g)) = 180 + 6(205.0) - 6(213.6) - 6(91.9) SÂ°(H2O(g)) = 180 + 1230 - 1281.6 - 551.4 SÂ°(H2O(g)) = -423 J/mol K

This is an approximation that may have some error due to the uncertainties in the estimates of SÂ° for C6H12O6(s) and H2O(l).

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