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Novosadov [1.4K]
3 years ago
11

Consider the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) Using standard thermodynamic data at 298K, calculate the entropy chang

e for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions. S°surroundings = J/K An error has been detected in your answer. Check
Chemistry
1 answer:
balandron [24]3 years ago
5 0

Answer:

the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.

Explanation:

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.

For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.

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The volume will be 568.89 mL.

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Boyle's law is expressed mathematically as:

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Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. That is, the pressure of the gas is directly proportional to its temperature. Gay-Lussac's law can be expressed mathematically as follows:

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Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T} =k

Studying an initial state 1 and a final state 2, it is fulfilled:

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In this case:

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Replacing:

\frac{960 mmHg*550 mL}{473K} =\frac{830 mmHg*V2}{423 K}

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V2=\frac{423 K}{830 mmHg} *\frac{960 mmHg*550 mL}{473K}

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