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Which of the following choices would have a positive entropy change?

A. 4NO2(g)+2H2O(l)+O2(g)−>4HNO3(aq)
B. Fe2O3(s)+3C(s)−>2Fe(s)+3CO(g)
C. 6CO2(g)+6H2O(g)−>C6H12O6(s)+6O2(g)
D. CaO(s)+CO2(g)−>CaCO3(s)

2 Answers

4 votes

Answer:

B.) Fe2O3(s)+3C(s)−>2Fe(s)+3CO(g)

Step-by-step explanation:

I got it correct on founders edtell

User Ben Kreeger
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2 votes

Answer:

  • The option B) Fe₂O₃ (s) + 3C(s) → 2Fe(s) + 3CO₂(g), because the reactants are only solid units and the products contain gas molecules.

Step-by-step explanation:

A positive entropy change means that the entropy of the products is greater than the entropy of the reactants.

Entropy in a measure of the radomness or disorder of the system.

Let's see every reaction:

A) 4NO₂ (g) + 2 H₂O (l) + O₂ (g) → 4 HNO₃ (aq)

Since 5 molecules of a gas (high disorder) combines with 2 molecules of liquid to produce 4 units of aqueous HNO₃ you may expect that the product is more ordered than the reactants, which means that the change in entropy is negative (the entropy decreases).

B) Fe₂O₃ (s) + 3C(s) → 2Fe(s) + 3CO₂(g)

The left side (reactants) show only solid substances which is a highly ordered arrangement while the right side (products) show the formation a solid (ordered arrangement) and a gas (highly disoredered arrangement), so you can predict the increase of the system entropy, i.e. a positive entropy change.

The equation C) shows the combination of 12 gas molecules to produce 1 solid and 6 gas molecules, so you can expect that the entropy will decrease, i.e. a negative entropy change.

For equation D) the products include solid and gas reactants while the product is just one unit of solid substance, letting you to predict a negative entropy change.

User Andi Keikha
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4.9k points