Question

The change in entropy, Δ????∘rxn , is related to the the change in the number of moles of gas molecules, Δ????gas . Determine the change in the moles of gas for each of the reactions and decide if the entropy increases, decreases, or has little or no change. A. K(s)+O2(g) ⟶ KO2(s) Δ????gas= mol The entropy, Δ????∘rxn , increases. has little or no change. decreases. B. CO(g)+3H2(g) ⟶ CH4(g)+H2(g) Δ????gas= mol The entropy, Δ????∘rxn , decreases. increases. has little or no change. C. CH4(g)+2O2(g) ⟶ CO2(g)+2H2O(g) Δ????gas= mol The entropy, Δ????∘rxn , has little or no change. decreases. increases. D. N2O2(g) ⟶ 2NO(g)+O2(g) Δ????gas= mol The entropy, Δ????∘rxn , decreases. increases. has little or no change.

Answer 1

Answer:

A. ΔS°rxn decreases

B. ΔS°rxn decreases

C. ΔS°rxn has little or no change

D. ΔS°rxn increases

Explanation:

The change in entropy (ΔS°rxn) is related to the change in the number of moles of gases (Δn(gas) = n(gas products) - n(gas reactants)). If the number of moles of gases increases, there are more possible microstates and entropy increases. The opposite happens when there are less gaseous moles. And little or no change in entropy is expected when the number of moles of reactants and products is the same.

A. K(s) + O₂(g) ⟶ KO₂(s)

Δn(gas) = 0 - 1 = -1

ΔS°rxn decreases

B. CO(g)+ 3 H₂(g) ⟶ CH₄(g) + H₂O(g)

Δn(gas) = 2 - 4 = -2

ΔS°rxn decreases

C. CH₄(g) + 2 O₂(g) ⟶ CO₂(g)+ 2 H₂O(g)

Δn(gas) = 3  - 3 = 0

ΔS°rxn has little or no change

D. N₂O₄(g) ⟶ 2 NO(g) + O₂(g)

Δn(gas) = 3 - 1 = 2

ΔS°rxn increases