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.

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

AFOKE88 AFOKE88 · Answer 2 · 2021-10-29T21:35:33+02:00

The change in the number of moles for each reaction and if its' change in entropy is increasing or decreasing is as follows;

A) Change in number of moles of gas = -1 and change in entropy decreases.

B)Change in number of moles of gas = -2 and change in entropy decreases.

C) Change in number of moles of gas = 1 and change in entropy increases.

D) Change in number of moles of gas = 1 and change in entropy increases.

The change in entropy (ΔS°rxn) in relation to the change in the number of moles of gases molecules is given as;

Δn(gas) = n(gas products) - n(gas reactants)).

This means that if the number of moles of gas products is more than that of gas reactants, then the change in entropy is positive and as such change in entropy is said to increase.

In contrast, if the number of moles of gas products is less than that of gas reactants, then the change in entropy is negative and as such change in entropy is said to decrease.

Let us look at the options;

Option A; K(s) + O₂(g) ⟶ KO₂(s)

Number of moles of gas reactants = 1

Number of moles of gas products = 0

Change in number of moles of gas = 0 - 1 = -1

Thus, change in entropy decreases

Option B; CO(g)+ 3 H₂(g) ⟶ CH₄(g) + H₂O(g)

Number of moles of gas reactants = 1 + 3 = 4

Number of moles of gas products = 1 + 1 = 2

Change in number of moles of gas = 2 - 4 = -2

Thus, change in entropy decreases

Option C; CH₄(g) + 2 O₂(g) ⟶ CO₂(g)+ 2 H₂O(g)

Number of moles of gas reactants = 1 + 1 = 2

Number of moles of gas products = 1 + 2 = 3

Change in number of moles of gas = 3 - 2 = 1

Thus, change in entropy increases

Option D; N₂O₄(g) ⟶ 2 NO(g) + O₂(g)

Number of moles of gas reactants = 1

Number of moles of gas products = 2 + 1 = 3

Change in number of moles of gas = 3 - 2 = 1

Thus, change in entropy increases

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