Question

Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.Drag the appropriate items to their respective bins: Positive, NegativeAg+(aq)+Br−(aq)→AgBr(s)CaCO3(s)→CaO(s)+CO2(g)2NH3(g)→N2(g)+3H2(g)2Na(s)+Cl2(g)→2NaCl(s)C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)I2(s)→I2(g)

Answer 1

Answer:

Ag+(aq)+Br−(aq)→AgBr(s)                                NEGATIVE

CaCO3(s)→CaO(s)+CO2(g)2                           POSITIVE

NH3(g)→N2(g)+3H2(g)                                    POSITIVE

2Na(s)+Cl2(g)→2NaCl(s)                                 NEGATIVE

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)           POSITIVE

I2(s)→I2(g)                                                        POSITIVE

Explanation:

We have to remember, to solve this problem, that the entropy of a gas is higher than that of a liquid which in turn  is higher than the solid. Therefore, comparing the reactants and products look for changes in the state of reactants and products. We also have to look for the increase or decrease of moles of each state based on the balanced chemical reaction.

Ag+(aq)+Br−(aq)→AgBr(s)

The reaction product is a single solid and the  the reactants were 2 species in solution. The change in entropy is negative.

CaCO3(s)→CaO(s)+CO2(g)2

Here we have a solid reactant and we have a solid product plus a gas product. The change in entropy is positive.

NH3(g)→N2(g)+3H2(g)

We have 4 mole gases as products starting from 1 mol reactant gas, the entropy has increased.

2Na(s)+Cl2(g)→2NaCl(s)

In this reaction 2 mol solid Na and 1 mol Cl₂ gas are converted into 2 mol solid NaCl, the entropy has decreased.

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)

The products are 7 mol of gas versus 6 mol of gas reactants and therefore entropy has increased.

I2(s)→I2(g)

1 mol solid I₂ goes into 1 mol gas making the change in  the entropy higher.