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

The predicted sign of the entropy change in each reaction are as follows;

• Ag+(aq). + Br−(aq) → AgBr(s). --Negative

• CaCO3(s) → CaO(s)+CO2(g) --Positive

• 2NH3(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

To solve this question, we must understand the meaning of entropy.

Entropy of a system or chemical entity is the degree of disorderliness in the substance.

The change in entropy, ΔS∘ is a measure of the difference between the entropy of products and the reactants and is given mathematically as;

• ΔS∘ = S(products) - S(reactants).

• We must also know that the entropy of a gas is greater than a liquid and that in turn is greater than a solid.

• The statement above therefore provides a basis for predicting the sign of the entropy change; ΔS∘ in each reaction.

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Answer 2

In the first reaction entropy sign is negative while for the second reaction the sign on entropy is positive.

The Entropy is defined as degree of disorderliness in the system.

• When entropy is negative the reaction will be non-spontaneous.
• When entropy id positive the reaction will be spontaneous.

The entropy of the reaction can be calculated by,

$\rm \bold{ \Delta S = S(products) - S(reactants)}$

Entropy for First reaction,

$\rm \bold{ Ag(aq) + Br(aq) \rightarrow AgBr(s)}$ will be negative

Entropy for second reaction,

$\rm \bold{ CaCO_3(s) \rightarrow CaO(s)+CO_2(g) }$ will be positive.

Hence, we can conclude that the the first reaction will be non- spontaneous while second reaction will spontaneous.

To know more about entropy, refer to the link:

https://brainly.com/question/13135498?referrer=searchResults