Chapter 17, Problem 17.1P

Interpretation Introduction

Interpretation:

The sign of $\Delta S$ is to predicted for the given physical or chemical changes.

Concept introduction: The entropy of gas is more than that of liquid or solid.

To determine The sign of $\Delta S$ for the given physical or chemical changes.

Answer to Problem 17.1P

Solution:

1. The change in entropy of boiling of water is positive.
2. The change in entropy of the given reaction is negative.
3. The change in entropy of the given reaction is positive.

(a)

Explanation of Solution

The intermolecular force of attraction of gas is less than that of liquid and solid.
Therefore, randomness of the gas is more than that of liquid or solid.

Entropy is the measurement of the randomness of a substance. Therefore, entropy of gas is more than that of liquid or solid.

The boiling of water causes the conversion of liquid water to turn into the gaseous water molecule. Therefore, the entropy of the system increases. Hence the change in entropy of boiling of water is positive.

(b)

The intermolecular force of attraction of gas is less than that of liquid and solid.
Therefore, randomness of the gas is more than that of liquid or solid.

Entropy is the measurement of the randomness of a substance. Therefore, entropy of gas is more than that of liquid or solid.

The given reaction is,

${\text{I}}_{\text{2}}\left(g\right)\underset{}{\rightleftharpoons }{\text{I}}_{\text{2}}\left(s\right)$

The species on the reactant side has gaseous molecule, while that on product side has solid molecules.
Therefore, the entropy of reactant is more than that of product and hence the change in entropy is negative.

(c)

The intermolecular force of attraction of gas is less than that of liquid and solid.
Therefore, randomness of the gas is more than that of liquid or solid.

Entropy is the measurement of the randomness of a substance. Therefore, entropy of gas is more than that of liquid or solid.

The given reaction is,

${\text{CaCO}}_3\left(s\right)\underset{}{\rightleftharpoons }\text{CaO}\left(s\right)+{\text{CO}}_2\left(g\right)$

The species on the reactant side has one mole of solid and on the product side, there is one mole of solid and one mole of gas.
Therefore, the entropy of product is more than that of reactant and hence the change in entropy is positive.