   Chapter 18, Problem 15PS

Chapter
Section
Textbook Problem

The third law of thermodynamics says that a perfect crystal at 0 K has zero entropy. The standard entropy of a substance, S° can be determined by evaluating the energy required to carry out conversion from 0 K to standard conditions. What information would be needed to calculate S° for liquid water at 298 K and 1 bar?

Interpretation Introduction

Interpretation: The information needed to calculate So for liquid water at 298 K and 1 bar should be identified.

Concept introduction:

Entropy is the measure of randomness in the system.  Standard entropy change in a reaction is the difference in entropy of the products and reactants. (ΔS°rxn) can be calculated by the following equation.

ΔS°rxn = S°Products- S°reactants

Where,

S°reactants is the standard entropy of the reactants

S°Products is the standard entropy of the products

The standard entropy S is the defined as the entropy that is gained after the conversion of perfect crystal at 0 K to the standard state conditions. According to third law of thermodynamics, the perfect crystal at 0 K has zero entropy.

Explanation

The entropy value is related to the heat absorbed by the system. The formula is,

ΔS = qT

The value of q is determined by using q = m c ΔT

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