Chapter 3, Problem 3A.1ST

Interpretation Introduction

Interpretation:

The change in entropy $(\Delta S)$ value has to be determined.

Concept Information:

Spontaneous process: A process which is initiated by itself, without the help of external energy source is called spontaneous process.  All spontaneous process is associated with the decrease in free energy in the system.

Entropy is the measure of randomness in the system. The entropy change in a reaction is the difference in entropy of the products and reactants. ${\text{(ΔS}}_{\text{rxn}}\text{)}$ can be calculated by the following equation.

${\text{ΔS}}_{\text{rxn}}{\text{ = S}}_{\text{Products}}-{\text{ S}}_{\text{reactants}}$

Where,

${\text{S}}_{\text{reactants}}$ is the entropy of the reactants

${\text{S}}_{\text{Products}}$ is the standard of the products

Standard entropy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate $\text{ΔS}$ for reaction of isolated system.

$\text{ΔS=}\frac{{\text{q}}_{\text{rev}}}{\text{T}}$

Explanation of Solution

The change in entropy is equal to the energy transferred as heat to the system reversibly divided by the temperature at which the transfer takes place.

Given: T $\text{=}\text{100}{}^{\text{o}}\text{C}\text{+}\text{273}\text{K}\text{=}\text{373}\text{K}$; and the heat transferred to the large mass of water at $0^{0}C$ $\left({\text{q}}_{\text{rev}}\right)\text{=100}\text{kJ}$,

Thus, change in entropy at ${100}^{0}C$ is,

$\begin{array}{c}\text{ΔS}\text{=}\frac{{\text{q}}_{\text{rev}}}{\text{T}}\\ \\ \text{=}\frac{\text{100}\text{kJ}}{\text{373}\text{K}}\\ \\ \text{=}268J.K^{-1}\end{array}$

Therefore, the change in entropy of the given system is $268J.K^{-1}$