Chapter 18, Problem 10PS

Interpretation Introduction

Interpretation: The information needed for calculating the entropy change for the given reaction which occurs when $2.25g$ of liquid water warmed from ${\text{20}}^{\text{o}}\text{C}\text{-}{\text{25}}^{\text{o}}\text{C}$ should be identified.

Concept introduction: The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  $\begin{array}{l}\Delta S^{\circ }\left(\text{universe}\right)=\Delta S^{\circ }\left(\text{system}\right)+\Delta S^{\circ }\left(\text{surroundings}\right)\\ \text{where}\Delta S^{\circ }\text{= Entropy change}\end{array}$

The $\Delta S^{\circ }\left(\text{universe}\right)$ should be greater than zero for a spontaneous process.

  $\begin{array}{l}\Delta S^{\circ }\left(\text{system}\right)={\text{Δ}}_{\text{r}}{S}^{\text{°}}=\sum n{S}^{\text{°}}\left(\text{products}\right)-\sum n{S}^{\text{°}}\left(\text{reactants}\right)\\ \text{where}\Delta S^{\circ }\text{= Entropy change}\end{array}$

  $\begin{array}{l}\Delta S^{\circ }\left(\text{surroundings}\right)=\frac{-{\Delta }_{\text{r}}{H}^o}{T}\\ \text{where}\Delta S^{\circ }\text{= Entropy change}\\ \text{T}=\text{Temperature}\\ {\Delta }_{\text{r}}{H}^o=\text{Enthalpy}\text{change}\end{array}$