Chapter 16, Problem 16.137MP

Interpretation Introduction

Interpretation:

$\Delta H°$ and $\Delta S°$ are assumed to be independent of temperature.  Molar solubility of $CaC{O}_3$ at $50°C$ in a solution prepared by dissolving $1.000L$ of $C{O}_2$ gas and $3.335g$ of solid Calcium hydroxide has to be calculated.  Solubility of $CaC{O}_3$ at $50°C$ is whether larger or smaller than at $25°C$ has to be explained.

Concept Introduction:

Standard free energy change of a reaction is related to molar solubility of a substance as,

$\Delta G°=-{\text{ RT ln K}}_{sp}$

Where,

$\begin{array}{l}\Delta G°=\text{ standard free energy change}\\ R=\text{ Universal gas constant}\\ \text{T}=\text{ Temperature}\\ {\text{K}}_{sp}\text{= molar solubility of salt}\end{array}$

Standard free energy change of a reaction is also related to standard enthalpy change and entropy change as follows,

$\Delta G°=\Delta H°-T\Delta S°$

$\Delta H°$ refers to standard enthalpy change.  Change in enthalpy in a reaction and heat of formation are related by the formula,

$\Delta H_{rxn}^{°}=\Sigma \Delta H_f^{°}(product)-\Sigma \Delta H_f^{°}(reac\tan t)$

Where,

$\begin{array}{l}\Delta H_{rxn}^{°}=\text{ enthalpy change in reaction}\\ \Delta H_f^{°}(product)=\text{ enthalpy or heat of formation of product}\\ \Delta H_f^{°}(reac\tan t)=\text{ enthalpy or heat of formation of }reac\tan t\end{array}$

$\Delta S°$ refers to standard entropy of the reaction which is termed as entropy change in a reaction under standard conditions of $298K$ and $1\text{ atm}$ pressure.  It is represented as,

$\Delta S°=S°\left(products\right)-S°\left(reac\tan ts\right)$