Chapter 8, Problem 8D.3E

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation of calcium oxide that dries organic solvent has to be given.

Answer to Problem 8D.3E

The balanced chemical equation of calcium oxide that dries organic solvent is,

  ${\text{CaO}}_{\left(\text{s}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }\text{Ca}{\left(\text{OH}\right)}_{\text{2(s)}}$

Explanation of Solution

Calcium oxide reacts with water to form calcium hydroxide.  The chemical reaction is,

  ${\text{CaO}}_{\left(\text{s}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }\text{Ca}{\left(\text{OH}\right)}_{\text{2(s)}}$

Calcium oxide and calcium hydroxide are insoluble in water.

The formation of calcium hydroxide from calcium oxide and water is self-balanced.

(b)

Interpretation Introduction

Interpretation:

The standard Gibbs free energy for the drying reaction at ${\text{25}}^{\text{o}}\text{C}$ has to be calculated.

Concept Introduction:

The standard Gibbs free energy can be calculated using the formula,

  ${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}$

Here, ${\text{ΔG}}^{\text{o}}\text{=}$ Change in Gibbs free energy

    ${\text{ΔH}}^{\text{o}}\text{=}$ Change in enthalpy

    $\text{T=}$ Temperature in kelvin

    ${\text{ΔS}}^{\text{o}}\text{=}$ Change in entropy

Answer to Problem 8D.3E

The standard Gibbs free energy for the drying reaction at ${\text{25}}^{\text{o}}\text{C}$ is $\text{-57}\text{.33}\text{kJ}{\text{mol}}^{\text{-1}}\text{.}$

Explanation of Solution

Given,

Temperature=${\text{25}}^{\text{o}}\text{C=298}\text{.15}\text{K}$

Calcium oxide reacts with water to form calcium hydroxide.  The chemical reaction is,

  ${\text{CaO}}_{\left(\text{s}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }\text{Ca}{\left(\text{OH}\right)}_{\text{2(s)}}$

The standard Gibbs free energy can be calculated using the formula,

  ${\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}$

The change in enthalpy $\left({\text{ΔH}}^{\text{o}}\right)$ is calculated as,

  $\begin{array}{l}{\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{Ca}{\left(\text{OH}\right)}_{\text{2(s)}}\text{=-986}\text{.09kJ/mole}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}{\text{CaO}}_{\left(\text{s}\right)}\text{=-635}\text{.09}\text{kJ/mole}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{=-285}\text{.83}\text{kJ/mole}\end{array}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{}_{\text{f(products)}}{\text{-ΔH}}^{\text{o}}{}_{\text{f(reactants)}}\\ {\text{ΔH}}^{\text{o}}\text{=}\left(\text{-986}\text{.09kJ/mole}\right)\text{-}\left(\text{-635}\text{.09kJ/mole+}\left(\text{-285}\text{.83kJ/mole}\right)\right)\\ {\text{ΔH}}^{\text{o}}\text{=-986}\text{.09kJ/mole-}\left(\text{-920}\text{.92kJ/mole}\right)\\ {\text{ΔH}}^{\text{o}}\text{=-65}\text{.17}\text{kJ/mole}\end{array}$

The change in enthalpy $\left({\text{ΔH}}^{\text{o}}\right)$ is $\text{-65}\text{.17}\text{kJ/mole}\text{.}$

The change in entropy $\left({\text{ΔS}}^{\text{o}}\right)$ is calculated as,

  $\begin{array}{l}{\text{ΔS}}^{\text{o}}\text{Ca}{\left(\text{OH}\right)}_{\text{2(s)}}\text{=83}\text{.39}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{ΔS}}^{\text{o}}{\text{CaO}}_{\left(\text{s}\right)}\text{=39}\text{.75}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{ΔS}}^{\text{o}}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{=69}{\text{.11JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

  $\begin{array}{l}{\text{ΔS}}^{\text{o}}{\text{=ΔS}}^{\text{o}}{}_{\text{(products)}}{\text{-ΔS}}^{\text{o}}{}_{\text{(reactants)}}\\ {\text{ΔS}}^{\text{o}}\text{=}\left(\text{83}{\text{.39 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{-}\left(\text{39}{\text{.75 JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{+69}{\text{.11JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ {\text{ΔS}}^{\text{o}}\text{=83}\text{.39}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{-}\left(\text{109}\text{.66}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ {\text{ΔS}}^{\text{o}}\text{=-26}\text{.27}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

The change in entropy $\left({\text{ΔS}}^{\text{o}}\right)$ is $\text{-0}\text{.02627}\text{kJ/mole/K}\text{.}$

  $\begin{array}{l}\text{TΔS=}\left(\text{298K}\right)\left(\text{-0}\text{.02657}\text{kJ/mole/K}\right)\\ \text{TΔS=-7}\text{.8324kJ/mole}\end{array}$

The standard Gibbs free energy is calculated as,

  $\begin{array}{l}{\text{ΔG}}^{\text{o}}{\text{=ΔH}}^{\text{o}}{\text{-TΔS}}^{\text{o}}\\ {\text{ΔG}}^{\text{o}}\text{=}\left(\text{-65}\text{.17}\text{kJ/mole}\right)\text{-}\left(\text{-7}\text{.8324kJ/mole}\right)\\ {\text{ΔG}}^{\text{o}}\text{=-65}\text{.17}\text{kJ/mole+7}\text{.8324}\text{kJ/mole}\\ {\text{ΔG}}^{\text{o}}\text{=-57}\text{.33kJ/mole}\end{array}$

The standard Gibbs free energy for the drying reaction at ${\text{25}}^{\text{o}}\text{C}$ is $\text{-57}\text{.33}\text{kJ}{\text{mol}}^{\text{-1}}\text{.}$