Chapter 5, Problem 5.6P

(a)

Interpretation Introduction

Interpretation:

Spontaneous decomposition temperature of ${\text{CaCO}}_{\text{3}}$ has to be estimated.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{Gibbs}\text{energy,}\\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{enthalpy,}\\ \text{T}\text{-}\text{temperature,}\\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{entropy}\text{.}\end{array}$

Explanation of Solution

Given compound,

  Calcium carbonate $\left({\text{CaCO}}_{\text{3}}\right)$.

From the relationship between Gibbs energy, enthalpy and entropy, the breakeven point in the spontaneity of a process is when ${\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{0}$.  If we assume ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ and ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ are independent of temperature, the temperature at which this occurs is obtained from

$\text{T}\text{=}\frac{{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}}{{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ for the decomposition of ${\text{CaCO}}_{\text{3}}$,

$\begin{array}{l}{\text{CaCO}}_{\text{3}}\stackrel{}{\to }\text{CaO}\text{+}{\text{CO}}_{\text{2}}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\text{-}\text{635}\text{.1}\text{+}\left(\text{-}\text{393}\text{.5}\right)\text{-}\left(\text{-}\text{1206}\text{.9}\right)\right]{\text{kJmol}}^{\text{-1}}\text{=}\text{178}\text{.3}{\text{kJmol}}^{\text{-1}}\end{array}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ for the decomposition of ${\text{CaCO}}_{\text{3}}$,

$\begin{array}{l}{\text{CaCO}}_{\text{3}}\stackrel{}{\to }\text{CaO}\text{+}{\text{CO}}_{\text{2}}\\ \\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\text{39}\text{.75}\text{+}\text{213}\text{.74}\text{-}\left(\text{92}\text{.9}\right)\right]{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\text{160}\text{.6}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the spontaneous decomposition temperature of ${\text{CaCO}}_{\text{3}}$,

$\begin{array}{l}\text{T}\text{=}\frac{{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}}{{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}}\text{=}\frac{\text{178}\text{.3}\text{×}{\text{10}}^{\text{3}}}{\text{160}\text{.6}}\text{K}\\ \\ \text{=}\text{1110}\text{K}\text{.}\end{array}$

Therefore, the spontaneous decomposition temperature of ${\text{CaCO}}_{\text{3}}$ is $\text{1110}\text{K}$.

(b)

Interpretation Introduction

Interpretation:

Dehydration temperature of ${\text{CuSO}}_{\text{4}}\text{.}\text{5}{\text{H}}_{\text{2}}\text{O}$ has to be estimated.

Concept Introduction:

Standard Gibbs energy:

The relationship between Gibbs energy, enthalpy and entropy is given by,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{T}{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\\ \text{where,}\\ {\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{Gibbs}\text{energy,}\\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{enthalpy,}\\ \text{T}\text{-}\text{temperature,}\\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{-}\text{standard}\text{reaction}\text{entropy}\text{.}\end{array}$

Explanation of Solution

Given compound,

  Copper sulphate pentahydrate $\left({\text{CuSO}}_{\text{4}}\text{.}\text{5}{\text{H}}_{\text{2}}\text{O}\right)$.

By using the relationship between Gibbs energy, enthalpy and entropy, the breakeven point in the spontaneity of a process is when ${\text{Δ}}_{\text{r}}{\text{G}}^{\sout{\text{0}}}\text{=}\text{0}$.  If we assume ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ and ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ are independent of temperature, the temperature at which this occurs is obtained from

$\text{T}\text{=}\frac{{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}}{{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}$ for the dehydration of ${\text{CuSO}}_{\text{4}}\text{.}\text{5}{\text{H}}_{\text{2}}\text{O}$,

$\begin{array}{l}{\text{CuSO}}_{\text{4}}\text{.}\text{5}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{CuSO}}_{\text{4}}{}_{\left(\text{s}\right)}\text{+}\text{5}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{Δ}}_{\text{f}}{\text{H}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\text{-}\text{771}\text{.36}\text{+}\text{5}\left(\text{-}\text{241}\text{.82}\right)\text{-}\left(\text{-}\text{2279}\text{.7}\right)\right]{\text{kJmol}}^{\text{-1}}\text{=}\text{299}\text{.24}{\text{kJmol}}^{\text{-1}}\end{array}$

Calculate the value of ${\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}$ for the dehydration of ${\text{CuSO}}_{\text{4}}\text{.}\text{5}{\text{H}}_{\text{2}}\text{O}$,

$\begin{array}{l}{\text{CuSO}}_{\text{4}}\text{.}\text{5}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{CuSO}}_{\text{4}}{}_{\left(\text{s}\right)}\text{+}\text{5}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ \\ {\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}\text{=}{\displaystyle \sum _{\text{products}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\text{-}{\displaystyle \sum _{\text{reactants}}\text{v}{\text{S}}_{\text{m}}^{\sout{\text{0}}}}\\ \\ \text{=}\left[\text{109}\text{+}\text{5}\left(\text{188}\text{.83}\right)\text{-}\left(\text{300}\text{.4}\right)\right]{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{=}\text{752}\text{.75}{\text{JK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Calculate the dehydration temperature of ${\text{CuSO}}_{\text{4}}\text{.}\text{5}{\text{H}}_{\text{2}}\text{O}$,

$\begin{array}{l}\text{T}\text{=}\frac{{\text{Δ}}_{\text{r}}{\text{H}}^{\sout{\text{0}}}}{{\text{Δ}}_{\text{r}}{\text{S}}^{\sout{\text{0}}}}\text{=}\frac{\text{299}\text{.24}\text{×}{\text{10}}^{\text{3}}}{\text{752}\text{.75}}\text{K}\\ \\ \text{=}\text{398}\text{K}\text{.}\end{array}$

Therefore, the dehydration temperature of ${\text{CuSO}}_{\text{4}}\text{.}\text{5}{\text{H}}_{\text{2}}\text{O}$ is $\text{398}\text{K}$.