Chapter 18, Problem 18.91QP

(a)

Interpretation Introduction

Interpretation:

For the reaction of tin oxide with either carbon or hydrogen, the value of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{,}$ the temperature at which these processes becomes spontaneous and industrially preferred process to obtain tin has to be given.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Spontaneous process:

The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.

Answer to Problem 18.91QP

For the reaction of tin oxide with hydrogen

• The value of enthalpy change $\left({\text{ΔH}}^{\text{o}}\right)$ is $\text{97}\text{.1}\text{×}{\text{10}}^{\text{3}}\text{J}\text{.}$
• The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ is $\text{115}\text{.45}\text{J/K}\text{.}$

For the reaction of tin oxide with carbon

• The value of enthalpy change $\left({\text{ΔH}}^{\text{o}}\right)$ is $\text{187}\text{.2}\text{×}{\text{10}}^{\text{3}}\text{J}\text{.}$
• The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ is $\text{207}\text{.21}\text{J/K}\text{.}$

Explanation of Solution

To calculate: The value of ${\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ for the given reactions

For the first reaction,

$\begin{array}{l}{\text{SnO}}_{\text{2}}{}_{\text{(s)}}\text{+ }{\text{2H}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ Sn}}_{\left(\text{s}\right)}\text{+}{\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{:}\text{-580}\text{.7}\text{0}\text{0}\text{2}\text{×}\text{-241}\text{.8}\text{kJ}\\ {\text{S}}^{\text{o}}\text{:}\text{52}\text{.3}\text{2}\text{×}\text{130}\text{.6}\text{51}\text{.55}\text{2}\text{×}\text{188}\text{.7}\text{J/K}\\ {\text{Temperature is 25 }}^{\text{o}}\text{C}\text{.}\end{array}$

Calculate the value of ${\text{ΔH}}^{\text{o}}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{2}\text{×}\text{-241}\text{.8}\right)\text{-}\left(\text{-580}\text{.7}\right)\right]\text{kJ}\\ \text{=}\left[\text{-483}\text{.6}\text{+}\text{580}\text{.7}\right]\text{kJ}\\ \text{=}\text{97}\text{.1}\text{×}{\text{10}}^{\text{3}}\text{J}\text{.}\end{array}$

Calculate the value of ${\text{ΔS}}^{\text{o}}$

$\begin{array}{l}{\text{ΔS}}^{\text{o}}\text{=}\left[\left(\text{2}\text{×}\text{188}\text{.7}\right)\text{+}\text{51}\text{.55}\text{-}\left(\text{2}\text{×}\text{130}\text{.6}\right)\text{-}\text{52}\text{.3}\right]\text{J/K}\\ \text{=}\left[\text{428}\text{.95}\text{-}\text{261}\text{.2}\text{-}\text{52}\text{.3}\right]\text{J/K}\\ \text{=}\text{115}\text{.45}\text{J/K}\text{.}\end{array}$

For the reaction of tin oxide with hydrogen

• The value of enthalpy change $\left({\text{ΔH}}^{\text{o}}\right)$ is $\text{97}\text{.1}\text{×}{\text{10}}^{\text{3}}\text{J}\text{.}$
• The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ is $\text{115}\text{.45}\text{J/K}\text{.}$

For the second reaction,

$\begin{array}{l}{\text{SnO}}_{\text{2}}{}_{\text{(s)}}\text{+ }{\text{C}}_{\text{(s)}}\stackrel{}{\to }{\text{ Sn}}_{\left(\text{s}\right)}\text{+}{\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{:}\text{-580}\text{.7}\text{0}\text{0}\text{-393}\text{.5}\text{kJ}\\ {\text{S}}^{\text{o}}\text{:}\text{52}\text{.3}\text{5}\text{.740}\text{51}\text{.55}\text{213}\text{.7}\text{J/K}\\ {\text{Temperature is 25 }}^{\text{o}}\text{C}\text{.}\end{array}$

Calculate the value of ${\text{ΔH}}^{\text{o}}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\text{-393}\text{.5}\text{-}\left(\text{-580}\text{.7}\right)\right]\text{kJ}\\ \text{=}\text{187}\text{.2}\text{kJ}\\ \text{=}\text{187}\text{.2}\text{×}{\text{10}}^{\text{3}}\text{J}\text{.}\end{array}$

Calculate the value of ${\text{ΔS}}^{\text{o}}$

$\begin{array}{l}{\text{ΔS}}^{\text{o}}\text{=}\left[\text{213}\text{.7}\text{+}\text{51}\text{.55}\text{-}\text{5}\text{.740}\text{-}\text{52}\text{.3}\right]\text{J/K}\\ \text{=}\left[\text{265}\text{.25}\text{-}\text{58}\text{.04}\right]\text{J/K}\\ \text{=}\text{207}\text{.21}\text{J/K}\text{.}\end{array}$

For the reaction of tin oxide with carbon

• The value of enthalpy change $\left({\text{ΔH}}^{\text{o}}\right)$ is $\text{187}\text{.2}\text{×}{\text{10}}^{\text{3}}\text{J}\text{.}$
• The value of entropy change $\left({\text{ΔS}}^{\text{o}}\right)$ is $\text{207}\text{.21}\text{J/K}\text{.}$

(b)

Interpretation Introduction

Interpretation:

For the reaction of tin oxide with either carbon or hydrogen, the value of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{,}$ the temperature at which these processes becomes spontaneous and industrially preferred process to obtain tin has to be given.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Spontaneous process:

The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.

Answer to Problem 18.91QP

At temperature higher than $\text{841}\text{K}$ the reaction of tin oxide with hydrogen will be spontaneous.

At temperature higher than $\text{904}\text{K}$ the reaction of tin oxide with carbon will be spontaneous.

Explanation of Solution

To calculate: At which temperature the given reaction will become spontaneous

The reaction of tin oxide with hydrogen is given.

Consider free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ is zero.  For spontaneous process the sign of free energy change should be negative $\left(\text{ΔG}\text{<}\text{0}\right)$.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{0}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{Substitute}\text{the}\text{values}\text{of}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}\\ \text{T}\text{=}\frac{{\text{ΔH}}^{\text{o}}}{{\text{ΔS}}^{\text{o}}}\text{=}\frac{\text{97}\text{.1}\text{×}{\text{10}}^{\text{3}}\text{J}}{\text{115}\text{.45}\text{J/K}}\\ \text{=}\text{841}\text{K}\text{.}\end{array}$

At temperature higher than $\text{841}\text{K}$ , the given reaction will be spontaneous.

The reaction of tin oxide with carbon is given.

Consider free energy change $\left({\text{ΔG}}^{\text{o}}\right)$ is zero.  For spontaneous process the sign of free energy change should be negative $\left(\text{ΔG}\text{<}\text{0}\right)$.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{0}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{Substitute}\text{the}\text{values}\text{of}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}\\ \text{T}\text{=}\frac{\text{ΔH}}{\text{ΔS}}\text{=}\frac{\text{187}\text{.2}\text{×}{\text{10}}^{\text{3}}\text{J}}{\text{207}\text{.21}\text{J/K}}\\ \text{=}\text{903}\text{K}\text{.}\end{array}$

At temperature higher than $\text{903}\text{K}$ the given reaction will be spontaneous.

(c)

Interpretation Introduction

Interpretation:

For the reaction of tin oxide with either carbon or hydrogen, the value of ${\text{ΔH}}^{\text{o}}\text{,}{\text{ΔS}}^{\text{o}}\text{,}$ the temperature at which these processes becomes spontaneous and industrially preferred process to obtain tin has to be given.

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

Spontaneous process:

The chemical or physical change can takes place by itself without the help of surroundings are called as spontaneous process.

Answer to Problem 18.91QP

The reaction with hydrogen is preferred if we consider temperature.  Carbon is less expensive compare to hydrogen.  On this basis the reaction of tin oxide with carbon is preferred.  Carbon is used as reducing agent for the production of tin.

Explanation of Solution

To give: The preferred industrial process for preparing tin

The reaction of tin oxide with hydrogen is preferred if we consider temperature.

Carbon is less expensive compare to hydrogen.  On this basis the reaction of tin oxide with carbon is preferred.

Carbon is used as reducing agent for the production of tin.