Chapter 20, Problem 20.110P

(a)

Interpretation Introduction

Interpretation:

The temperature in which reaction becomes spontaneous and corresponding temperature has to be calculated

Concept introduction:

Spontaneous process: A process which is initiated by itself, without the help of external energy source is called spontaneous process.  All spontaneous process is associated with the decrease in free energy in the system.

Entropy is a thermodynamic quantity, which is the measure of randomness in a system.  The term entropy is useful in explaining the spontaneity of a process.  For all spontaneous process in an isolated system there will be an increase in entropy.  Entropy is represented by the letter ‘S’.  It is a state function.  The change in entropy gives information about the magnitude and direction of a process.  Factors like temperature, molar mass, molecular complexity and phase transition occurring in a reaction influences the entropy in a system.

Free energy (or) entropy change is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ${\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}$

Where,

  ${\text{ΔG}}^{\text{o}}$ is the standard change in free energy of the system

  $\Delta {{\rm H}}^{\text{o}}$ is the standard change in enthalpy of the system

  $\text{T}$ is the absolute value of the temperature

  $\Delta {\text{S}}^{\text{o}}$ is the change in entropy in the system

Answer to Problem 20.110P

Given reaction the calculated temperature value is $766K.$

Explanation of Solution

Given $C_2{H}_2$ formation reaction is,

  $2C{H}_4\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\stackrel{}{\to }{C}_2{H}_4\left(g\right)+2H_2\left(g\right)+H_{2}O\left(g\right)$

In this reaction two mole of $C{H}_4$ reacted with half mole $O_2$ produced to one mole of acetylene.

Standard enthalpy change is,

The enthalpy change for the $C_2{H}_4$ formation reaction is calculated as follows,

  ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}\Delta {\text{H}}^{\text{°}}{}_{\text{f(Products)}}-{\displaystyle \sum \text{n}}\Delta {\text{H}}^{\text{°}}{}_{\text{f(reactants)}}$

  $\begin{array}{l}{\text{ΔH}}_{rxn}^o\text{ =}\left[\begin{array}{l}\left(1\text{mol}{C}_2{H}_2\right)\left({\text{ΔH}}_f^{o}of{C}_2{H}_2\right)+\left(2\text{mol}{H}_2\right)\left({\text{ΔH}}_f^{o}of{H}_2\right)\\ +\left(1\text{mol}{H}_{2}O\right)\left({\text{ΔH}}_f^{o}of{H}_{2}O\right)\end{array}\right]\\ \\ -\left[\left(2molC{H}_4\right)\left({\text{ΔH}}_f^{o}ofC{H}_4\right)+\left(\text{1/2}mol{O}_2\right)\left({\text{ΔH}}_f^{o}of{O}_2\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}\left[\begin{array}{l}\left(1\text{mol}\right)\left(227kJ/mol\right)+\left(2\text{mol}\right)\left(0kJ/mol\right)\\ +\left(1\text{mol}\right)\left(-241.826kJ/mol\right)\end{array}\right]-\\ \\ \left[\left(2mol\right)\left(-74.87kJ/mol\right)+\left(\text{1/2}mol{O}_2\right)\left(0kJ/mol\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}134.941kJ.\end{array}$

Hence, the enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ changes $\underset{\_}{134.941kJ}.$

Entropy change  ${\text{ΔS}}^{\text{°}}{}_{system}$

Calculate the change in entropy for this reaction as follows,

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

Where, $\left(m\right)and\left(n\right)$ are the stoichiometric co-efficient.

  $\begin{array}{l}{\text{ΔS}}_{rxn}^o\text{ =}\left[\begin{array}{l}\left(1\text{mol}{C}_2{H}_2\right)\left(S^{o}of{C}_2{H}_2\right)+\left(2\text{mol}{H}_2\right)\left(S^{o}of{H}_2\right)\\ +\left(1\text{mol}{H}_{2}O\right)\left(S^{o}of{H}_{2}O\right)\end{array}\right]\\ \\ -\left[\left(2molC{H}_4\right)\left(S^{o}ofC{H}_4\right)+\left(\text{1/2}mol{O}_2\right)\left(S^{o}of{O}_2\right)\right]\\ \\ {\text{ΔS}}_{rxn}^o\text{ =}\left[\begin{array}{l}\left(1\text{mol}\right)\left(200.85\text{J/mol}\cdot K\right)+\left(2\text{mol}\right)\left(\text{130}\text{.6}\text{J/mol}\cdot K\right)\\ +\left(1\text{mol}\right)\left(188.72\text{J/mol}\cdot K\right)\end{array}\right]-\\ \\ \left[\left(2mol\right)\left(186.1\text{J/mol}\cdot K\right)+\left(\text{1/2}mol\right)\left(205.0\text{J/mol}\cdot K\right)\right]\\ \\ {\text{ΔS}}_{rxn}^o\text{ =}176.07=176.1J/K.\end{array}$

The ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{176.1J/K}.$

The entropy change is positive sign for ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$ indicates the formation of $C_2{H}_4$ is product favored.

Determination for temperature (T)

The reaction will become spontaneous when $\Delta G$ change from being positve to being negative. This point occuers when ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=0$.

Standared Free energy change equation is,

  $\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}\\ \\ {\text{ΔG}}^{\text{o}}=0=\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}-------\left[1\right]\\ \\ \Delta {{\rm H}}^{\text{o}}=\text{T}\Delta {\text{S}}^{\text{o}}------------\left[2\right]\end{array}$

Rearrange the equation (2) to calculate temprature T,

$\text{T}=\frac{\Delta {\rm H}}{\Delta S}------\left[3\right]$

Hence,

Calculated enthalpy $\Delta H$ and entropy $\Delta S$ values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=1\text{34}\text{.941}\text{kJ}\\ \\ \Delta S_{\text{rxn}}^{\text{o}}=176.07\text{J/K}\end{array}$

  $\begin{array}{c}T=\frac{134.914kJ}{176.07J/K}\left(\frac{{10}^3}{1kJ}\right)=766.25206\\ \\ T=766K.\end{array}$

At temprature above $766K$, reaction is spontaneous. Because both enthalpy $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}$ and entropy $\Delta S_{\text{rxn}}^{\text{o}}$ values are positve, the reaction becomes spontaneous above this temprature.

(b)

Interpretation Introduction

Interpretation:

The temperature for the formation of acetylene from carbon and hydrogen has to be calculated.

Concept introduction:

Free energy (or) entropy change is the term that is used to explain the total energy content in a thermodynamic system that can be converted into work.  The free energy is represented by the letter G.  All spontaneous process is associated with the decrease of free energy in the system.  The equation given below helps us to calculate the change in free energy in a system.

  ${\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}$

Entropy is a thermodynamic quantity, which is the measure of randomness in a system.  The term entropy is useful in explaining the spontaneity of a process. For all spontaneous process in an isolated system there will be an increase in entropy. Entropy is represented by the letter ‘S’.  It is a state function.  The change in entropy gives information about the magnitude and direction of a process.  The entropy changes associated with a phase transition reaction can be found by the following equation.

  $\text{Τ }\text{}\text{=}\text{}\text{}\frac{{\text{ΔΗ}}^{\text{o}}}{{\text{ΔS}}^{\text{o}}}$

Where,

  $\Delta {{\rm H}}^{\text{o}}$ is the change in enthalpy of the system

  ${\text{ΔG}}^{\text{o}}$ is the standard change in free energy of the system

  $\text{T}$ is the absolute value of the temperature

Answer to Problem 20.110P

Acetylene formation reaction calculated temperature value is $3860K$.

Explanation of Solution

Given $C_2{H}_2$ formation reaction is,

  $2C\left(s\right)+H_2\left(g\right)\stackrel{}{\to }{C}_2{H}_4\left(g\right)$

In this reaction, elements of carbon (graphite) reacted with hydrogen produced a one mole of acetylene.

Standard enthalpy change is,

The enthalpy change for the $C_2{H}_4$ formation reaction is calculated as follows,

  ${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}\Delta {\text{H}}^{\text{°}}{}_{\text{f(Products)}}-{\displaystyle \sum \text{n}}\Delta {\text{H}}^{\text{°}}{}_{\text{f(reactants)}}$

  $\begin{array}{l}{\text{ΔH}}_{rxn}^o\text{ =}\left[\left(1\text{mol}{C}_2{H}_2\right)\left({\text{ΔH}}_f^{o}of{C}_2{H}_2\right)\right]\\ \\ -\left[\left(2molC\right)\left({\text{ΔH}}_f^{o}ofC\right)+\left(\text{1}mol{h}_2\right)\left({\text{ΔH}}_f^{o}of{O}_2\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}\left[\left(1\text{mol}\right)\left(227kJ/mol\right)\right]-\\ \\ \left[\left(2mol\right)\left(0kJ/mol\right)+\left(\text{1}mol{O}_2\right)\left(0kJ/mol\right)\right]\\ \\ {\text{ΔH}}_{rxn}^o\text{ =}227kJ.\end{array}$

Hence, the enthalpy ${\text{(ΔH}}^{\text{°}}{}_{\text{rxn}}\text{)}$ changes $\underset{\_}{227kJ}.$

Entropy change  ${\text{ΔS}}^{\text{°}}{}_{system}$

Calculate the change in entropy for this reaction as follows,

  ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}\text{ = }{\displaystyle \sum \text{m }}{\text{S}}^{\text{°}}{}_{\text{Products}}-{\displaystyle \sum \text{n}}{\text{S}}^{\text{°}}{}_{\text{reactants}}$

Where, $\left(m\right)and\left(n\right)$ are the stoichiometric co-efficient.

  $\begin{array}{l}{\text{ΔS}}_{rxn}^o\text{ =}\left[\left(1\text{mol}{C}_2{H}_2\right)\left(S^{o}of{C}_2{H}_2\right)\right]\\ \\ -\left[\left(2molC\right)\left(S^{o}ofC\right)+\left(\text{1}mol{H}_2\right)\left(S^{o}of{H}_2\right)\right]\\ \\ {\text{ΔS}}_{rxn}^o\text{ =}\left[\left(1\text{mol}\right)\left(200.85\text{J/mol}\cdot K\right)\right]-\\ \\ \left[\left(2mol\right)\left(5.686\text{J/mol}\cdot K\right)+\left(\text{1}mol\right)\left(130.6\text{J/mol}\cdot K\right)\right]\\ \\ {\text{ΔS}}_{rxn}^o\text{ =}58.878=58.9J/K.\end{array}$

The ${\text{(ΔS}}^{\text{°}}{}_{\text{rxn}}\text{)}$ of the reaction is $\underset{\_}{58.9J/K}.$

The entropy change is positive sign for ${\text{ΔS}}^{\text{°}}{}_{\text{rxn}}$ indicates the formation of $C_2{H}_4$ is product favored.

Determination for temperature (T)

The reaction will become spontaneous when $\Delta G$ change from being positve to being negative. This point occuers when ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=0$.

Standared Free energy change equation is,

  $\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{ = }\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}\\ \\ {\text{ΔG}}^{\text{o}}=0=\Delta {{\rm H}}^{\text{o}}\text{- T}\Delta {\text{S}}^{\text{o}}-------\left[1\right]\\ \\ \Delta {{\rm H}}^{\text{o}}=\text{T}\Delta {\text{S}}^{\text{o}}------------\left[2\right]\end{array}$

Rearrange the equation (2) to calculate temprature T,

$\text{T}=\frac{\Delta {\rm H}}{\Delta S}------\left[3\right]$

Hence,

Calculated enthalpy $\Delta H$ and entropy $\Delta S$ values are

  $\begin{array}{l}\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}=227\text{kJ}\\ \\ \Delta S_{\text{rxn}}^{\text{o}}=58.878\text{J/K}\end{array}$

  $\begin{array}{c}T=\frac{227kJ}{58.878J/K}\left(\frac{{10}^3}{1kJ}\right)=3855.43\\ \\ T=3860K\end{array}$

At temprature will become above $3860K$, this also reaction spontaneous. Because both enthalpy $\Delta {{\rm H}}_{\text{rxn}}^{\text{o}}$ and entropy $\Delta S_{\text{rxn}}^{\text{o}}$ values are positve, the reaction becomes spontaneous above this temprature.

(c)

Interpretation Introduction

Interpretation:

The reason for the immediate cooling of reaction mixture has to be identified.

Concept introduction:

Chemical equilibrium: The term applied to reversible chemical reactions. It is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. The equilibrium is achieved; the concentrations of reactant and products become constant.

Forward Reaction: This type of reaction has involved irreversible, if obtained product cannot be converted back in to respective reactants under the same conditions. Backward Reaction: This type of reaction process involved a reversible, if the products can be converted into a back to reactants.

Thermal decomposition reaction: This reaction caused by heat or decomposition of starting substance is the temperature at which the substance chemically decomposes. In other words large molecules being broken down into single elements (or) compounds.

Answer to Problem 20.110P

Acetylene formation is,

  $2C{H}_4\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\stackrel{}{\to }{C}_2{H}_4\left(g\right)+2H_2\left(g\right)+H_{2}O\left(g\right)$

Explanation of Solution

Given $C_2{H}_2$ formation reaction is,

  $2C{H}_4\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\stackrel{}{\to }{C}_2{H}_4\left(g\right)+2H_2\left(g\right)+H_{2}O\left(g\right)$

Considering the revrese reaction of its formation. The acetylene is product under conditions at which acetylene is unstable aand can decompose back into its elements. It must be quickely cooled to a temprature where its thermal decomposition rate is slow.

The reaction rate is higher at the temperature. The time required (kinetics) overshadows the lower yield (thermodynamics).