Q20.58 P - Chemistry: Molecular Nature Of Matter And Change

Question

One reaction used to produce small quantities of pure ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$ is

 ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}\mathbf{\text{OH(g)}}\mathbf{\rightleftharpoons }{\mathbf{\text{CO(g)+2H}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

 (a) Determine ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}$ and ${\mathbf{\text{ΔS}}}^{\mathbf{\text{°}}}$ for the reaction at $\mathbf{\text{298 K}}$ .

(b) Assuming that these values are relatively independent of temperature, calculate  ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$at${\mathbf{\text{28}}}^{\mathbf{\text{°}}}{\mathbf{\text{C,128}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}$ , and ${\mathbf{\text{228}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}$.

(c) What is the significance of the different values of   ${\mathbf{\text{ΔG}}}^{\mathbf{\text{°}}}$?

Step-by-Step Solution

Answer

1. The enthalpy  $\left({\text{ΔH}}_{\text{rxn }}^{\text{°}}\right)$ value is $\text{90.7kJ}$ and entropy $\left({\text{ΔS}}_{\text{rxn }}^{\text{°}}\right)$   value is $\text{221J/K}$ .

    The enthalpy and entropy changes values are a positive sign for  ${\text{ΔS}}_{\text{rxn}}^{\text{°}}$ indicates the formation of methanol is favored at a given temperature.

1. The standard free energy values are  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{24.3kJ,2.2kJ}}$ and $\text{-19.9kJ}$ .
2. Given the reaction the enthalpy and entropy values are positive at high temperatures.

Step 1: Definition of Concept

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

Step 2: Determine ΔH ° and ΔS ° for the reaction

(a)

Considering the given decomposition reaction:

 ${\text{CH}}_{\text{3}}\text{OH(g)}\rightleftharpoons {\text{CO(g)+2H}}_{\text{2}}\text{( g)}$

Calculate the change in Gibb's free energy at 298 K using the following formula:

Standard enthalpy change is,

${\text{ΔH}}^{\text{o}}$ Formation of values,

 $\begin{array}{l}\text{CO(g)=-110.5 kJ/mol}\\ {\text{H}}_{\text{2}}\text{( g)=0 kJ/mol}\\ {\text{CH}}_{\text{3}}\text{OH(g)=-201.2 kJ/mol}\end{array}$

The reaction's enthalpy change is calculated as follows:

 $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{\text{ΔH}}_{\text{f( Products) }}^{\text{°}}\text{-}\sum \text{n}{\text{ΔH}}_{\text{f( (reactants) }}^{\text{°}}\text{-}[\left({\text{1 molCH}}_{\text{3}}\text{OH}\right)({\text{ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{CH}}_{\text{3}}\text{OH})]\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{(1 molCO)(-110.5 kJ/mol)+}\left({\text{2 molHH}}_{\text{2}}\right)\text{(0 kJ/mol)}\right]\\ \text{-}\left[\left({\text{1 molCHCOH}}_{\text{3}}\right)\text{(-201.2 kJ/mol)}\right]\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=90.7kJ}\end{array}$

The reaction's enthalpy change is positive.

Hence, the enthalpy  $\left({\text{ΔH}}_{\text{rxn}}^{\text{°}}\right)$ value is  $\text{90.7kJ}$.

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

Calculate the entropy change in this reaction as follows:

 ${\text{DS}}_{\text{rxn }}^{\text{0}}\text{=}\sum \text{m}{\text{S}}_{{\text{P}}_{\text{roducts }}^{\text{0}}}^{\text{0}}\text{-}\sum \text{n}{\text{S}}_{\text{reactants }}^{\text{0}}$

The stoichiometric co-efficient are (m) and (n), respectively.

 $\begin{array}{c}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}[\text{(1 molCO)}({\text{S}}^{\text{o}}\text{of}\text{CO}){\text{+(2 molH)}}_{\text{2}})({\text{S}}^{\text{o}}\text{of}{\text{H}}_{\text{2}})]\\ \text{-}[\left({\text{1 molCH}}_{\text{3}}\text{OH}\right)({\text{S}}^{\text{o}}\text{of}{\text{CH}}_{\text{3}}\text{OH})]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=[(1 mol)(197.5 J/mol×K)+(2 mol)(130.6 J/mol×K)}\\ \text{-[(1 mol)(238.0 J/mol×K)]}\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=220.7=221J/K}\end{array}$

The  $\left({\text{ΔS}}_{\text{rxn}}^{\text{0}}\right)$ of the reaction is $\underset{\_}{\text{221J/K}}$ .

For ${\text{ΔS}}_{\text{rxn }}^{\text{°}}$ , the enthalpy and entropy changes are positive, indicating that the formation of methanol is favored at the given temperature.

 

Step 3: Calculate ΔG °

(b)

Considering the given decomposition reaction:

 ${\text{CH}}_{\text{3}}\text{OH(g)}\rightleftharpoons {\text{CO(g)+2H}}_{\text{2}}\text{( g)}$

Calculate the change in free energy  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$.

Standard Free energy change equation is,

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

Temperature$\left({\text{T}}_{\text{1}}\right)$ :

The enthalpy and entropy values calculated are,

 $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=90.7k.J}\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=220.7 J/K}\\ {\text{T}}_{\text{1}}\text{=28+273=301K}\end{array}$

These numbers are plugging into the standard free energy equation,

 $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=90.7 kJ-}\left[\text{(301 K)(220.7 J/K)}\left({\text{1 kJ/10}}^{\text{3}}\right)\right]\\ \text{ =24.2693 kJ}\\ {\text{ΔG}}_{\text{roo}}^{\text{o}}\text{=24.3kJ}\end{array}$

Temperature $\left({\text{T}}_{\text{2}}\right)$  :

The enthalpy and entropy values calculated are,

 $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=90.7kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=220.7J/K}\\ {\text{T}}_{\text{1}}\text{=128+273=401K}\end{array}$

These figures are used to fill in the blanks in the standard free energy equation.

 $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=90.7 kJ-}\left[\text{(401 K)(220.7 J/K)}\left({\text{1 kJ/10}}^{\text{3}}\right)\right]\\ {\text{ =2.1993 kJΔG}}_{\text{rxn}}^{\text{o}}\\ \text{ =2.2 kJ}\end{array}$

Temperature $\left({\text{T}}_{\text{3}}\right)$:

The enthalpy and entropy values calculated are,

 $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=90.7kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=220.7J/K}\\ {\text{T}}_{\text{1}}\text{=228+273=501K}\end{array}$

These values are plugging above standard free energy equation,

 $\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=90.7 kJ-}\left[\text{(501 K)(220.7 J/K)}\left({\text{1 kJ/10}}^{\text{3}}\right)\right]\\ \text{ =-19.8707 kJ}\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-19.9 kJ}\end{array}$

Independent standard free energy values are ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{24.3kJ,2.2kJand-19.9kJ}$ .

Step 4: Find the significance of the different values of ΔG °

(c)

Considering the given decomposition reaction:

 ${\text{CH}}_{\text{3}}\text{OH(g)}\rightleftharpoons {\text{CO(g)+2H}}_{\text{2}}\text{( g)}$

The enthalpy and entropy value calculated are:

 $\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=90.7 kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=220.7 J/K}\end{array}$

At ${\text{28}}^{\text{°}}\text{C}$ , the reaction is non-spontaneous, near equilibrium at  ${\text{128}}^{\text{°}}\text{C}$, and spontaneous at ${\text{228}}^{\text{°}}\text{C}$  for the substances in their standard states.

At high temperatures, reactions involving positive  ${\text{ΔH}}_{\text{rxn}}^{\text{°}}$ and ${\text{ΔS}}_{\text{rxn}}^{\text{°}}$  become spontaneous.