Q20.54 P - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the oxidation of carbon monoxide:

 $\mathbf{\text{CO(g)+}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

(a) Predict the signs of ${\mathbf{\text{ΔS}}}^{\mathbf{\text{°}}}$ and ${\mathbf{\text{ΔH}}}^{\mathbf{\text{°}}}$ . Explain.

(b) Calculate ${\text{ΔG}}^{\text{°}}$ by two different methods

Step-by-Step Solution

Answer

1. Given reaction the entropy  ${\text{ΔS}}^{\text{°}}$ is negative and enthalpy  ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$  is also negative.
2. The standard free energy value is  $\text{-257.2kJ}$.

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: Predict the signs of ΔS ° , ΔH ° and Explain

(a)

Considering the given information:

The chemical equation for the formation of ${\text{CO}}_{\text{2}}\text{( g)}$  from its elements is shown below:  

$\text{CO(g)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{CO}}_{\text{2}}\text{(g)}$

According to the above equation, one mole of   ${\text{CO}}_{\text{2}}$is formed from its elements ${\text{CO}}_{\text{2}}$  in their standard states.

Predicting entropy  ${{\text{ΔS}}^{\text{°}}}_{\text{system }}$:

 

 $\text{1,}\frac{\text{1}}{\text{2}}$ moles of gaseous reactants are converted into one mole of gaseous product in this reaction.

 $\begin{array}{c}{\text{ΔS}}_{\text{rxn }}^{\text{0}}{\text{=S}}_{\text{Products }}^{\text{0}}{\text{-S}}_{\text{reactants }}^{\text{0}}\\ {\text{ΔS}}_{\text{rxn }}^{\text{°}}\text{=1-2}\\ {\text{ΔS}}_{\text{rxn }}^{\text{°}}\text{=-1}\end{array}$

Because the number of moles of gas decreases from reactants to products, the reaction entropy   ${\text{ΔS}}^{\text{°}}$is negative.

Enthalpy as a predictor of ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ :

The oxidation of carbon monoxide CO requires an initial energy input to start the reaction but then releases energy that is exothermic and has a negative enthalpy ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ .

Step 3: Calculate ΔG ° by two different methods

(b)

Considering the given information:

The chemical equation for the formation of  ${\text{CO}}_{\text{2}}\text{( g)}$ from its elements is shown below:  

$\text{CO(g)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{CO}}_{\text{2}}\text{(g)}$

1st method:

Calculate ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$  from  ${\text{ΔG}}_{\text{f}}^{\text{o}}$ values of products and reactants,

The Gibbs free energy equation is as follows:

 ${\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{{\text{ΔG}}_{\text{f}}}^{\text{°}}\text{(Products)-}\sum \text{n}{{\text{ΔG}}_{\text{f}}}^{\text{°}}\text{(Reactants)}$

The reaction's free energy change is calculated as follows:

 $\begin{array}{l}{\text{ΔG}}_{\text{rxn }}^{\text{o}}\text{=}[\left({\text{1 molCO}}_{\text{2}}\right)({\text{ΔG}}_{\text{f}}^{\text{o}}\text{of}{\text{CO}}_{\text{2}})]\text{-}[\text{(1molCO)}({\text{ΔG}}_{\text{f}}^{\text{o}}\text{ofCO)+}\left({\text{1/2molO}}_{\text{2}}\right)({\text{ΔG}}_{\text{f}}^{\text{o}}\text{of}{\text{O}}_{\text{2}})]\\ {\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=[(2 mol)(-394.4 kJ/mol)]-[(2 mol)(-137.2 kJ/mol)}\text{+}\left({\text{1/2 molO}}_{\text{2}}\right)\text{(0 kJ/mol)}]\end{array}$

${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-257.2 kJ}$

As a result, the standard free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{-257.2}}\text{kJ}$.

2nd Method

Given the reaction,

 $\text{CO(g)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{CO}}_{\text{2}}\text{(g)}$

Calculate  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$ from  ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ and ${\text{ΔS}}_{\text{rxn}}^{\text{o}}$  at  $\text{298K}$

The change in standard enthalpy is,

The enthalpy change for the reaction is calculated as follows:

 $\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{\text{ΔH}}_{\text{f( Products )}}^{\text{°}}\text{-}\sum {\text{nΔH}}_{\text{f(reactants )}}^{\text{°}}\\ {\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}\left[\left({\text{1 molCOO}}_{\text{2}}\right)\left({\text{ΔH}}_{\text{rxn}}^{\text{o}}{\text{ of CO}}_{\text{2}}\right)\right]\text{-}\left[\text{(1 molCO)}\left({\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{ of CO }\right)\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\left({\text{ΔH}}_{\text{rxn}}^{\text{o}}{\text{ of O}}_{\text{2}}\right)\right]\\ {\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}\left[{\text{(1 molCO)}}_{\text{2}}\right)\text{(-393.5 kJ/mol)}]\text{-[(1 molCO)(-110.5 kJ/mol)}\text{+}\left({\text{1/2molOO}}_{\text{2}}\right)\text{(0 kJ/mol)}]\\ {\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=-283.0kJ}\end{array}$

The enthalpy change is negative.

As a result, the enthalpy  $\left({\text{ΔH}}_{\text{rxn}}^{\text{°}}\right)\text{value is -283.0kJ}$

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

The standard entropy change equation is,

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

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

 $\begin{array}{l}{\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=}\left[\left({\text{2 molCO}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ of CO}}_{\text{2}}\right)\right]\text{-}\left[\text{(1 molCO)}\left({\text{S}}^{\text{o}}\text{ of CO }\right)\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ of O}}_{\text{2}}\right)\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=[(1 molCO2)(213.7 J/mol×K)]-[(1 molCO)(197.5 J/mol×K)}\text{+}\left({\text{1/2molO}}_{\text{2}}\right)\text{(205.0 J/mol×K)}]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=-86.3 J/K}\end{array}$

Therefore, the  $\left({\text{ΔS}}_{\text{rxn }}^{\text{°}}\right)$of the reaction is  $\text{-86.3JJ/K}$

Next, compute the change in free energy  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}$

Standard The equation for free energy change is,

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

Free energy change  ${\text{ΔG}}_{\text{f}}^{\text{o}}$

Enthalpy and entropy values calculated are

$\begin{array}{l}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=-283.0kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=-283.0 kJ}\end{array}$

These values are entered into the standard free energy equation,

 $\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-283.0 kJ-}\left[\text{(298 K)(-86.3 J/K)}\left({\text{1kJ/10}}^{\text{3}}\text{J}\right)\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=-257.2826 kJ}\end{array}$

Therefore, the standard free energy value is ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=}\underset{\_}{\text{-257.3kJ}}$ .