Q20.59P - Chemistry: Molecular Nature Of Matter And Change

Question

A reaction that occurs in the internal combustion engine is

 ${\mathbf{\text{N}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g)+O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\rightleftharpoons }\mathbf{\text{2NO(g)}}$

 (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  ${\mathbf{\text{100.}}}^{\mathbf{\text{°}}}{\mathbf{\text{C, 2560.}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}$, and  ${\mathbf{\text{3540.}}}^{\mathbf{\text{°}}}\mathbf{\text{C}}$.

(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 $\text{180.58kJ}$  and entropy $\left({\text{ΔS}}_{\text{rxn }}^{\text{°}}\right)$   value is $\text{24.8./K}$ .
2. The standard free energy values are ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=171.33kJ, 110.3kJ}$  and  $\text{86.0kJ}$.
3. Given reaction values of   $\text{ΔG}$ become smaller at higher 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 NO formation reaction:

 ${\text{N}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2NO(g)}$

One mole of nitrogen monoxide was produced when an equimolar of  ${\text{N}}_{\text{2}}$ reacted with ${\text{O}}_{\text{2}}$ .

Standard enthalpy change is,

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

 $\begin{array}{l}{\text{N}}_{\text{2}}\text{(g)=0 kJ/mol}\\ {\text{O}}_{\text{2}}\text{(g)=0 kJ/mol}\\ \text{NO(g)=92.29 kJ/mol}\end{array}$

The reaction's enthalpy change is calculated as follows:

 $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\sum \text{m}{\text{ΔH}}_{\text{f( Products )}}^{\text{0}}\text{-}\sum \text{n}{\text{ΔH}}_{\text{f(reactants )}}^{\text{°}}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}[\text{(2 molNO)}({\text{ΔH}}_{\text{f}}^{\text{0}}\text{of}\text{NO})]\\ \text{-}[\left({\text{1molN}}_{\text{2}}\right)({\text{ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{N}}_{\text{2}})\text{+}\left({\text{1molO}}_{\text{2}}\right)({\text{ΔH}}_{\text{f}}^{\text{°}}\text{of}{\text{O}}_{\text{2}})]\\ {\text{ΔH}}_{\text{rxn }}^{\text{0}}\text{=[(2molNO)(90.29 kJ/mol)]}\\ \text{-}[\left({\text{1molN}}_{\text{2}}\right)\text{(0 kJ/mol)}\text{+}\left({\text{1molO}}_{\text{2}}\right)\text{(0 kJ/mol)}]\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=180.58 kJ}\end{array}$

The reaction's enthalpy change is positive.

Hence, the enthalpy $\left({\text{ΔH}}_{\text{rxn}}^{\text{°}}\right)$  value is  $\text{180.58kJ}$.

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.

 $\begin{array}{c}{\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=}\left[\text{(2 molNO)}\left({\text{S}}^{\text{o}}\text{ of CO }\right)\right]\text{-}\left[\left({\text{1 molN}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ of N}}_{\text{2}}\right)\text{+}\left({\text{1 molO}}_{\text{2}}\right)\left({\text{S}}^{\text{o}}{\text{ of O}}_{\text{2}}\right)\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=[(2 molNO)(210.65 J/mol×K)]}\\ \text{-}\left[\left({\text{1 mol N}}_{\text{2}}\right)\text{(238.0 J/mol×K)+}\left({\text{1 molO}}_{\text{2}}\right)\text{(205.0 J/mol×K)}\right]\\ {\text{ΔS}}_{\text{rxn}}^{\text{°}}\text{=24.8J/K}\end{array}$

The $\left({\text{ΔS}}_{\text{rxn}}^{\text{0}}\right)$  of the reaction is $\underset{\_}{\text{24.8J/K}}$ .

For ${\text{ΔS}}_{\text{rxn}}^{\text{°}}$, the enthalpy and entropy changes are positive, indicating that the formation of NO is favored at the given temperature.

 

Step 3: Calculate ΔG °

(b)

Considering the given NO formation reaction:

 ${\text{N}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2NO(g)}$

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

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,

 $\begin{array}{c}{\text{ΔH}}_{\text{r×n}}^{\text{o}}\text{=180.58kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=24.8J/K}\\ {\text{T}}_{\text{1}}\text{=273+100=373K}\end{array}$

These numbers are plugging into the standard free energy equation,

 $\begin{array}{c}{\text{ΔG}}_{\text{373}}^{\text{o}}\text{=180.58 kJ-}\left[\text{(373 K)(24.8 J/K)}\left({\text{1 kJ/10}}^{\text{3}}\right)\right]\\ \text{ =171.3299 kJ}\\ {\text{ΔG}}_{\text{373}}^{\text{o}}\text{=171.33kJ}\end{array}$

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

The enthalpy and entropy values calculated are,

 $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=180.58kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=24.8J/K}\\ {\text{T}}_{\text{I}}\text{=273+2560=2833K}\end{array}$

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

 $\begin{array}{c}{\text{ΔG}}_{\text{2833}}^{\text{o}}\text{=180.58 kJ-}\left[\text{(2833 K)(24.8 J/K)}\left({\text{1 kJ/10}}^{\text{3}}\right)\right]\\ \text{ =110.3216 kJ}\\ {\text{ ΔG}}_{\text{rxn}}^{\text{o}}\text{=110.3kJ}\end{array}$

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

The enthalpy and entropy values calculated are,

 $\begin{array}{c}{\text{ΔH}}_{\text{rxn }}^{\text{o}}\text{=180.58kJ}\\ {\text{ΔS}}_{\text{rxn}}^{\text{o}}\text{=24.8J/K}\\ {\text{ T}}_{\text{I}}\text{=273+3540=3813K}\end{array}$

These values are plugging above standard free energy equation,

 $\begin{array}{c}{\text{ΔG}}_{\text{3813}}^{\text{o}}\text{=180.58 kJ-}\left[\text{(3813 K)(24.8 J/K)}\left({\text{1 kJ/10}}^{\text{3}}\right)\right]\text{=86.0176 kJ}\\ {\text{ ΔG}}_{\text{rxn}}^{\text{o}}\text{=86.0kJ}\end{array}$

Independent standard free energy values are  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}\text{=171.33kJ,110.3kJ}$ and $\text{86.0kJ}$ .

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

(c)

Considering the given NO formation reaction:

 ${\text{N}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons \text{2NO(g)}$

At higher temperatures, the value of  $\text{ΔG}$ decreases. At any of these temperatures, the reaction is not spontaneous; however, as the temperature rises, the reaction becomes less non-spontaneous.