Chapter 18, Problem 18.50QP

(a)

Interpretation Introduction

Interpretation:

The equation has to be determined given carbon monoxide and nitric oxide atmospheric reactions.

Explanation of Solution

The atmospheric equilibrium reaction of given the different terms of process (a) is shown below.

  $\text{a)}\text{.}\text{2CO(g)}\text{+}\text{2NO(g)}\to {\text{2CO}}_{\text{2}}(g)\text{+}{\text{N}}_{\text{2}}(g)\text{-----[1]}$

The equal mole ratio of carbon monoxide $\text{CO}$ and nitric oxide $\text{NO}$ reacted in gas phase conditions to produce a two moles of ${\text{CO}}_{\text{2}}$ and nitrogen (${\text{N}}_{\text{2}}$) gas, the balanced equilibrium reactions shown above.

(b)

Interpretation Introduction

Interpretation:

The oxidizing and reducing agent has to be identified given atmospheric reactions.

Concept Introduction:

An oxidizing agent is a substance that causes oxidation by accepting electron. The oxidizing agent is reduced.

A reducing agent is a substance that causes reduction by donating electrons. The reducing agent is oxidized.

Explanation of Solution

The atmospheric equilibrium reaction is,

$\begin{array}{l}\text{i)}\text{.}\text{2CO(g)}\text{+}\text{2NO(g)}\to {\text{2CO}}_{\text{2}}(g)\text{+}{\text{N}}_{\text{2}}(g)\text{-----[1]}\\ \text{ii)}\text{.}\text{NO---Oxidizing}\text{agent}\\ \text{CO---Reducing agent}\text{(Gains of}\text{electrons)}\text{. }\end{array}$ 

Given statement (b) nitric oxide ($\text{NO}$) is a very good oxidizing agent it is gains a one electrons form carbon monoxide ($\text{CO}$), will get (${\text{CO}}_{\text{2}}$) $(N_2)$ gas.

Examination above reaction clearly shows that $(NO)$ is a oxidizing agent since it added oxygen $(O)$ to $(CO)$ in reactant side in which means it oxidized the $(CO)$ to $(C{O}_2)$

The $(CO)$ is reactant since it accepts one oxygen $(O)$ from $(NO)$ and reduces it hence $(CO)$ acts as reducing agent which gets oxidized finally.

(c)

Interpretation Introduction

Interpretation:

The equilibrium pressure $(\text{Kp})$ value has to be calculated given gases phase equilibrium reaction at ${\text{25}}^{°}\text{C}$. 

Concept Introduction:

Free energy (Gibbs free energy) 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 $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$

    $\begin{array}{l}{\text{ΔG}}^{\text{0}}\text{=-RTln}\text{K}\\ \text{ΔG}=Freeenergy\\ {\text{ΔG}}^{\text{0}}=S\tan dard-statefreeenergy\\ \text{R}\text{=}\text{Gas}\text{Constant}(\text{0}\text{.0826}\text{l}\text{.atm/K}\text{.atm})\\ T=\text{Temprature}\text{273}\text{K}\\ \text{K=}\text{Equlibrium}\text{Constant}{\text{(K}}_{\text{P}}\text{and}{\text{K}}_{\text{C}}\text{)}\end{array}$

Explanation of Solution

The atmospheric reaction fallow as,

$\begin{array}{l}\text{2CO}\text{+}\text{2NO}\to {\text{2CO}}_{\text{2}}\text{+}{\text{N}}_{\text{2}}\\ {\text{ΔG}}_{\text{f}}^{\text{0}}={\text{CO}}_{\text{2}}\text{=}394.4\text{KJ/mol}\\ {\text{ΔG}}_{\text{f}}^{\text{0}}\text{=}{\text{N}}_{\text{2}}\text{=}137.3\text{KJ/mol}\text{(This}\text{values}\text{from}\text{equilibrium}\text{table-3)}\\ {\text{ΔG}}_{\text{f}}^{\text{0}}\text{=}\text{CO=}86.1\text{KJ/mol,}\text{NO=}\text{86}\text{.7}\text{KJ/mol}\\ {\text{ΔG}}_{\text{rxn}}^0={\displaystyle \sum n{\text{ΔGf}}^{\text{0}}\text{(Products)}}{\displaystyle \sum m{\text{Δf}}^0(Reac\tan ts)}\\ {\text{ΔG}}^{\text{0}}{}_{\text{rxn}}=2{\text{ΔG}}_{\text{f}}^0(C{O}_2)+{\text{ΔG}}_{\text{f}}^0(N_2)-2{\text{ΔG}}_{\text{f}}^0(CO)-2{\text{ΔG}}_{\text{f}}^0(NO)\end{array}$

$\begin{array}{l}\text{InK}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{6.876\times {10}^5\text{KJ/mol})}{(8.314\text{J/K}\cdot \text{mol)}(298K)}\\ \text{InK}\text{=}\frac{{\text{ΔG}}^{\text{0}}}{\text{-RT}}=\frac{6.876\times {10}^5\text{KJ/mol})}{(2477.572)}\\ \text{InK}\text{=}\text{277}\text{.5}(\because e^{277.5}=3.286)\\ \text{Kp=}\text{3}{\text{.29×10}}^{\text{120}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Given equilibrium reaction, the $(Qp)$ and reaction direction has to be determined.

Concept Introduction:

Free energy (Gibbs free energy) 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 $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

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

Where, $\text{"n"}$ is the number of moles

Reaction quotient: This type of chemical equilibrium reaction proceeds likely to produced, given either the pressure (or) the concentration of the reactants and the products. The value can be compared to the equilibrium constant, to determine the direction of the reaction that is take place. Then reaction quotient (Qc) the indication of Q can be used to determine which direction will shift to reach of chemical equilibrium process.

Explanation of Solution

$\begin{array}{l}\text{a)}\text{.}\text{2CO}\text{+}\text{2NO}\to {\text{2CO}}_{\text{2}}\text{+}{\text{N}}_{\text{2}}\text{-----[1]}\\ \text{Apply for partial pressure values for reaction (1)}\\ \text{2CO(g)}\text{+}\text{2NO(g)}\to {\text{2CO}}_{\text{2}}(g)\text{+}{\text{N}}_{\text{2}}(g)\\ \text{Qp=}\frac{(P_{N_2}){(P_{C{O}_2})}^2}{{(P_{CO})}^2{(P_{NO})}^2}=\frac{\text{Product}}{\text{Reactant}}-----[2]\\ \text{Qp=}\frac{(0.80){(3.0\times {10}^{-4})}^2}{{(5.0\times {10}^{-5})}^2{(5.0\times {10}^{-7})}^2}\\ \text{Qp=}1.2\times {10}^{14}\end{array}$

The $(Qp)$ value is $<<Kp$ so, reaction will proceeds from left to right side. 

(e)

Interpretation Introduction

Interpretation:

The raising temperatures favor the formation of $N_2$ and $C{O}_2$ should be determined.

Concept Introduction:

Standard enthalpy:  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 $\text{G}$.  All spontaneous process is associated with the decrease of free energy in the system.  The standard free energy change ${\text{(ΔG}}^{\text{°}}{}_{\text{rxn}})$ is the difference in free energy of the reactants and products in their standard state.

${\text{ΔH}}^{\text{°}}{}_{\text{rxn}}\text{=}\sum {\text{nΔH}}_{\text{f}}{}^{\text{°}}\text{(Products)-}\sum {\text{nΔH}}_{\text{f}}{}^{\text{°}}\text{(Reactants)}$

Where, $\text{"n"}$ is the number of moles

Explanation of Solution

Let us consider a statement (e).

$\begin{array}{l}\Delta H^0={\text{2ΔH}}_{\text{f}}^{\text{o}}{\text{(CO}}_{\text{2}}\text{)+}{\text{2ΔH}}_{\text{f}}^{\text{o}}{\text{(N}}_{\text{2}}\text{)-}{\text{2ΔH}}_{\text{f}}^{\text{o}}{\text{(CO)-2ΔH}}_{\text{f}}^{\text{o}}\text{(NO)}\\ \text{The respactive values are substituted for above equlibrium reaction}\\ {\text{ΔH}}^{\text{0}}=(2)(-393.5\text{KJ/mol})+(0)-(2)(-110.5\text{KJ/mol})-(2)(90.4\text{KJ/mol})\\ {\text{ΔH}}^{\text{0}}=-746.8\text{KJ/mol}\end{array}$

Since $\Delta H°$ value is negative, raising the temperature will decrease $Kp$ thereby increasing amount of reactants and decreasing amount of products.

No the formation of $N_{2}andCO$ it is not favoured by raising the temperature.