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

Question

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Answer 1

Question

Chapter 18, Problem 18.50QP

(a)

Interpretation Introduction

Interpretation:

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

(a)

Expert Solution

Explanation of Solution

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

$a). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]$

The equal mole ratio of carbon monoxide $CO$ and nitric oxide $NO$ reacted in gas phase conditions to produce a two moles of $CO2$ and nitrogen ( $N2$ ) 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.

(b)

Expert Solution

Explanation of Solution

The atmospheric equilibrium reaction is,

$i). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]ii). NO---Oxidizing agent CO---Reducing agent (Gains of electrons).$

Given statement (b) nitric oxide ( $NO$ ) is a very good oxidizing agent it is gains a one electrons form carbon monoxide ( $CO$ ), will get ( $CO2$ ) $(N2)$ 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 $(CO2)$

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 $(Kp)$ value has to be calculated given gases phase equilibrium reaction at $25°C$ .

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

(c)

Expert Solution

Explanation of Solution

The atmospheric reaction fallow as,

$2CO + 2NO → 2CO2 + N2 ΔGf0=CO2 = 394.4 KJ/molΔGf0 = N2 = 137.3 KJ/mol (This values from equilibrium table-3)ΔGf0= CO= 86.1KJ/mol, NO= 86.7 KJ/molΔGrxn0 = ∑nΔGf0(Products)∑mΔf0(Reactants)ΔG0rxn = 2ΔGf0 (CO2)+ ΔGf0(N2)− 2ΔGf0 (CO)−2ΔGf0(NO)$

$InK =ΔG0-RT= 6.876×105 KJ/mol)(8.314 J/K ⋅mol)(298 K) InK =ΔG0-RT = 6.876×105 KJ/mol)(2477.572)InK = 277.5 (∵ e277.5=3.286)Kp= 3.29×10120$

(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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Where, $"n"$ 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.

(d)

Expert Solution

Explanation of Solution

$a). 2CO + 2NO → 2CO2 + N2 -----[1]Apply for partial pressure values for reaction (1) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)Qp= (PN2)(PCO2)2(PCO)2(PNO)2=ProductReactant−−−−−[2] Qp= (0.80)(3.0×10−4)2(5.0×10−5)2(5.0×10−7)2Qp= 1.2×1014$

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 $N2$ and $CO2$ 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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔH°rxn=∑nΔHf°(Products)-∑nΔHf°(Reactants)$

Where, $"n"$ is the number of moles

(e)

Expert Solution

Explanation of Solution

Let us consider a statement (e).

$ΔH0= 2ΔHfo(CO2)+ 2ΔHfo(N2)- 2ΔHfo(CO)-2ΔHfo(NO)The respactive values are substituted for above equlibrium reactionΔH0 = (2)(−393.5 KJ/mol) + (0)−(2)(−110.5 KJ/mol)−(2)(90.4 KJ/mol)ΔH0 = −746.8 KJ/mol$

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

No the formation of $N2 and CO$ it is not favoured by raising the temperature.

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Answer 2

Question

Chapter 18, Problem 18.50QP

(a)

Interpretation Introduction

Interpretation:

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

(a)

Expert Solution

Explanation of Solution

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

$a). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]$

The equal mole ratio of carbon monoxide $CO$ and nitric oxide $NO$ reacted in gas phase conditions to produce a two moles of $CO2$ and nitrogen ( $N2$ ) 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.

(b)

Expert Solution

Explanation of Solution

The atmospheric equilibrium reaction is,

$i). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]ii). NO---Oxidizing agent CO---Reducing agent (Gains of electrons).$

Given statement (b) nitric oxide ( $NO$ ) is a very good oxidizing agent it is gains a one electrons form carbon monoxide ( $CO$ ), will get ( $CO2$ ) $(N2)$ 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 $(CO2)$

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 $(Kp)$ value has to be calculated given gases phase equilibrium reaction at $25°C$ .

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

(c)

Expert Solution

Explanation of Solution

The atmospheric reaction fallow as,

$2CO + 2NO → 2CO2 + N2 ΔGf0=CO2 = 394.4 KJ/molΔGf0 = N2 = 137.3 KJ/mol (This values from equilibrium table-3)ΔGf0= CO= 86.1KJ/mol, NO= 86.7 KJ/molΔGrxn0 = ∑nΔGf0(Products)∑mΔf0(Reactants)ΔG0rxn = 2ΔGf0 (CO2)+ ΔGf0(N2)− 2ΔGf0 (CO)−2ΔGf0(NO)$

$InK =ΔG0-RT= 6.876×105 KJ/mol)(8.314 J/K ⋅mol)(298 K) InK =ΔG0-RT = 6.876×105 KJ/mol)(2477.572)InK = 277.5 (∵ e277.5=3.286)Kp= 3.29×10120$

(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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Where, $"n"$ 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.

(d)

Expert Solution

Explanation of Solution

$a). 2CO + 2NO → 2CO2 + N2 -----[1]Apply for partial pressure values for reaction (1) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)Qp= (PN2)(PCO2)2(PCO)2(PNO)2=ProductReactant−−−−−[2] Qp= (0.80)(3.0×10−4)2(5.0×10−5)2(5.0×10−7)2Qp= 1.2×1014$

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 $N2$ and $CO2$ 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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔH°rxn=∑nΔHf°(Products)-∑nΔHf°(Reactants)$

Where, $"n"$ is the number of moles

(e)

Expert Solution

Explanation of Solution

Let us consider a statement (e).

$ΔH0= 2ΔHfo(CO2)+ 2ΔHfo(N2)- 2ΔHfo(CO)-2ΔHfo(NO)The respactive values are substituted for above equlibrium reactionΔH0 = (2)(−393.5 KJ/mol) + (0)−(2)(−110.5 KJ/mol)−(2)(90.4 KJ/mol)ΔH0 = −746.8 KJ/mol$

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

No the formation of $N2 and CO$ it is not favoured by raising the temperature.

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Answer 3

Question

Chapter 18, Problem 18.50QP

(a)

Interpretation Introduction

Interpretation:

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

(a)

Expert Solution

Explanation of Solution

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

$a). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]$

The equal mole ratio of carbon monoxide $CO$ and nitric oxide $NO$ reacted in gas phase conditions to produce a two moles of $CO2$ and nitrogen ( $N2$ ) 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.

(b)

Expert Solution

Explanation of Solution

The atmospheric equilibrium reaction is,

$i). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]ii). NO---Oxidizing agent CO---Reducing agent (Gains of electrons).$

Given statement (b) nitric oxide ( $NO$ ) is a very good oxidizing agent it is gains a one electrons form carbon monoxide ( $CO$ ), will get ( $CO2$ ) $(N2)$ 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 $(CO2)$

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 $(Kp)$ value has to be calculated given gases phase equilibrium reaction at $25°C$ .

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

(c)

Expert Solution

Explanation of Solution

The atmospheric reaction fallow as,

$2CO + 2NO → 2CO2 + N2 ΔGf0=CO2 = 394.4 KJ/molΔGf0 = N2 = 137.3 KJ/mol (This values from equilibrium table-3)ΔGf0= CO= 86.1KJ/mol, NO= 86.7 KJ/molΔGrxn0 = ∑nΔGf0(Products)∑mΔf0(Reactants)ΔG0rxn = 2ΔGf0 (CO2)+ ΔGf0(N2)− 2ΔGf0 (CO)−2ΔGf0(NO)$

$InK =ΔG0-RT= 6.876×105 KJ/mol)(8.314 J/K ⋅mol)(298 K) InK =ΔG0-RT = 6.876×105 KJ/mol)(2477.572)InK = 277.5 (∵ e277.5=3.286)Kp= 3.29×10120$

(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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Where, $"n"$ 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.

(d)

Expert Solution

Explanation of Solution

$a). 2CO + 2NO → 2CO2 + N2 -----[1]Apply for partial pressure values for reaction (1) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)Qp= (PN2)(PCO2)2(PCO)2(PNO)2=ProductReactant−−−−−[2] Qp= (0.80)(3.0×10−4)2(5.0×10−5)2(5.0×10−7)2Qp= 1.2×1014$

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 $N2$ and $CO2$ 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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔH°rxn=∑nΔHf°(Products)-∑nΔHf°(Reactants)$

Where, $"n"$ is the number of moles

(e)

Expert Solution

Explanation of Solution

Let us consider a statement (e).

$ΔH0= 2ΔHfo(CO2)+ 2ΔHfo(N2)- 2ΔHfo(CO)-2ΔHfo(NO)The respactive values are substituted for above equlibrium reactionΔH0 = (2)(−393.5 KJ/mol) + (0)−(2)(−110.5 KJ/mol)−(2)(90.4 KJ/mol)ΔH0 = −746.8 KJ/mol$

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

No the formation of $N2 and CO$ it is not favoured by raising the temperature.

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Answer 4

Question

Chapter 18, Problem 18.50QP

(a)

Interpretation Introduction

Interpretation:

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

(a)

Expert Solution

Explanation of Solution

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

$a). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]$

The equal mole ratio of carbon monoxide $CO$ and nitric oxide $NO$ reacted in gas phase conditions to produce a two moles of $CO2$ and nitrogen ( $N2$ ) 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.

(b)

Expert Solution

Explanation of Solution

The atmospheric equilibrium reaction is,

$i). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]ii). NO---Oxidizing agent CO---Reducing agent (Gains of electrons).$

Given statement (b) nitric oxide ( $NO$ ) is a very good oxidizing agent it is gains a one electrons form carbon monoxide ( $CO$ ), will get ( $CO2$ ) $(N2)$ 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 $(CO2)$

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 $(Kp)$ value has to be calculated given gases phase equilibrium reaction at $25°C$ .

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

(c)

Expert Solution

Explanation of Solution

The atmospheric reaction fallow as,

$2CO + 2NO → 2CO2 + N2 ΔGf0=CO2 = 394.4 KJ/molΔGf0 = N2 = 137.3 KJ/mol (This values from equilibrium table-3)ΔGf0= CO= 86.1KJ/mol, NO= 86.7 KJ/molΔGrxn0 = ∑nΔGf0(Products)∑mΔf0(Reactants)ΔG0rxn = 2ΔGf0 (CO2)+ ΔGf0(N2)− 2ΔGf0 (CO)−2ΔGf0(NO)$

$InK =ΔG0-RT= 6.876×105 KJ/mol)(8.314 J/K ⋅mol)(298 K) InK =ΔG0-RT = 6.876×105 KJ/mol)(2477.572)InK = 277.5 (∵ e277.5=3.286)Kp= 3.29×10120$

(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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Where, $"n"$ 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.

(d)

Expert Solution

Explanation of Solution

$a). 2CO + 2NO → 2CO2 + N2 -----[1]Apply for partial pressure values for reaction (1) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)Qp= (PN2)(PCO2)2(PCO)2(PNO)2=ProductReactant−−−−−[2] Qp= (0.80)(3.0×10−4)2(5.0×10−5)2(5.0×10−7)2Qp= 1.2×1014$

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 $N2$ and $CO2$ 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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔH°rxn=∑nΔHf°(Products)-∑nΔHf°(Reactants)$

Where, $"n"$ is the number of moles

(e)

Expert Solution

Explanation of Solution

Let us consider a statement (e).

$ΔH0= 2ΔHfo(CO2)+ 2ΔHfo(N2)- 2ΔHfo(CO)-2ΔHfo(NO)The respactive values are substituted for above equlibrium reactionΔH0 = (2)(−393.5 KJ/mol) + (0)−(2)(−110.5 KJ/mol)−(2)(90.4 KJ/mol)ΔH0 = −746.8 KJ/mol$

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

No the formation of $N2 and CO$ it is not favoured by raising the temperature.

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Answer 5

Chapter 18, Problem 18.50QP

(a)

Interpretation Introduction

Interpretation:

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

(a)

Expert Solution

Explanation of Solution

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

$a). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]$

The equal mole ratio of carbon monoxide $CO$ and nitric oxide $NO$ reacted in gas phase conditions to produce a two moles of $CO2$ and nitrogen ( $N2$ ) 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.

(b)

Expert Solution

Explanation of Solution

The atmospheric equilibrium reaction is,

$i). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]ii). NO---Oxidizing agent CO---Reducing agent (Gains of electrons).$

Given statement (b) nitric oxide ( $NO$ ) is a very good oxidizing agent it is gains a one electrons form carbon monoxide ( $CO$ ), will get ( $CO2$ ) $(N2)$ 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 $(CO2)$

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 $(Kp)$ value has to be calculated given gases phase equilibrium reaction at $25°C$ .

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

(c)

Expert Solution

Explanation of Solution

The atmospheric reaction fallow as,

$2CO + 2NO → 2CO2 + N2 ΔGf0=CO2 = 394.4 KJ/molΔGf0 = N2 = 137.3 KJ/mol (This values from equilibrium table-3)ΔGf0= CO= 86.1KJ/mol, NO= 86.7 KJ/molΔGrxn0 = ∑nΔGf0(Products)∑mΔf0(Reactants)ΔG0rxn = 2ΔGf0 (CO2)+ ΔGf0(N2)− 2ΔGf0 (CO)−2ΔGf0(NO)$

$InK =ΔG0-RT= 6.876×105 KJ/mol)(8.314 J/K ⋅mol)(298 K) InK =ΔG0-RT = 6.876×105 KJ/mol)(2477.572)InK = 277.5 (∵ e277.5=3.286)Kp= 3.29×10120$

(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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Where, $"n"$ 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.

(d)

Expert Solution

Explanation of Solution

$a). 2CO + 2NO → 2CO2 + N2 -----[1]Apply for partial pressure values for reaction (1) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)Qp= (PN2)(PCO2)2(PCO)2(PNO)2=ProductReactant−−−−−[2] Qp= (0.80)(3.0×10−4)2(5.0×10−5)2(5.0×10−7)2Qp= 1.2×1014$

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 $N2$ and $CO2$ 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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔH°rxn=∑nΔHf°(Products)-∑nΔHf°(Reactants)$

Where, $"n"$ is the number of moles

(e)

Expert Solution

Explanation of Solution

Let us consider a statement (e).

$ΔH0= 2ΔHfo(CO2)+ 2ΔHfo(N2)- 2ΔHfo(CO)-2ΔHfo(NO)The respactive values are substituted for above equlibrium reactionΔH0 = (2)(−393.5 KJ/mol) + (0)−(2)(−110.5 KJ/mol)−(2)(90.4 KJ/mol)ΔH0 = −746.8 KJ/mol$

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

No the formation of $N2 and CO$ it is not favoured by raising the temperature.

Answer 6

Chapter 18, Problem 18.50QP

(a)

Interpretation Introduction

Interpretation:

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

(a)

Expert Solution

Explanation of Solution

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

$a). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]$

The equal mole ratio of carbon monoxide $CO$ and nitric oxide $NO$ reacted in gas phase conditions to produce a two moles of $CO2$ and nitrogen ( $N2$ ) gas, the balanced equilibrium reactions shown above.

Answer 7

Chapter 18, Problem 18.50QP

(a)

Interpretation Introduction

Interpretation:

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

Answer 8

(a)

Expert Solution

Explanation of Solution

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

$a). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]$

The equal mole ratio of carbon monoxide $CO$ and nitric oxide $NO$ reacted in gas phase conditions to produce a two moles of $CO2$ and nitrogen ( $N2$ ) gas, the balanced equilibrium reactions shown above.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(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.

(b)

Expert Solution

Explanation of Solution

The atmospheric equilibrium reaction is,

$i). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]ii). NO---Oxidizing agent CO---Reducing agent (Gains of electrons).$

Given statement (b) nitric oxide ( $NO$ ) is a very good oxidizing agent it is gains a one electrons form carbon monoxide ( $CO$ ), will get ( $CO2$ ) $(N2)$ 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 $(CO2)$

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.

Answer 13

(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.

Answer 14

(b)

Expert Solution

Explanation of Solution

The atmospheric equilibrium reaction is,

$i). 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) -----[1]ii). NO---Oxidizing agent CO---Reducing agent (Gains of electrons).$

Given statement (b) nitric oxide ( $NO$ ) is a very good oxidizing agent it is gains a one electrons form carbon monoxide ( $CO$ ), will get ( $CO2$ ) $(N2)$ 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 $(CO2)$

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.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

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

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

(c)

Expert Solution

Explanation of Solution

The atmospheric reaction fallow as,

$2CO + 2NO → 2CO2 + N2 ΔGf0=CO2 = 394.4 KJ/molΔGf0 = N2 = 137.3 KJ/mol (This values from equilibrium table-3)ΔGf0= CO= 86.1KJ/mol, NO= 86.7 KJ/molΔGrxn0 = ∑nΔGf0(Products)∑mΔf0(Reactants)ΔG0rxn = 2ΔGf0 (CO2)+ ΔGf0(N2)− 2ΔGf0 (CO)−2ΔGf0(NO)$

$InK =ΔG0-RT= 6.876×105 KJ/mol)(8.314 J/K ⋅mol)(298 K) InK =ΔG0-RT = 6.876×105 KJ/mol)(2477.572)InK = 277.5 (∵ e277.5=3.286)Kp= 3.29×10120$

Answer 19

(c)

Interpretation Introduction

Interpretation:

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

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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$

$ΔG0 =-RTln KΔG = Free energyΔG0 = Standard−state free energyR = Gas Constant (0.0826 l .atm/K.atm)T = Temprature 273 KK= Equlibrium Constant (KP and KC)$

Answer 20

(c)

Expert Solution

Explanation of Solution

The atmospheric reaction fallow as,

$2CO + 2NO → 2CO2 + N2 ΔGf0=CO2 = 394.4 KJ/molΔGf0 = N2 = 137.3 KJ/mol (This values from equilibrium table-3)ΔGf0= CO= 86.1KJ/mol, NO= 86.7 KJ/molΔGrxn0 = ∑nΔGf0(Products)∑mΔf0(Reactants)ΔG0rxn = 2ΔGf0 (CO2)+ ΔGf0(N2)− 2ΔGf0 (CO)−2ΔGf0(NO)$

$InK =ΔG0-RT= 6.876×105 KJ/mol)(8.314 J/K ⋅mol)(298 K) InK =ΔG0-RT = 6.876×105 KJ/mol)(2477.572)InK = 277.5 (∵ e277.5=3.286)Kp= 3.29×10120$

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Where, $"n"$ 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.

(d)

Expert Solution

Explanation of Solution

$a). 2CO + 2NO → 2CO2 + N2 -----[1]Apply for partial pressure values for reaction (1) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)Qp= (PN2)(PCO2)2(PCO)2(PNO)2=ProductReactant−−−−−[2] Qp= (0.80)(3.0×10−4)2(5.0×10−5)2(5.0×10−7)2Qp= 1.2×1014$

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

Answer 25

(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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔG°rxn=∑nΔGf°(Products)-∑nΔGf°(Reactants)$

Where, $"n"$ 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.

Answer 26

(d)

Expert Solution

Explanation of Solution

$a). 2CO + 2NO → 2CO2 + N2 -----[1]Apply for partial pressure values for reaction (1) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g)Qp= (PN2)(PCO2)2(PCO)2(PNO)2=ProductReactant−−−−−[2] Qp= (0.80)(3.0×10−4)2(5.0×10−5)2(5.0×10−7)2Qp= 1.2×1014$

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

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

(e)

Interpretation Introduction

Interpretation:

The raising temperatures favor the formation of $N2$ and $CO2$ 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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔH°rxn=∑nΔHf°(Products)-∑nΔHf°(Reactants)$

Where, $"n"$ is the number of moles

(e)

Expert Solution

Explanation of Solution

Let us consider a statement (e).

$ΔH0= 2ΔHfo(CO2)+ 2ΔHfo(N2)- 2ΔHfo(CO)-2ΔHfo(NO)The respactive values are substituted for above equlibrium reactionΔH0 = (2)(−393.5 KJ/mol) + (0)−(2)(−110.5 KJ/mol)−(2)(90.4 KJ/mol)ΔH0 = −746.8 KJ/mol$

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

No the formation of $N2 and CO$ it is not favoured by raising the temperature.

Answer 31

(e)

Interpretation Introduction

Interpretation:

The raising temperatures favor the formation of $N2$ and $CO2$ 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 $G$ . All spontaneous process is associated with the decrease of free energy in the system. The standard free energy change $(ΔG°rxn)$ is the difference in free energy of the reactants and products in their standard state.

$ΔH°rxn=∑nΔHf°(Products)-∑nΔHf°(Reactants)$

Where, $"n"$ is the number of moles

Answer 32

(e)

Expert Solution

Explanation of Solution

Let us consider a statement (e).

$ΔH0= 2ΔHfo(CO2)+ 2ΔHfo(N2)- 2ΔHfo(CO)-2ΔHfo(NO)The respactive values are substituted for above equlibrium reactionΔH0 = (2)(−393.5 KJ/mol) + (0)−(2)(−110.5 KJ/mol)−(2)(90.4 KJ/mol)ΔH0 = −746.8 KJ/mol$