General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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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
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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
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Explanation of Solution

The atmospheric equilibrium reaction is,

i).2CO(g)+2NO(g)2CO2(g)+N2(g)-----[1]ii).NO---OxidizingagentCO---Reducing agent(Gains ofelectrons) 

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=-RTlnKΔG=Free energyΔG0=Standardstate free energyR=GasConstant(0.0826l.atm/K.atm)T=Temprature273KK=EqulibriumConstant(KPandKC)

(c)

Expert Solution
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Explanation of Solution

The atmospheric reaction fallow as,

2CO+2NO2CO2+N2ΔGf0=CO2=394.4KJ/molΔGf0=N2=137.3KJ/mol(Thisvaluesfromequilibriumtable-3)ΔGf0=CO=86.1KJ/mol,NO=86.7KJ/molΔGrxn0=nΔGf0(Products)mΔf0(Reactants)ΔG0rxn=2ΔGf0(CO2)+ΔGf0(N2)2ΔGf0(CO)2ΔGf0(NO)

InK=ΔG0-RT=6.876×105KJ/mol)(8.314J/Kmol)(298K)InK=ΔG0-RT=6.876×105KJ/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
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Explanation of Solution

a).2CO+2NO2CO2+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)(3.0×104)2(5.0×105)2(5.0×107)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
Check Mark

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)(393.5KJ/mol)+(0)(2)(110.5KJ/mol)(2)(90.4KJ/mol)ΔH0=746.8KJ/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 N2andCO it is not favoured by raising the temperature. 

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Chapter 18 Solutions

General Chemistry

Ch. 18.6 - Practice Exercise Calculate the equilibrium...Ch. 18.6 - Prob. 2PECh. 18.6 - Prob. 3PECh. 18.6 - Prob. 1RCCh. 18 - Prob. 18.1QPCh. 18 - Prob. 18.2QPCh. 18 - Prob. 18.3QPCh. 18 - Prob. 18.4QPCh. 18 - Prob. 18.5QPCh. 18 - Prob. 18.7QPCh. 18 - Prob. 18.8QPCh. 18 - Prob. 18.9QPCh. 18 - 18.10 Arrange the following substances (1 mole...Ch. 18 - Prob. 18.11QPCh. 18 - Prob. 18.12QPCh. 18 - Prob. 18.13QPCh. 18 - 18.14 State whether the sign of the entropy...Ch. 18 - 18.15 Define free energy. What are its units? Ch. 18 - 18.16 Why is it more convenient to predict the...Ch. 18 - 18.17 Calculate ΔG° for the following reactions at...Ch. 18 - 18.18 Calculate ΔG° for the following reactions at...Ch. 18 - Prob. 18.19QPCh. 18 - Prob. 18.20QPCh. 18 - Prob. 18.21QPCh. 18 - Prob. 18.22QPCh. 18 - Prob. 18.23QPCh. 18 - 18.24 For the autoionization of water at...Ch. 18 - Prob. 18.25QPCh. 18 - Prob. 18.26QPCh. 18 - Prob. 18.27QPCh. 18 - Prob. 18.28QPCh. 18 - Prob. 18.29QPCh. 18 - Prob. 18.30QPCh. 18 - Prob. 18.31QPCh. 18 - Prob. 18.32QPCh. 18 - Prob. 18.33QPCh. 18 - Prob. 18.34QPCh. 18 - Prob. 18.35QPCh. 18 - Prob. 18.36QPCh. 18 - Prob. 18.37QPCh. 18 - Prob. 18.38QPCh. 18 - Prob. 18.39QPCh. 18 - Prob. 18.40QPCh. 18 - Prob. 18.41QPCh. 18 - Prob. 18.42QPCh. 18 - Prob. 18.43QPCh. 18 - Prob. 18.44QPCh. 18 - Prob. 18.45QPCh. 18 - Prob. 18.46QPCh. 18 - 18.47 Calculate the equilibrium pressure of CO2...Ch. 18 - Prob. 18.48QPCh. 18 - 18.49 Referring to Problem 18.48, explain why the...Ch. 18 - Prob. 18.50QPCh. 18 - Prob. 18.51QPCh. 18 - Prob. 18.52QPCh. 18 - Prob. 18.53QPCh. 18 - Prob. 18.54QPCh. 18 - Prob. 18.55QPCh. 18 - 18.56 Crystallization of sodium acetate from a...Ch. 18 - Prob. 18.57QPCh. 18 - Prob. 18.58QPCh. 18 - Prob. 18.59QPCh. 18 - Prob. 18.60QPCh. 18 - Prob. 18.61QPCh. 18 - Prob. 18.62QPCh. 18 - Prob. 18.63QPCh. 18 - Prob. 18.64QPCh. 18 - Prob. 18.65QPCh. 18 - Prob. 18.66QPCh. 18 - Prob. 18.67QPCh. 18 - Prob. 18.68QPCh. 18 - Prob. 18.69QPCh. 18 - Prob. 18.70QPCh. 18 - Prob. 18.71QPCh. 18 - Prob. 18.72QPCh. 18 - 18.73 (a) Over the years there have been numerous...Ch. 18 - Prob. 18.74QPCh. 18 - 18.75 Shown here are the thermodynamic data for...Ch. 18 - Prob. 18.76QPCh. 18 - Prob. 18.77QPCh. 18 - Prob. 18.78QPCh. 18 - Prob. 18.79QPCh. 18 - Prob. 18.80QPCh. 18 - Prob. 18.81QPCh. 18 - Prob. 18.82QPCh. 18 - Prob. 18.83QPCh. 18 - 18.84 Large quantities of hydrogen are needed for...Ch. 18 - Prob. 18.85QPCh. 18 - Prob. 18.86QPCh. 18 - Prob. 18.87QPCh. 18 - Prob. 18.88QPCh. 18 - Prob. 18.89QPCh. 18 - Prob. 18.90QPCh. 18 - Prob. 18.91QPCh. 18 - Prob. 18.92QPCh. 18 - Prob. 18.93QPCh. 18 - Prob. 18.94QPCh. 18 - Prob. 18.95QPCh. 18 - Prob. 18.96QPCh. 18 - Prob. 18.98QPCh. 18 - Prob. 18.100SPCh. 18 - Prob. 18.101SPCh. 18 - Prob. 18.102SPCh. 18 - Prob. 18.103SPCh. 18 - Prob. 18.104SPCh. 18 - Prob. 18.105SPCh. 18 - Prob. 18.106SPCh. 18 - Prob. 18.107SPCh. 18 - Prob. 18.108SPCh. 18 - 18.109 The boiling point of diethyl ether is...
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