General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 18, Problem 18.69QP

(a)

Interpretation Introduction

Interpretation:

The standard free energy ΔG° value should be derived given nitric oxide formation reaction at 25°C.

Concept Introduction:

Thermodynamics is the branch of science that relates heat and energy in a system.  The four laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy.  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.  In non-spontaneous reaction there is an increase of free energy in the system.

Free energy(ΔG0): In thermodynamics free energy or Gibbs free energy is the energy that is used to express the total energy content of a system.  According to second law of thermodynamics, in all spontaneous process is associated with the decrease in free energy of the system.  That is the change in free energy will be negative.

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)

(a)

Expert Solution
Check Mark

Explanation of Solution

The equilibrium constant is related to the to the standard free energy change by the given equation (1).

  N2(g)+O2(g)2NO(g)ΔG=-RTlnK-------------[1]We rewrite the above equationΔG=GivenvaluesProduct=173.4KJ/mol2(2NOProduct)ΔG=86.7kJ/mol

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant Kp value should be calculated for given reaction.

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

The calculate the standard free (ΔGf0) value for given reaction is shown below,

  ΔG0=-RTlnK---------[1]ΔG=Free energyΔG0=Standardstate free energyR=GasConstant(0.08314 (or) 8.314.atm/K.atm)T=Temprature273KK=EqulibriumConstant(KPandKC)

(b)

Expert Solution
Check Mark

Explanation of Solution

First calculate the equilibrium constant (Kp) value for given reaction

N2(g)+O2(g)2NO(g)ΔGf0=-RTlnK-------------[1]HereR=0.08314J/K.mol (or) -8.314J/K.molSubstitutedfor(Kp,RandT)valuesareequation(1)Given the values 173.4KJ/mol173.4×103J/mol=(8.314J/K.mol)(298K)lnKpRewrite the above equation (1)lnKp=ΔG0-RT=173.4×103J/mol)(8.314J/Kmol)(298K)lnKp=ΔG0-RT=173.4×103)(2477.572)lnKp = -69.99Kp=e69.99Kp=4.0×1031

Given NO formation reaction equilibrium constant value is 4.0×1031

(c)

Interpretation Introduction

Interpretation:

The nitric oxide starting formation has to determine.

Concept Introduction:

Thermodynamics is the branch of science that relates heat and energy in a system.  The four laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy. 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 equation given below helps us to calculate the change in free energy in a system.

ΔG = ΔΗ- TΔS

Where,

  ΔG  is the change in free energy of the system

  ΔΗ is the change in enthalpy of the system

  T is the absolute value of the temperature

  ΔS is the change in entropy in the system

(c)

Expert Solution
Check Mark

Explanation of Solution

 We calculate the enthalpy changes for given reaction

N2(g)+O2(g)2NO(g)HereStandard enthalpy changes equation InK2K1=ΔH0R(T2-T1T1×T2)[1]ΔH°totalno.ofproductformationΔHfo(NO)The equilibrium values of NO= 86.7kJ/molΔHfo(NO)=2(86.7kJ/mol)=173.4kJ/molTherespactivevaluesaresubstitutedequation(1)InK24.0×10-31=173.4×103J/mol8.314J/molK(1373K-298K(1373)(298K))InK24.0×10-31=173.4J/mol8.314J/molK(1075K409154K)InK24.0×10-31=(208563.86)(2.672×10-3K)InK24.0×10-31=5572.82KK2=3×10-7

(d)

Interpretation Introduction

Interpretation:

The lighting helps to produce a better crop, the reason behind this has to explained. 

Concept Introduction:

Thermodynamics is the branch of science that relates heat and energy in a system.  The four laws of thermodynamics explain the fundamental quantities such as temperature, energy and randomness in a system.  Entropy is the measure of randomness in a system.  For a spontaneous process there is always a positive change in entropy.  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.  In non-spontaneous reaction there is an increase of free energy in the system.

Non-spontaneous reaction: This type of reaction explain as, endergonic reaction (mean by heat absorption non-spontaneous process) or unfavourable reaction in a chemical reaction in which the standard change in free energy is positive and energy is absorbed.

(d)

Expert Solution
Check Mark

Explanation of Solution

The lighting promotes the formation of NO (from N2 and O2 in the air), which eventually leads to the formation of nitrate ions (NO3) as essential nutrition for plants.

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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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