General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 18, Problem 18.29QP

(a)

Interpretation Introduction

Interpretation:

The equilibrium pressure (Kp) values has to be calculated given the calcium carbonate CaCO3 decomposition reaction at 25°C.

Concept Information:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant.  If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Kp:  The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.

Kp=Kc(RT)Δn(or)Kc=Kp(RT)Δn(or)Kp=(RT)ΔnΔn=Product - Reactant

Thermodynamics is the branch of science that relates heat and energy in a system.  The 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

(a)

Expert Solution
Check Mark

Explanation of Solution

The calcium carbonate decomposition equilibrium expression given by,

CaCO3(s)CaO(s)+CO2(g)Kp= CO2[Considergasesmoleculeonly]ΔH°=177.8KJ/molKp=Kc[8.314atm/K.atm×T]Δn   Kp=Kc(RT)Δn(or)Kc=Kp(RT)Δn(or)Kp=(RT)ΔnHereΔn=1-0=(Product - Reactant)andgivenTemprature250CT=(25+ 273)K=298KΔG=ΔH-TΔS    ΔG0=ΔH0-TΔS0[1]Calculate the 250The respactive values are substituted in equation (1)ΔG0=ΔH0-TΔS0=(177.8×103J/mol)(298K)(160.5J/K.mol)ΔG0=130.0×103J/mol(PCO2)=Kp=e-ΔG0RT=e130.0×103J/mol(8.314J/K.mol)(298K)(PCO2)=Kp=e52.47Takingforantilogbothsides,  we getKp= 1.6×10-23atm_

(b)

Interpretation Introduction

Interpretation:

The equilibrium pressure (Kp) values has to be calculated given the enthalpy (ΔH°)  and entropy (ΔS°) calcium carbonate decomposition reaction at 800°C.

Concept Information:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant.  If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Kp:  The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.

Kp=Kc(RT)Δn(or)Kc=Kp(RT)Δn(or)Kp=(RT)ΔnΔn=Product - Reactant

Thermodynamics is the branch of science that relates heat and energy in a system.  The 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

(b)

Expert Solution
Check Mark

Explanation of Solution

First we calculate the enthalpy and entropy values for given decomposition reaction at 800°C.

CaCO3(s)CaO(s)+CO2(g)Kp= CO2ΔH=177.8KJ/molKp=Kc[8.314.atm/K.atm×T]Δn   Kp=Kc(RT)Δn(or)Kc=Kp(RT)Δn(or)Kp=(RT)Δn(PCO2)=Kp=e-ΔG0RTHereΔn=1-0=(Product - Reactant)andgivenTemprature3500CT=(800+ 273)K=1073KΔG=ΔH-TΔS    ΔG°=ΔH°-TΔS°[2]The respactive values are substituted for equation (2)ΔG°=ΔH°-TΔS°=(177.8×103J/mol)(1073K)(160.5J/K.mol)ΔG0=5.584×103J/mol(PCO2)=Kp=e-ΔG0RT=5.53×103J/mol(8.314J/Kmol)(1073K)(PCO2)=Kp=e0.625Kp= 0.535 atm 

Calcium carbonate, partial pressure (Kp) value is 0.535 atm_

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Chapter 18 Solutions

General Chemistry

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