General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 18, Problem 18.106SP

(1)

Interpretation Introduction

Interpretation:

The equilibrium process should be determined from the given equilibrium reaction.

Concept Introduction:

Free energy(ΔG): 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)

    ΔG=ΔG°+RTlnQΔG=Free energyΔG0=Standardstate free energyR=GasConstant(8.314.atm/K.atm)T=Temprature273KQ=EqulibriumConstant(KPandKC)

(1)

Expert Solution
Check Mark

Explanation of Solution

We calculate the equilibrium constant (Kp) from the ΔG° value

  A2(g)+B2(g)2AB(g)ΔG°=-RTlnK-3.4×103J/mol=-(8.314J/molK)(298K)lnKplnKp=-3.4×103J/mol-(8.314J/K.mol)(298K)lnK=1.37roundedvalue1.40K=e-1.40K=3.9(or)K=4 

Reaction-cA2(g)+B2(g)2AB(g)Q=[AB]2[A2][B2]Given the partial pressure values 0.10atmNo. reactants(A=2)&(B=2) and (C=4)productsQ=[0.40]2[0.20][0.20]=4.0

ΔG=ΔG°+RTlnQ 

The reaction mixture (c) is at equilibrium (Q=K) and ΔG=0

(2)

Interpretation Introduction

Interpretation:

The negative entropy value has to be calculated and identified for the given equilibrium reaction.

Concept Introduction:

Free energy(ΔG): 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)

    ΔG=ΔG°+RTlnQΔG=Free energyΔG0=Standardstate free energyR=GasConstant(8.314.atm/K.atm)T=Temprature273KQ=EqulibriumConstant(KPandKC)

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.

(2)

Expert Solution
Check Mark

Explanation of Solution

Reaction-aA2(g)+B2(g)2AB(g)QP=[AB]2[A2][B2]Given the partial pressure values 0.10atmQP=[0.30]2[0.30][0.20]=1.5

The value of ΔG could also be calculated using the equation,

ΔG=ΔG°+RTlnQ The (Q) values is substituted equationΔG=3400J/mol+(8.314J/K.mol)(298K)In(1.5)=-3400J/mol+(8.314J/K.mol)(298K)(0.4054)ΔG=2.4×103J/mol

Reaction mixture (a) has a negative ΔG value, because Q<K the system of equilibrium will shift into right towards products, to reach the equilibrium.

(3)

Interpretation Introduction

Interpretation:

The positive entropy value has to be calculate and identified given equilibrium reaction. 

Concept Information:

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)

    ΔG=ΔG°+RTlnQΔG=Free energyΔG0=Standardstate free energyR=GasConstant(8.314.atm/K.atm)T=Temprature273KQ=EqulibriumConstant(KPandKC)ln=(ve(log)State Function)

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.

(3)

Expert Solution
Check Mark

Explanation of Solution

Reaction-cA2(g)+B2(g)2AB(g)QP=[AB]2[A2][B2]QP=[0.60]2[0.20][0.30]=6.0

ΔG+ΔG°+RTlnQ The respative (Q) value is substituted equationΔG=3400J/mol+(8.314J/K.mol)(298K)In(6)=-3400J/mol+(8.314J/K.mol)(298K)(1.7917)ΔG=1.0×103kJ/mol

The reaction mixture (b) has positive ΔG value, because Q>K the system of equilibrium will shift left, toward reactants, to reach the equilibrium. This shift corresponds to a positive ΔG value.

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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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