Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 16, Problem 119QRT

(a)

Interpretation Introduction

Interpretation:

The value of ΔrG° for the reaction at 25°C, CH4(g)+H2O(g)CO(g)+3H2(g), has to be calculated.

Concept Introduction:

The Gibbs free energy of a system is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system. The Gibbs free energy of the system is a state function as it is defined in terms of thermodynamic properties that are state functions. The Gibbs free energy of a system is directly related to the equilibrium constant of a reaction. The symbol for equilibrium constant in terms of partial pressure is Kp.

(a)

Expert Solution
Check Mark

Answer to Problem 119QRT

The value of ΔrG° for the reaction at 25°C, CH4(g)+H2O(g)CO(g)+3H2(g), is 142.124kJmol1_.

Explanation of Solution

The value of standard Gibbs free energy of the formation H2(g) is 0kJmol1.

The value of standard Gibbs free energy of the formation CO(g) is 137.168kJmol1.

The value of standard Gibbs free energy of the formation of CH4(g) is 50.72kJmol1.

The value of standard Gibbs free energy of the formation of H2O(g) is 228.572kJmol1.

The given reaction is shown below.

  CH4(g)+H2O(g)CO(g)+3H2(g)

The formula to calculate the standard enthalpy change of the reaction is shown below.

    ΔrG°=ΔfG°(CO(g))+3×ΔfG°(H2(g))(ΔfG°(CH4(g))+ΔfG°(H2O(g)))

Where,

  • ΔrG° is the Gibbs free energy change of reaction.
  • ΔfG°(H2(g)) is the standard Gibbs free energy of formation of H2(g).
  • ΔfG°(CO(g)) is the standard Gibbs free energy of formation of CO(g).
  • ΔfG°(CH4(g)) is the standard Gibbs free energy of formation of CH4(g).
  • ΔfG°(H2O(g)) is the standard Gibbs free energy of formation of H2O(g).

Substitute the value of ΔfG°(H2(g)), ΔfG°(CH4(g)), ΔfG°(H2O(g)) and ΔfG°(CO(g)) in the above equation.

    ΔrG°=(137.168kJmol1)+3×(0kJmol1)(50.72kJmol1+(228.572kJmol1))=(137.168kJmol1)+279.292kJmol1=142.124kJmol1_

Therefore, the value of ΔrG° for the reaction at 25°C is 142.124kJmol1_.

(b)

Interpretation Introduction

Interpretation:

The value of equilibrium costant KP for the reaction at 25°C, CH4(g)+H2O(g)CO(g)+3H2(g), has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 119QRT

The value of equilibrium costant KP for the reaction at 25°C, CH4(g)+H2O(g)CO(g)+3H2(g) is 1.25×1025_.

Explanation of Solution

The given reaction is shown below.

  CH4(g)+H2O(g)CO(g)+3H2(g)

The value of ΔrG° for the reaction at 25°C is 142.124kJmol1.

The conversion of ΔrG° from kJmol1 to Jmol1 is shown below.

    ΔrG°=(142.124kJmol1)(1000Jmol11kJmol1)=142.124×103Jmol1

The temperature of the system is 25°C.

The conversion of temperature from °C to K is shown below.

    T=(25°C+273.15)K=298.15K

The relation between the standard Gibbs free energy change of the reaction and standard equilibrium constant of the reaction is given by the expression as shown below.

  ΔrG°=RTlnK°

Where,

  • ΔrG° is the standard Gibbs free energy change of the reaction.
  • K° is the standard equilibrium constant of the reaction.
  • R is the gas constant (8.314JK1mol1).
  • T is the temperature.

Rearrange the equation for the value of K°.

    K°=eΔrG°RT

Substitute the value of ΔrG°, R and T in the above equation.

    K°=e(142.124×103Jmol1)(8.314JK1mol1)(298.15K)=e(40.684×103)(2478.82)=e57.34=1.25×1025_

For the complete gaseous system. The value of K° corresponds to KP.

Therefore, the value of equilibrium costant KP for the given reaction is 1.25×1025_.

(c)

Interpretation Introduction

Interpretation:

Whether the reaction at 25°C, CH4(g)+H2O(g)CO(g)+3H2(g), is product favored or not has to be stated. The temperature at which the reaction would changes from reactant-favored to product-favored has to be stated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 119QRT

The given reaction is reactant favored. the temperature at which the reaction would changes from reactant-favored to product-favored is 960.23K_.

Explanation of Solution

The given reaction is shown below.

  CH4(g)+H2O(g)CO(g)+3H2(g)

The value of ΔrG° for the reaction at 25°C is 142.124kJmol1.

The value of ΔrG° of the reaction reaction is positive. The sign of positive sign of ΔrG° of the reaction indicates that the reaction reactant favored. Therefore, the given reaction is reactant favored.

The value of standard entropy of H2(g) is 130.684JK1mol1.

The value of standard entropy of CO(g) is 197.674JK1mol1.

The value of standard entropy of CH4(g) is 186.264JK1mol1.

The value of standard entropy of H2O(g) is 188.825JK1mol1.

The formula to calculate the standard enthalpy change of the reaction is shown below.

    ΔrS°=S°(CO(g))+3×S°(H2(g))(S°(CH4(g))+S°(H2O(g)))

Where,

  • ΔrS° is the entropy change of reaction.
  • S°(H2(g)) is the standard entropy of H2(g).
  • S°(CO(g)) is the standard entropy of CO(g).
  • S°(CH4(g)) is the standard entropy of CH4(g).
  • S°(H2O(g)) is the standard entropy of H2O(g).

Substitute the value of S°(H2(g)), S°(CO(g)), S°(CH4(g)) and S°(H2O(g)) in the above equation.

    ΔrS°=197.674JK1mol1+3×(130.684JK1mol1)(186.264JK1mol1+188.825JK1mol1)=197.674JK1mol1+(392.052JK1mol1)(375.089JK1mol1)=214.637JK1mol1

Therefore, the entropy change of the reaction is 214.637JK1mol1.

The value of standard enthalpy of the formation H2(g) is 0kJmol1.

The value of standard enthalpy of the formation CO(g) is 110.525kJmol1.

The value of standard enthalpy of the formation of CH4(g) is 74.81kJmol1.

The value of standard enthalpy of the formation of H2O(g) is 241.818kJmol1.

The given reaction is shown below.

  CH4(g)+H2O(g)CO(g)+3H2(g)

The formula to calculate the standard enthalpy change of the reaction is shown below.

    ΔrH°=ΔfH°(CO(g))+3×ΔfH°(H2(g))(ΔfH°(CH4(g))+ΔfH°(H2O(g)))

Where,

  • ΔrH° is the enthalpy change of reaction.
  • ΔfH°(H2(g)) is the standard enthalpy of formation of H2(g).
  • ΔfH°(CO(g)) is the standard enthalpy of formation of CO(g).
  • ΔfH°(CH4(g)) is the standard enthalpy of formation of CH4(g).
  • ΔfH°(H2O(g)) is the standard enthalpy of formation of H2O(g).

Substitute the value of ΔfH°(H2(g)), ΔfH°(CO(g)), ΔfH°(CH4(g)) and ΔfH°(H2O(g)) in the above equation.

    ΔrH°=110.525kJmol1+3×(0kJmol1)(74.81kJmol1+(241.818kJmol1))=110.525kJmol1(316.628kJmol1)=206.103kJmol1

The conversion of ΔrH° from kJmol1 to Jmol1 is shown below.

    ΔrH°=(206.103kJmol1)(1000Jmol11kJmol1)=206.103×103Jmol1

The temperature at which the reactant favored reaction becomes product favored reaction is calculated by the formula as shown below.

    T=ΔrH°ΔrS°

Where,

  • ΔrH° is the enthalpy of change of reaction.
  • ΔrS° is the entropy change of reaction.
  • T is the temperature.

Substitute the value of ΔrH° and ΔrS° in the equation.

    T=206.103×103Jmol1214.637JK1mol1=960.23K_

Therefore, the temperature at which the reaction would changes from reactant-favored to product-favored is 960.23K_.

(d)

Interpretation Introduction

Interpretation:

The value of equilibrium costant Kc for the reaction at 1000K, CH4(g)+H2O(g)CO(g)+3H2(g), has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 119QRT

The value of equilibrium costant Kc for the reaction at 25°C, CH4(g)+H2O(g)CO(g)+3H2(g), at 1000K is 4.139×104mol2/L2atm2_.

Explanation of Solution

The given reaction is shown  below.

  CH4(g)+H2O(g)CO(g)+3H2(g)

The entropy change of the reaction is 214.637JK1mol1.

The enthalpy of change of reaction is 206.103×103Jmol1.

The given temperature is 1000K.

The relation between ΔrH° and ΔrS° is given by the expression as shown below.

  ΔrG°=ΔrH°TΔrS°        (1)

Where,

  • ΔrG° is the standard Gibbs free energy change of the reaction.
  • ΔrH° is the standard enthalpy change of the reaction.
  • ΔrS° is the standard entropy change of the reaction.
  • T is the temperature.

Substitute the values of ΔrH°, ΔrS° and T in the above equation.

    ΔrG°=206.103×103Jmol1(1000K)(214.637JK1mol1)=206.103×103Jmol1214.637×103Jmol1=8.534×103Jmol1

The relation between the standard Gibbs free energy change of the reaction and standard equilibrium constant of the reaction is given by the expression as shown below.

  ΔrG°=RTlnK°

Where,

  • ΔrG° is the standard Gibbs free energy change of the reaction.
  • K° is the standard equilibrium constant of the reaction.
  • R is the gas constant (8.314JK1mol1).
  • T is the temperature.

Rearrange the equation for the value of K°.

    K°=eΔrG°RT

Substitute the value of ΔrG°, R and T in the above equation.

    K°=e(8.534×103Jmol1)(8.314JK1mol1)(1000K)=e(8.534×103)(8314)=e1.026=2.79

For the complete gaseous system. The value of K° corresponds to KP.

Therefore, the value of equilibrium costant KP for the given reaction is 2.79.

The number of moles of the gaseous molecule on the product side is 4.

The number of moles of the gaseous molecule on the reactant side is 2.

The relation between equilibrium constants Kc and KP is shown below.

  Kc=KP(RT)(npnr)

Where,

  • R is the gas constant (0.0821Latm/molK).
  • T is the temperature.
  • np is the number of moles of the gaseous molecule on the product side.
  • nr is the number of moles of the gaseous molecule on the reactant side.
  • KP is the equilibrium constant in terms of partial pressure.
  • Kc is the equilibrium constant in terms of concentration.

Substitute the value of R, T, np, and nr in the above equation.

  Kc=2.79((0.0821Latm/molK)(1000K))(42)=2.79((82.1Latm/mol))(2)=2.796740.41L2atm2/mol2=4.139×104mol2/L2atm2_

The equilibrium constants are unitless quantities.

Therefore, the value of Kc is 4.139×104mol2/L2atm2_.

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Chapter 16 Solutions

Chemistry: The Molecular Science

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