Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 20, Problem 11Q

(a)

Interpretation Introduction

Interpretation: Reaction of commercial production of hydrogen is given. The value of ΔH° and ΔS° is to be calculated for the given reaction. The temperature at which this reaction is favored is to be calculated.

Concept introduction: The expression to calculate ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

The expression to calculate ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

A reaction is said to be favored if the value of ΔG° is negative.

To determine: The value of ΔH° and ΔS° for the given reaction.

(a)

Expert Solution
Check Mark

Explanation of Solution

The stated reaction is,

CH4(g)+H2O(g)CO(g)+3H2(g)

Refer to Appendix 4

The value of ΔH°(kJ/mol) for the given reactant and product is,

Molecules ΔH°(kJ/mol)
CH4(g) 75
H2O(g) 242
CO(g) 110.5
H2(g) 0

The formula of ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

Where,

  • ΔH° the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH°(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔH°(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all the values from the table in the above equation.

ΔH°=npΔH°(product)nfΔH°(reactant)=[3(0)+(110.5){(75)+(242)}]kJ=206.5kJ_

The value of ΔS° for the given reaction is 216J/K_ .

Refer to Appendix 4

The value of ΔS°(J/Kmol) for the given reactant and product is,

Molecules ΔS°(kJ/mol)
CH4(g) 186
H2O(g) 189
CO(g) 198
H2(g) 131

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

Where,

  • ΔS° is the standard entropy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔS°(product) is the standard entropy of product at a pressure of 1atm .
  • ΔS°(reactant) is the standard entropy of reactant at a pressure of 1atm .

Substitute all the values from the table in the above equation.

ΔS°=npΔS°(product)nfΔS°(reactant)=[3(131)+(198){(186)+(189)}]J/K=216J/K_

(b)

Interpretation Introduction

Interpretation: Reaction of commercial production of hydrogen is given. The value of ΔH° and ΔS° is to be calculated for the given reaction. The temperature at which this reaction is favored is to be calculated.

Concept introduction: The expression to calculate ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

The expression to calculate ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

A reaction is said to be favored if the value of ΔG° is negative.

To determine: The temperature at which this reaction is favored.

(b)

Expert Solution
Check Mark

Explanation of Solution

The value of ΔH is 206.5kJ .

The value of ΔS is 216J/K .

The conversion of kilo-joule (kJ) into joule (J) is done as,

1kJ=103J

Hence,

The conversion of 206.5kJ into joule is,

206.5kJ=(206.5×103)J=206.5×103J

Formula

The formula of ΔG is,

ΔG=ΔH°TΔS°

Where,

  • ΔH° is the standard enthalpy of reaction.
  • ΔG° is the free energy change.
  • T is the given temperature.
  • ΔS° is the standard entropy of reaction.

At equilibrium, the value of ΔG is zero.

Substitute the values of ΔG°,ΔH° and ΔS° in the above equation.

ΔG=ΔH°TΔS°0=(206.5×103J)T(216J/K)T=956K_

The reaction will be favored if the temperature is greater than 956K .

Conclusion

The required value of ΔS° for the given reaction is 216J/K_ and ΔH° is 206.5kJ_ . The temperature at which this reaction is favored is 956K_

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Consider the following reaction: 2 HBr( g)----> H 2( g) + Br 2( g)                                                                   <------ with K = 5.3 × 10 −20. a. Write the expression for the equilibrium constant for this reaction. b. Are the reactants or products favored at equilibrium? c. Would you predict ΔH to be positive or negative? d. Are the reactants or products lower in energy? e. Would you predict this reaction to be fast or slow? Explain your choice.
TRUE OR FALSE 1. Lowering the temperature of the system will lead to an increase in entropy. 2. Spontaneous process occurs from an ordered state to a more random arrangement. 3. The increase in the number of particles can lead to an increase in entropy. 4. A negative ΔS indicates an increase in randomness of a system 5. Lowering the temperature of the system will lead to a decrease in entropy. 6. The second law of thermodynamics states that the total entropy of a system decreases in any spontaneous process. 7. Free energy is a portion of the total energy of a system that is available to do useful work. 8. If change of G is positive, then it's non-spontaneous. 9. Spontaneity can predict the rate of a process (i.e., how fast or slow it occurs). 10. The increase in the number of particles can lead to a decrease in entropy.
Part 1 and part 2 go together For part 2, Then use the data below to calculate ∆G rxn using free energies of formation.

Chapter 20 Solutions

Chemistry

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