Chapter 17, Problem 70E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# Hydrogen sulfide can be removed from natural gas by the reaction 2 H 2 S ( g ) + SO 2 ( g ) ⇌ 3 S ( s ) + 2 H 2 O ( g ) Calculate ∆G° and K (at 298 K) for this reaction. Would this reaction be favored at a high or low temperature?

Interpretation Introduction

Interpretation: The reaction between H2S and SO2 is given. The value of ΔG° and K at a given temperature is to be calculated. The temperature conditions at which the stated reaction will be favored is to be stated.

Concept introduction: Equilibrium constant K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, then the free energy change will be,

ΔG=0Q=K

The expression for free energy change is,

ΔG°=RTln(K)

The expression for standard Gibbs free energy, ΔG° , is,

ΔG°=npΔG°(product)nfΔG°(reactant)

Explanation

Explanation

The reaction that takes place is,

2H2S(g)+SO2(g)3S(s)+2H2O(g)

Refer Appendix 4 .

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

 Molecules ΔG°(kJ/mol) SO2(g) −300 S(g) 0 H2S(g) −34 H2O(g) −229

The formula of ΔG° is,

ΔG°=npΔG°(product)nfΔG°(reactant)

Where,

• np is the number of moles of each product.
• nr is the number of moles each reactant.
• ΔG°(product) is the free energy of product at a pressure of 1atm .
• ΔG°(reactant) is the free energy of reactant at a pressure of 1atm .

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

ΔG°=npΔG°(product)nfΔG°(reactant)=[3(0)+2(229){2(34)+(300)}]kJ=90kJ_

The value of ΔG° is 90kJ .

Temperature is 298K .

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

1kJ=103J

Hence,

The conversion of 90kJ into joule is,

198kJ=(90kJ×103)J=90kJ×103J

Formula

The expression for free energy change is,

ΔG=ΔG°+RTln(Q)

Where,

• ΔG is the free energy change for a reaction at specified pressure.
• R is the gas law constant (8.3145J/Kmol) .
• T is the absolute temperature.
• Q is the reaction constant.

Since, the reaction is at equilibrium, the free energy change,

ΔG=0Q=K

Where,

K is the equilibrium constant.

Substitute these values in the free energy change expression.

ΔG=ΔG°+RTln(Q)ΔG°=RTln(K)ln(K)=ΔG°RT

Substitute the values of R,ΔG° and T in the above expression.

ln(K)=ΔG°RT=90×103J(8.3145J/K)(298K)K=5

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