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2. Using the data in Appendix A of Benjamin's Water Chemistry, answer the following questions a. What is the Henry's law constant (KH) at 25 °C for the exchange of hydrogen sulfide between the gas and aqueous phases? H₂S (g) →H₂S (aq) b. What is the Henry's law constant for the reaction in part a) at a temperature of 5 °C?

2. Using the data in Appendix A of Benjamin's Water Chemistry, answer the following questions a. What is the Henry's law constant (KH) at 25 °C for the exchange of hydrogen sulfide between the gas and aqueous phases? H₂S (g) →H₂S (aq) b. What is the Henry's law constant for the reaction in part a) at a temperature of 5 °C?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Question 2: Chemistry of Hydrogen Sulfide** Using the data in Appendix A of Benjamin’s *Water Chemistry*, answer the following questions: a. What is the Henry’s law constant (K_H) at 25 °C for the exchange of hydrogen sulfide between the gas and aqueous phases? \[ \text{H}_2\text{S (g)} \leftrightarrow \text{H}_2\text{S (aq)} \] b. What is the Henry’s law constant for the reaction in part a) at a temperature of 5 °C?
Transcribed Image Text:**Question 2: Chemistry of Hydrogen Sulfide** Using the data in Appendix A of Benjamin’s *Water Chemistry*, answer the following questions: a. What is the Henry’s law constant (K_H) at 25 °C for the exchange of hydrogen sulfide between the gas and aqueous phases? \[ \text{H}_2\text{S (g)} \leftrightarrow \text{H}_2\text{S (aq)} \] b. What is the Henry’s law constant for the reaction in part a) at a temperature of 5 °C?
Expert Solution
Step 1

The Answer to the Following question is-

 

b) Henry's Law Constant KH = e-∆GRT
Now calculating ∆G for the given reaction at 25° C.

Converting 0°C into Kelin T= 273 + 25 = 298K

H2S (g)↔H2S (aq)

∆G°=∆H°-T∆S°

∆H°=∆H°f(Product)-∆H°f(reactant)=(-39.25KJ/mol)-(-20.63KJ/mol)=-19.12KJ/mol

∆S°=∆S°f(product)-∆S°f(reactant)=121.3J/mol.K-205.7 J/mol.K=-84.4 J/mol.K

∆S°=-84.4 J/mol.K=-0.0844KJ/mol.K

T∆S=298×(-0.0844 KJ/mol.K)=-25.15 KJ/mol

∆G°=∆H°-T∆S°

∆G°= -19.12 KJ/mol - (-25.15 kJ/mol)= 6.03 KJ/mol

Putting the value in the KH we get:

KH=e-6.030.008314×298=e-2.43=0.088

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