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
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
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?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ebe91de-3090-46ac-907f-d0393c4d1db8%2Fcf93a40f-7b62-4733-a2bf-a5c65132c88a%2Fldvh1be_processed.jpeg&w=3840&q=75)
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?
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Step 1
The Answer to the Following question is-
b) Henry's Law Constant KH =
Now calculating for the given reaction at 25° C.
Converting 0°C into Kelin T= 273 + 25 = 298K
-19.12 KJ/mol - (-25.15 kJ/mol)= 6.03 KJ/mol
Putting the value in the KH we get:
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