5. The Henry's law constant for O2 in water at 25 °C is 1.26 x 10-3 M/atm. What partial pressure of O, is necessary to achieve an equilibrium concentration of 1.5 x 10-3 M 0,? a. 1.9 x 10-6 atm b. 0.24 atm с. 0.84 atm d. 1.2 atm e. 5.3 x 105 atm
5. The Henry's law constant for O2 in water at 25 °C is 1.26 x 10-3 M/atm. What partial pressure of O, is necessary to achieve an equilibrium concentration of 1.5 x 10-3 M 0,? a. 1.9 x 10-6 atm b. 0.24 atm с. 0.84 atm d. 1.2 atm e. 5.3 x 105 atm
Chemistry
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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
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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![### Henry's Law and Partial Pressure
**Question:**
The Henry's law constant for \( O_2 \) in water at 25°C is \( 1.26 \times 10^{-3} \, \text{M/atm} \). What partial pressure of \( O_2 \) is necessary to achieve an equilibrium concentration of \( 1.5 \times 10^{-3} \, \text{M} \) \( O_2 \)?
**Options:**
a. \( 1.9 \times 10^{-3} \, \text{atm} \)
b. \( 0.24 \, \text{atm} \)
c. \( 0.84 \, \text{atm} \)
d. \( 1.2 \, \text{atm} \)
e. \( 5.3 \times 10^5 \, \text{atm} \)
**Explanation:**
To find the partial pressure of \( O_2 \) required to achieve the given concentration, we use Henry's Law, which is stated as:
\[ C = k_H \cdot P \]
Where:
- \( C \) is the concentration of the gas in the solution (in this case, \( 1.5 \times 10^{-3} \, \text{M} \)),
- \( k_H \) is the Henry's law constant (given as \( 1.26 \times 10^{-3} \, \text{M/atm} \)),
- \( P \) is the partial pressure of the gas.
Rearranging the formula to solve for \( P \):
\[ P = \frac{C}{k_H} \]
Substitute the given values:
\[ P = \frac{1.5 \times 10^{-3} \, \text{M}}{1.26 \times 10^{-3} \, \text{M/atm}} \]
\[ P \approx 1.19 \, \text{atm} \]
Therefore, the answer closest to the calculated pressure is:
\[ \boxed{d. \, 1.2 \, \text{atm}} \]
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#### No Graphs or Diagrams
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Transcribed Image Text:### Henry's Law and Partial Pressure
**Question:**
The Henry's law constant for \( O_2 \) in water at 25°C is \( 1.26 \times 10^{-3} \, \text{M/atm} \). What partial pressure of \( O_2 \) is necessary to achieve an equilibrium concentration of \( 1.5 \times 10^{-3} \, \text{M} \) \( O_2 \)?
**Options:**
a. \( 1.9 \times 10^{-3} \, \text{atm} \)
b. \( 0.24 \, \text{atm} \)
c. \( 0.84 \, \text{atm} \)
d. \( 1.2 \, \text{atm} \)
e. \( 5.3 \times 10^5 \, \text{atm} \)
**Explanation:**
To find the partial pressure of \( O_2 \) required to achieve the given concentration, we use Henry's Law, which is stated as:
\[ C = k_H \cdot P \]
Where:
- \( C \) is the concentration of the gas in the solution (in this case, \( 1.5 \times 10^{-3} \, \text{M} \)),
- \( k_H \) is the Henry's law constant (given as \( 1.26 \times 10^{-3} \, \text{M/atm} \)),
- \( P \) is the partial pressure of the gas.
Rearranging the formula to solve for \( P \):
\[ P = \frac{C}{k_H} \]
Substitute the given values:
\[ P = \frac{1.5 \times 10^{-3} \, \text{M}}{1.26 \times 10^{-3} \, \text{M/atm}} \]
\[ P \approx 1.19 \, \text{atm} \]
Therefore, the answer closest to the calculated pressure is:
\[ \boxed{d. \, 1.2 \, \text{atm}} \]
---
#### No Graphs or Diagrams
The provided image contains only text with a mathematical problem and options. There are no graphs or diagrams to explain further.
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