Problem 53
Question
At 740 torr and \(20.0^{\circ} \mathrm{C}\), nitrogen has a solubility in water of \(0.018 \mathrm{~g} \mathrm{~L}^{-1}\). At 620 torr and \(20.0{ }^{\circ} \mathrm{C}\), its solubility is \(0.015 \mathrm{~g} \mathrm{~L}^{-1}\). Show that nitrogen obeys Henry's law.
Step-by-Step Solution
Verified Answer
To show that nitrogen obeys Henry's Law, calculate the Henry's Law constant for each condition. If they are consistent, nitrogen obeils the law.
1Step 1: Understanding Henry's Law
Henry's Law states that at a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid. This can be expressed with the formula: \( S = k_H \times P \), where \( S \) is the solubility of the gas, \( k_H \) is the Henry's law constant for the gas-liquid pair, and \( P \) is the partial pressure of the gas.
2Step 2: Calculating Henry's Law Constant for First Condition
Using the solubility and pressure for the first condition, calculate Henry's law constant, \( k_{H1} \), using the formula from Henry's Law. \( k_{H1} = \frac{S_1}{P_1} \), where \( S_1 = 0.018 \, \mathrm{g} \, \mathrm{L}^{-1} \) and \( P_1 = 740 \, \mathrm{torr} \).
3Step 3: Calculating the Henry's Law Constant for Second Condition
Similarly, calculate Henry's law constant, \( k_{H2} \), using the solubility and pressure for the second condition. \( k_{H2} = \frac{S_2}{P_2} \), where \( S_2 = 0.015 \, \mathrm{g} \, \mathrm{L}^{-1} \) and \( P_2 = 620 \, \mathrm{torr} \).
4Step 4: Comparing Henry's Law Constants
Compare the values of \( k_{H1} \) and \( k_{H2} \). If Henry's Law is obeyed, the values should be consistent.
5Step 5: Conclusion
If the values of \( k_{H1} \) and \( k_{H2} \) are indeed consistent, it means that nitrogen obeys Henry's Law at \(20.0^{\text{\circ}} C\).
Key Concepts
Gas Solubility in LiquidsGas-Liquid EquilibriumPartial Pressure
Gas Solubility in Liquids
Understanding how gases dissolve in liquids is essential for fully grasping many phenomena in chemistry, including the behavior of carbonated drinks, preserving foods, and even our own respiration. When a gas comes in contact with a liquid, it can enter and become dissolved within the liquid to form a solution. The solubility of a gas in a liquid is influenced by temperature, pressure, and the nature of the gas and the solvent. As a general rule, the solubility of a gas increases with pressure and decreases with temperature.
The behavior of gases dissolving in liquids can be elegantly described by Henry's Law. This principle allows us to predict and quantify the solubility of a gas within a liquid under various conditions. To illustrate, nitrogen's solubility in water changes when the pressure above the water is altered, as seen in the exercise. By comparing solubility at two different pressures, we directly apply the concept of gas solubility and Henry's Law.
The behavior of gases dissolving in liquids can be elegantly described by Henry's Law. This principle allows us to predict and quantify the solubility of a gas within a liquid under various conditions. To illustrate, nitrogen's solubility in water changes when the pressure above the water is altered, as seen in the exercise. By comparing solubility at two different pressures, we directly apply the concept of gas solubility and Henry's Law.
Gas-Liquid Equilibrium
The gas-liquid equilibrium is an essential concept to understand when discussing the solubility of gases in liquids. It refers to the state where the rate at which gas molecules enter the liquid is equal to the rate at which they escape back into the gas phase. This dynamic equilibrium is what Henry's Law describes. At this point, the concentration of the gas in the liquid remains constant as long as the temperature and the gas's partial pressure do not change.
When discussing gas solubility, knowing that equilibrium implies a steady state helps us to comprehend why and how conditions such as temperature and pressure influence a gas's solubility in a liquid. In the exercise provided, nitrogen's solubility in water at two different pressures, while at the same temperature, reflected the gas-liquid equilibrium under varying conditions, demonstrating that the system obeys Henry's Law.
When discussing gas solubility, knowing that equilibrium implies a steady state helps us to comprehend why and how conditions such as temperature and pressure influence a gas's solubility in a liquid. In the exercise provided, nitrogen's solubility in water at two different pressures, while at the same temperature, reflected the gas-liquid equilibrium under varying conditions, demonstrating that the system obeys Henry's Law.
Partial Pressure
Partial pressure is a term that describes the pressure a single gas contributes to a mixture of gases and plays a vital role in determining gas solubility according to Henry's Law. In a mixture, each gas exerts pressure as if it alone occupied the entire volume, and the total pressure is the sum of the individual partial pressures. This concept directly affects the amount of the gas that will dissolve when in contact with a liquid.
To understand how partial pressure influences gas solubility, let's refer back to our exercise. When the partial pressure of nitrogen is higher at 740 torr, more nitrogen dissolves in the water compared to when the pressure is lower at 620 torr. Adjusting the partial pressure essentially allows control over how much gas can be dissolved in the liquid, which is crucial for industrial applications such as beverage carbonation and manufacturing processes requiring precise gas concentrations.
To understand how partial pressure influences gas solubility, let's refer back to our exercise. When the partial pressure of nitrogen is higher at 740 torr, more nitrogen dissolves in the water compared to when the pressure is lower at 620 torr. Adjusting the partial pressure essentially allows control over how much gas can be dissolved in the liquid, which is crucial for industrial applications such as beverage carbonation and manufacturing processes requiring precise gas concentrations.
Other exercises in this chapter
Problem 50
Consider the formation of a solution of aqueous potassium chloride. Write the thermochemical equations for (a) the conversion of solid KCl into its gaseous ions
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