Which of the following graph correctly represents the plots of \(K_H\) at 1 bar gases in water versus temperature?
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The Correct Option is D
Approach Solution - 1
Approach Solution -2
The solution requires an understanding of Henry's Law and the factors that influence the Henry's Law constant (\( K_H \)).
- Henry's Law: This law relates the partial pressure (\( p \)) of a gas above a liquid to its mole fraction (\( x \)) in the liquid phase: \[ p = K_H \cdot x \] From this equation, the solubility of a gas (represented by \( x \)) is inversely proportional to \( K_H \) for a given partial pressure. A higher \( K_H \) value means lower solubility. \[ \text{Solubility} \propto \frac{1}{K_H} \]
- Effect of Temperature on \( K_H \): The dissolution of gases in liquids is typically an exothermic process. A simplified model suggests that increasing the temperature decreases solubility, which would mean \( K_H \) monotonically increases with temperature. However, the actual behavior for many nonpolar gases in water is more complex. The value of \( K_H \) first increases with temperature, reaches a maximum (corresponding to a minimum in solubility), and then begins to decrease at higher temperatures. This non-monotonic behavior is a more accurate representation.
- Dependence of \( K_H \) on the Nature of the Gas: The value of \( K_H \) is specific to the gas-solvent pair. For nonpolar gases in a polar solvent like water, the solubility is governed by London dispersion forces. Gases that are smaller and less polarizable interact more weakly with water, leading to lower solubility and a higher \( K_H \) value.
Step-by-Step Solution:
Step 1: Analyze the Temperature Dependence shown in the graphs.
The graphs show different trends for \( K_H \) versus temperature (\( t/^\circ\text{C} \)).
- Graphs (2) and (3) show a simple, monotonic increase of \( K_H \) with temperature. This represents the introductory model where gas solubility always decreases with increasing temperature.
- Graphs (1) and (4) show a more complex, non-monotonic relationship: \( K_H \) first increases, reaches a maximum value, and then decreases. This is a more accurate depiction of the behavior of many real gases (like He, N\(_2\), O\(_2\), CH\(_4\)) in water.
Given the options, we should consider the more accurate, non-monotonic trend as the correct representation of the physical phenomenon.
Step 2: Analyze the relative order of \( K_H \) values for the gases in option (4).
Option (4) plots \( K_H \) for Helium (He), Nitrogen (N\(_2\)), and Methane (CH\(_4\)). We need to determine the correct order of their \( K_H \) values.
- Helium (He): A very small noble gas with extremely low polarizability. It has very weak intermolecular interactions with water.
- Nitrogen (N\(_2\)): A nonpolar diatomic molecule, larger and more polarizable than He.
- Methane (CH\(_4\)): A nonpolar molecule, larger and more polarizable than N\(_2\).
Greater polarizability leads to stronger London dispersion forces with water molecules, resulting in higher solubility.
Therefore, the order of solubility is: \( \text{Solubility(CH}_4\text{)} > \text{Solubility(N}_2\text{)} > \text{Solubility(He)} \).
Since \( K_H \) is inversely proportional to solubility, the order of the Henry's Law constants must be the reverse:
\[ K_H(\text{He}) > K_H(\text{N}_2) > K_H(\text{CH}_4) \]
Step 3: Evaluate Graph (4) based on the analysis.
Let's check if Graph (4) is consistent with our findings.
- Shape of the Curves: The graph correctly shows the non-monotonic behavior where \( K_H \) first increases and then decreases with temperature.
- Relative Positions: At any given temperature on the graph, the curve for He is the highest, followed by N\(_2\), and then CH\(_4\) is the lowest. This corresponds to the order \( K_H(\text{He}) > K_H(\text{N}_2) > K_H(\text{CH}_4) \), which we determined to be correct.
Conclusion:
Graph (4) correctly represents both the sophisticated temperature dependence and the relative magnitudes of the Henry's Law constants for He, N\(_2\), and CH\(_4\) in water. The other graphs are incorrect because they either show a simplified temperature dependence (3) or an incorrect relative ordering of the \( K_H \) values (1, 2).
Therefore, the correct representation is given by option (4).
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