Which of the following graph correctly represents the plots of \(K_H\) at 1 bar gases in water versus temperature?
Show Hint
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).
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main The Kinetic Theory of Gases Questions
- Consider the transition metal ions \( \text{Mn}^{3+}, \text{Cr}^{3+}, \text{Fe}^{3+} \) and \( \text{Co}^{3+} \) and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective \(d\)-orbitals of the complexes is
- Statement I: For an ideal gas, heat capacity at constant volume is always greater than the heat capacity at constant pressure.
Statement II: In a constant volume process, no work is produced and all the heat withdrawn goes into the chaotic motion and is reflected by a temperature increase of the ideal gas.
In the light of the above statements, choose the correct answer: - At \( 27^\circ \text{C} \), 0.1 M, 1 L \( K_4[Fe(CN)_6] \) aqueous solution and 0.1 M, 1 L \( FeCl_3 \) aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the solutes. Which of the following statement is correct?
- The correct order of molar heat capacities measured at 298 K and 1 bar is :
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,