Q.Value of Henry's constant $K_H$ ____________.

The key idea here is Henry's Law, which describes the solubility of a gas in a liquid.

1. Henry's Law states that the partial pressure of a gas ($P$) above a liquid is directly proportional to its mole fraction ($x$) in the solution: $P = K_H \cdot x$. Here, $K_H$ is Henry's constant.
2. The dissolution of a gas in a liquid is an exothermic process, meaning it releases heat.
3. According to Le Chatelier's principle, an increase in temperature shifts the equilibrium of an exothermic process towards the reactants (undissolved gas), thereby decreasing the solubility ($x$) of the gas in the liquid. …

Henry's Law describes the relationship between the partial pressure of a gas above a liquid and its solubility in that liquid. Understanding how temperature affects gas solubility is key to determining the behavior of Henry's constant.

When a gas dissolves in a liquid, it's an equilibrium process: gas molecules from the headspace enter the liquid phase, and dissolved gas molecules escape back into the headspace.

For most gases, the dissolution process is exothermic, meaning it releases heat. This is because when gas molecules enter the liquid phase, they form intermolecular attractions with the solvent molecules, which is an energy-releasing process.

Let's break down the reasoning step-by-step.

1. Recall Henry's Law:

   Henry's Law states that the partial pressure of a gas ($P$) above a liquid is directly proportional to its mole fraction ($x$) in the solution.

   $P = K_H \cdot x$

   Here, $P$ is the partial pressure of the gas, $x$ is the mole fraction of the gas in the solution (a measure of its solubility), and $K_H$ is Henry's constant.

2. Understand the physical meaning of $K_H$:

   From the formula $P = K_H \cdot x$, we can rearrange it to $x = P/K_H$. This shows that for a given partial pressure $P$, the mole fraction $x$ (which represents solubility) is inversely proportional to $K_H$.

   • A higher $K_H$ value means that for a given partial pressure, the gas has a lower solubility (smaller $x$).
   • A lower $K_H$ value means that for a given partial pressure, the gas has a higher solubility (larger $x$). Essentially, $K_H$ is a measure of the tendency of a gas to escape from the solution; a higher $K_H$ means a greater tendency to escape, hence lower solubility.

3. Consider the effect of temperature on gas solubility:

   As mentioned, the dissolution of a gas in a liquid is generally an exothermic process. This means that heat is released when the gas dissolves.

When the temperature of the system is increased, we are essentially adding heat to the system.

4. Apply Le Chatelier's Principle:

Le Chatelier's Principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. …

Method: Le Chatelier’s Principle applied to Gas Solubility

Step 1 — Recall Henry’s Law

Henry’s Law states:

where:

• $p$ = partial pressure of the gas above the liquid
• $x$ = mole fraction of the gas dissolved in the liquid
• $K_H$ = Henry’s constant (specific to gas–solvent pair)

Step 2 — Understand the physical process

Dissolution of a gas in a liquid is exothermic (releases heat):

Step 3 — Apply Le Chatelier’s Principle

• Increase in temperature → system shifts to absorb heat → equilibrium shifts backward (towards gas phase).
• This means solubility decreases with temperature rise.

Step 4 — Relate solubility change to $K_H$ …

Here’s a breakdown of the common mistakes students make on this Henry’s Law question, and how to avoid each.

1. Confusing Solubility with Henry’s Constant

The Mistake:

Students think: “If temperature increases, gases dissolve less, so $K_H$ should decrease.”

They pick option (ii).

Why it’s wrong:

Henry’s Law is:

where $x$ is the mole fraction of the gas in the liquid.

When temperature rises, solubility ($x$) decreases for a fixed pressure $P$.

From the equation, if $x$ goes down and $P$ is constant, $K_H$ must increase.

How to avoid:

• Always write the formula: $P = K_H \cdot x$
• Remember: $K_H$ is inversely proportional to solubility at constant pressure.
• If solubility decreases → $K_H$ increases.

✓ Correct answer: (i) increases with increase in temperature.

2. Treating $K_H$ Like a Universal Constant

The Mistake:

Students assume “constant” in the name means it never changes, so they pick (iii).

Why it’s wrong:

$K_H$ is a temperature-dependent constant for a given gas-solvent pair. It is not a universal constant like $R$ or $N_A$.

How to avoid:

• Know that $K_H$ changes with temperature (and also with the nature of the gas and solvent).
• Only “constant” means constant for a fixed temperature.

3. Misremembering the Trend from Graphs

The Mistake:

Students recall a graph where solubility vs. temperature first rises then falls for some liquids, and apply that to gases. They pick (iv).

Why it’s wrong:

For gases in liquids, solubility always decreases with temperature (exothermic dissolution). There is no initial increase.

How to avoid:

• Separate gas-in-liquid behaviour from solid-in-liquid behaviour.
• For gases: solubility decreases monotonically with temperature → $K_H$ increases monotonically.

4. Forgetting the “Pressure Constant” Condition

The Mistake:

Students think about $K_H$ changing without specifying that pressure is held constant.

Why it’s wrong: …