A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isother

Q: Why is the exponent of pressure 0.5 in the given equation?

In the Freundlich isotherm $\frac{x}{m} = Kp^{1/n}$, the value of $1/n$ ranges from 0 to 1. An exponent of 0.5 indicates that the adsorption is occurring at a moderate pressure, where the extent of adsorption depends fractionally on the pressure.

Q: Does chemical adsorption (chemisorption) behave the same way with temperature?

No! Chemisorption involves the formation of strong chemical bonds, which often requires activation energy. Therefore, chemisorption initially increases with an increase in temperature (to overcome the activation barrier) before eventually decreasing at very high temperatures.

Q: What happens to the adsorption if we increase the pressure to a very high value?

At very high pressures, the solid surface becomes completely saturated with gas molecules. At this point, further increasing the pressure has no effect on adsorption, and the exponent $1/n$ approaches 0 ($\frac{x}{m} = Kp^0 = K$).

Q: Why is physical adsorption always exothermic?

Adsorption involves gas molecules losing a degree of freedom as they get restricted to a 2D surface, meaning entropy decreases ($\Delta S < 0$). For the process to be spontaneous ($\Delta G < 0$), the enthalpy change ($\Delta H$) must be negative, making it an exothermic process.

Q: What is the difference between an adsorption isotherm and an adsorption isobar?

An adsorption isotherm plots the extent of adsorption ($\frac{x}{m}$) against pressure ($p$) while keeping the temperature constant. An adsorption isobar plots $\frac{x}{m}$ against temperature ($T$) while keeping the pressure constant.

Question

A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation $\frac{x}{m} = Kp^{0.5}$. Adsorption of the gas increases with

Options:

(A) increase in $p$ and increase in $T$

(B) increase in $p$ and decrease in $T$

(C) decrease in $p$ and decrease in $T$

(D) decrease in $p$ and increase in $T$

Sigma Percentile · Accepted Answer

The phenomenon of adsorption is all around us, from the activated charcoal in water filters to the catalytic converters in our cars. But how do we control how much gas sticks to a solid surface? This problem beautifully illustrates the dual control we have over physical adsorption: pressure and temperature. Let's break down the mechanics of this process.

### Decoding the Freundlich Isotherm

The problem provides us with the Freundlich adsorption isotherm equation: 
$$ \frac{x}{m} = Kp^{0.5} $$

Here, $x$ is the mass of the gas adsorbed, $m$ is the mass of the solid adsorbent, and $p$ is the pressure of the gas. The term $\frac{x}{m}$ represents the **extent of adsorption**.

The crucial part of this equation is the exponent of pressure, which is $0.5$. Because this exponent is positive, it tells us that the extent of adsorption is directly proportional to the square root of the pressure ($\frac{x}{m} \propto \sqrt{p}$).

Physically, this makes perfect sense. If you increase the pressure of a gas above a solid surface, you are essentially forcing more gas molecules to collide with and stick to the available sites on the surface. Therefore, **an increase in pressure ($p$) leads to an increase in adsorption**.

### The Thermodynamics of Physisorption

Now, let's turn our attention to temperature. The problem specifically mentions **physical adsorption** (or physisorption).

Physisorption occurs due to weak van der Waals forces between the gas molecules and the solid surface. When these bonds form, energy is released. This means that physical adsorption is an **exothermic process**. We can represent this equilibrium as:

$$ 	ext{Gas} + 	ext{Solid} ightleftharpoons 	ext{Gas adsorbed} + 	ext{Heat} $$

To understand how temperature affects this equilibrium, we invoke **Le Chatelier's Principle**. If we increase the temperature of a system at equilibrium, the system will try to counteract that change by shifting in the direction that absorbs heat.

Since the forward reaction (adsorption) releases heat, increasing the temperature will drive the reaction backward (desorption). Conversely, if we want to drive the reaction forward to increase adsorption, we must remove heat from the system. Therefore, **a decrease in temperature ($T$) leads to an increase in adsorption**.

### The Final Verdict

We have established two independent conditions to maximize the physical adsorption of the gas:
1. We must **increase the pressure ($p$)** to force more molecules onto the surface.
2. We must **decrease the temperature ($T$)** to thermodynamically favor the exothermic adsorption process.

Combining these two findings, the adsorption of the gas increases with an increase in $p$ and a decrease in $T$. This perfectly aligns with option (b).

Animated Solution for Chemistry - Surface Chemistry: A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation $\frac{x}{m} = K p^{0.5}$ . Adsorption of the gas increases with

Select Answer:

Visualized Solution

$Analyzing Pressure Dependence$

$Analyzing Temperature Dependence$

$Conclusion$

The Sigma Insight: Adsorption

Decoding the Freundlich Isotherm

The Thermodynamics of Physisorption

The Final Verdict

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