Animated Solution for Chemistry - Surface Chemistry: ${\text{CH}}_4$ is adsorbed on $1\text{ g}$ charcoal at $0^{\circ }\text{C}$ following the Freundlich adsorption isotherm. $10.0\text{ mL}$ of ${\text{CH}}_4$ is adsorbed at $100\text{ mm}$ of $\text{Hg}$, whereas $15.0\text{ mL}$ is adsorbed at $200\text{ mm}$ of $\text{Hg}$. The volume of ${\text{CH}}_4$ adsorbed at $300\text{ mm}$ of $\text{Hg}$ is $10^x\text{ mL}$. The value of $x$ is ....... $\times 10^{-2}$. (Nearest integer) [Use ${\log }_{10}2=0.3010$, ${\log }_{10}3=0.4771$]

Imagine a piece of charcoal acting like a microscopic sponge, but instead of soaking up water into its bulk, it traps gas molecules strictly on its surface. This surface phenomenon is known as adsorption. In our problem, methane gas (${\text{CH}}_4$) is being adsorbed onto the surface of $1\text{ g}$ of charcoal at a constant temperature of $0^{\circ }\text{C}$.

Because the mass of the charcoal is fixed at $1\text{ g}$, the extent of adsorption—typically denoted as $\frac{x}{m}$ (mass of adsorbate per unit mass of adsorbent)—can be directly represented by the volume $v$ of the methane gas adsorbed. This is a direct application of Avogadro's Law, where volume is proportional to the number of moles, and thus the mass, of the gas at constant temperature and pressure.

To mathematically model how the volume of adsorbed gas changes with the applied pressure, we use the Freundlich Adsorption Isotherm. This empirical relationship is given by the equation:

Here, $v$ is the volume of gas adsorbed, $p$ is the pressure of the gas, and $k$ and $n$ are constants that depend on the nature of the adsorbate and adsorbent at a particular temperature. The exponent $\frac{1}{n}$ is particularly interesting; it usually lies between $0$ and $1$, indicating that while adsorption increases with pressure, the rate of this increase slows down as the surface of the charcoal becomes saturated with gas molecules.

We are provided with two specific data points from an experiment: 1. At a pressure of $100\text{ mm Hg}$, the volume adsorbed is $10.0\text{ mL}$. 2. At a pressure of $200\text{ mm Hg}$, the volume adsorbed is $15.0\text{ mL}$.

Let's plug these physical observations into our mathematical model to create a system of equations:

We now have a system of two equations with two unknowns ($k$ and $\frac{1}{n}$). The most elegant way to solve this is not by isolating $k$ and substituting, but by dividing the two equations. This technique instantly eliminates the constant $k$, leaving us with a clean ratio.

Dividing Equation (ii) by Equation (i):

The Key to Exponents

Our unknown variable, $\frac{1}{n}$, is trapped in the exponent. To bring it down to the base level where we can solve for it algebraically, we must employ logarithms. Taking the base-10 logarithm on both sides yields:

Now, we substitute the given logarithmic values ($\log 2=0.3010$ and $\log 3=0.4771$):

This value of $0.585$ confirms our theoretical understanding that $\frac{1}{n}$ should lie between $0$ and $1$.

Armed with the value of $\frac{1}{n}$, we can now tackle the core question: What is the volume $v$ when the pressure is increased to $300\text{ mm Hg}$? We set up our third equation:

Once again, we use the power of ratios. Dividing Equation (iii) by Equation (i) eliminates $k$:

Taking the logarithm on both sides again:

Substitute our calculated value of $\frac{1}{n}$:

To isolate $v$, we convert this logarithmic expression back into its exponential form:

The problem states that the volume adsorbed at $300\text{ mm Hg}$ is $10^x\text{ mL}$. By comparing our result with this format, we can clearly see that:

However, the question asks for the value of $x$ in the specific format of an integer multiplied by $10^{-2}$. To achieve this, we shift the decimal point two places to the right:

Rounding to the nearest integer, we arrive at our final answer:

The Effect of Temperature

While we have solved the math, it is crucial to understand the physics. What if the experiment was conducted at $25^{\circ }\text{C}$ instead of $0^{\circ }\text{C}$?

Physical adsorption involves weak van der Waals forces and is inherently an exothermic process ($\Delta H<0$). According to Le Chatelier's principle, increasing the temperature of an exothermic process shifts the equilibrium backward. Therefore, at a higher temperature, the kinetic energy of the gas molecules increases, making it harder for them to stay trapped on the surface. The entire Freundlich isotherm curve would shift downwards, meaning less volume of gas would be adsorbed at any given pressure.