Problem 15 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(20 \%\) of surface sites are occupied by \(\mathrm{N}_{2}\) molecules. The density of surface site is \(6.023 \times 10^{14} \mathrm{~cm}^{-2}\) and total surface area is \(1000 \mathrm{~cm}^{2}\). The catalyst is heated to \(300 \mathrm{~K}\) while \(\mathrm{N}_{2}\) is completely desorbed into a pressure of \(0.001\) atm and volume of \(2.46 \mathrm{~cm}^{3}\). Find the number of active sites occupied by each \(\mathrm{N}_{2}\) molecule.

Step-by-Step Solution

Answer

Each \(\mathrm{N}_2\) molecule occupies about 2 active sites.

Step 1: Calculate Total Number of Surface Sites

To find the total number of surface sites, multiply the density of surface sites by the total surface area. Number of surface sites \( = 6.023 \times 10^{14} \text{ cm}^{-2} \times 1000 \text{ cm}^2 \). Thus, Number of surface sites \( = 6.023 \times 10^{17} \).

Step 2: Calculate Number of Surface Sites Occupied by \(\mathrm{N}_2\)

Since 20% of surface sites are occupied by \(\mathrm{N}_2\), we calculate this as follows: Number of occupied sites \( = 0.20 \times 6.023 \times 10^{17} \). Thus, number of occupied sites \( = 1.2046 \times 10^{17} \).

Step 3: Calculate Total Number of \(\mathrm{N}_2\) Molecules Desorbed

Using the ideal gas law \(PV = nRT\), we calculate the number of moles of \(\mathrm{N}_2\) desorbed. \(P = 0.001 \text{ atm},\ V = 2.46 \text{ cm}^3 = 2.46 \times 10^{-3} \text{ L},\ R = 0.0821 \ \text{L atm K}^{-1} \text{ mol}^{-1},\ T = 300 \text{ K}\). \(n = \frac{PV}{RT} = \frac{0.001 \times 2.46 \times 10^{-3}}{0.0821 \times 300}\). Calculate \(n = 1.0 \times 10^{-7} \text{ mol}\).

Step 4: Find the Number of \(\mathrm{N}_2\) Molecules Desorbed

The number of \(\mathrm{N}_2\) molecules can be found by multiplying the number of moles by Avogadro's number \(6.022 \times 10^{23}\). Number of molecules \( = 1.0 \times 10^{-7} \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} \). Thus, number of molecules \( = 6.022 \times 10^{16} \).

Step 5: Calculate Active Sites per \(\mathrm{N}_2\) Molecule

Now, to find the number of active sites occupied by each \(\mathrm{N}_2\) molecule, divide the total occupied sites by the number of molecules. Number of active sites per molecule \( = \frac{1.2046 \times 10^{17}}{6.022 \times 10^{16}}\). Calculate to get approximately 2 active sites per molecule.

Key Concepts

Surface Site Density

Surface site density is a crucial term used in surface chemistry and catalysis. It refers to the number of reactive sites available per unit area on the surface of a catalyst. These are the spots where chemical reactions can occur. In our case, the surface site density is given as \(6.023 \times 10^{14} \text{ cm}^{-2}\).
To understand how many sites we are dealing with, we calculate the total number of surface sites by multiplying the surface site density by the total surface area. Here, the equation is:\[\text{Total number of surface sites} = \text{density} \times \text{area}\]Thus making use of the given numbers:\[6.023 \times 10^{14} \text{ cm}^{-2} \times 1000 \text{ cm}^2 = 6.023 \times 10^{17} \text{ sites}\]This large number represents the vast array of potential spots for \(\mathrm{N}_2\) molecules to temporarily adhere during reactions.

Desorption

Desorption is the process where molecules detach from a solid surface and enter the gaseous phase. In this exercise, desorption occurs when \(\mathrm{N}_2\) molecules leave the surface of a catalyst.Initially, it's given that \(20\%\) of the surface sites are occupied by \(\mathrm{N}_2\) molecules. To determine how many \(\mathrm{N}_2\) molecules desorb, we first calculate the number of occupied surface sites:\[\text{Number of occupied sites} = 0.20 \times 6.023 \times 10^{17} = 1.2046 \times 10^{17} \]These sites are eventually freed as the \(\mathrm{N}_2\) molecules detach. Understanding desorption helps us calculate how many active sites are occupied by each molecule, highlighting the importance of surface interactions in catalysis.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that describes the state of a gas under a set of conditions. It combines several gas properties into one expression:\[\ PV = nRT \]where:- \(P\) is the pressure of the gas- \(V\) is the volume- \(n\) is the number of moles- \(R\) is the ideal gas constant \((0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})\)- \(T\) is the temperature in Kelvin.In this case, we use this equation to find the number of moles of \(\mathrm{N}_2\) desorbed. The given values are:\[P = 0.001 \, \text{atm}, \ V = 2.46 \times 10^{-3} \, \text{L}, \ R = 0.0821, \ T = 300 \, \text{K}\]Plugging these into the equation:\[n = \frac{PV}{RT} = \frac{0.001 \times 2.46 \times 10^{-3}}{0.0821 \times 300} = 1.0 \times 10^{-7} \, \text{mol}\]Calculating the number of moles allows us to determine the total number of \(\mathrm{N}_2\) molecules that desorbed. The understanding of the ideal gas law is pivotal for linking conditions such as pressure and volume with the quantity of gas, which is vital in chemical engineering and surface catalysis contexts.