Problem 73 - A Complete Resource Book in Chemistry for JEE Main

Question

At what temperature, the rate of diffusion of \(\mathrm{N}_{2}\) would be \(1.625\) times the rate of effusion of \(\mathrm{SO}_{2}\) at \(50^{\circ} \mathrm{C}\) ? (a) \(110 \mathrm{~K}\) (b) \(173 \mathrm{~K}\) (c) \(373 \mathrm{~K}\) (d) \(273 \mathrm{~K}\)

Step-by-Step Solution

Answer

The temperature is approximately \(373 \mathrm{~K}\) (option c).

Step 1: Understand the Problem

We need to find the temperature at which the rate of diffusion of \(N_2\) is 1.625 times the rate of effusion of \(SO_2\) at \(50^{\circ} \mathrm{C}\). According to Graham's law, the rate of effusion/diffusion is inversely proportional to the square root of molar mass of the gas.

Step 2: Apply Graham's Law

According to Graham’s law, \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\), where \(r\) is the rate of diffusion/effusion and \(M\) is the molar mass. Let \(r_{N_2}\) be the rate of diffusion of \(N_2\) and \(r_{SO_2}\) be the rate of effusion of \(SO_2\). Thus, \(\frac{r_{N_2}}{r_{SO_2}} = \frac{1.625}{1}\).

Step 3: Write the Equation Using Given Information

Graham’s law modified for temperature is \(\frac{r_{N_2}}{r_{SO_2}} = \frac{1.625}{1} = \sqrt{\frac{M_{SO_2} \cdot T_{N_2}}{M_{N_2} \cdot T_{SO_2}}}\). Here, \(M_{SO_2} = 64\) and \(M_{N_2} = 28\), with \(T_{SO_2} = 323\, \mathrm{K}\). We need to find \(T_{N_2}\).

Step 4: Rearrange the Equation

Rearrange the equation to find \(T_{N_2}\): \(1.625^2 = \frac{M_{SO_2} \cdot T_{N_2}}{M_{N_2} \cdot T_{SO_2}}\). This simplifies to \(T_{N_2} = \frac{1.625^2 \cdot M_{N_2} \cdot T_{SO_2}}{M_{SO_2}}\).

Step 5: Calculate

Substitute the values into the equation: \(T_{N_2} = \frac{1.625^2 \cdot 28 \cdot 323}{64}\). Calculate step by step: \(1.625^2 = 2.640625\), then \(2.640625 \cdot 28 \cdot 323 = 23883.7125\) and finally divide by \(64\) to get \(T_{N_2} \approx 373.2\).

Step 6: Select Closest Answer

The calculated temperature \(T_{N_2} \approx 373.2 \mathrm{K}\) fits option (c) \(373 \mathrm{~K}\).

Key Concepts

Rate of Diffusion

Diffusion is a fascinating phenomenon where gas particles spread from an area of high concentration to an area of low concentration. This process happens because gas molecules are moving randomly in all directions. The rate at which this spreading or mixing happens is called the rate of diffusion.
It's important to understand that lighter gas molecules move faster, which means they diffuse more quickly compared to heavier gas molecules. Here, the rate of diffusion is affected by the molar mass (or molecular weight) of the gas. According to Graham's Law, the rate of diffusion is inversely proportional to the square root of the molar mass.

• For example, a lighter gas like hydrogen will diffuse faster than a heavier gas like carbon dioxide.
• This means if you have a gas like \( ext{N}_2\) (lighter) and another like \( ext{SO}_2\) (heavier), the \( ext{N}_2\) diffuses faster.

In the context of our problem, knowing this principle helped us to solve for the temperature where the diffusion rate of nitrogen was 1.625 times the rate of \( ext{SO}_2\). This comparison makes understanding diffusion essential when considering different gases' behaviors.

Rate of Effusion

Effusion is closely related to diffusion but slightly different. It's the process where gas molecules escape through a tiny opening into a vacuum or an area with lower pressure. The rate of effusion, like diffusion, is also governed by the speed of gas molecules.
According to Graham's Law,%20the%20rate%20of%20effusion%20is%20aff%Z%209/Z%20sically%20similar%20to%20the%20rate%20of%20diffusion,%20given%20by%20the%20equation%20\(\frac{r_1}{r_2}%20=%20\sqrt{\frac{M_2}{M_1}}\).
Consider these points to understand it better:

• The lighter the molecule, the faster it effuses through the pinhole.
• Like diffusion, if you compare \( ext{N}_2\) and \( ext{SO}_2\), the lighter \( ext{N}_2\) will effuse faster.

In our original problem, Graham's Law helped establish a relationship between the rate of effusion and molar mass, enabling us to compute the desired temperature condition by converting rates of diffusion and effusion.
This demonstrates why understanding effusion is crucial when dealing with gases in real-world scenarios.

Molar Mass Calculation

Calculating molar mass is a key concept in chemistry, especially when you're dealing with gases and applying principles like Graham's Law. Molar mass is essentially the weight of one mole of a substance, usually expressed in grams per mole. In problems involving diffusion and effusion, having accurate molar mass values allows us to predict gas behavior.
Here's how you can find and use molar mass:

• Look at the periodic table to find the atomic weights of elements in a compound.
• Add the atomic weights together to get the compound's molar mass.

For instance, in our problem:

• Molar mass of \( ext{N}_2\) = 28 (since N is about 14 and there are two Ns).
• Molar mass of \( ext{SO}_2\) = 64 (S is about 32 and O is about 16, and there are two Os).

Understanding the calculation of molar masses enabled us to effectively use Graham's Law. It helped in setting up the equation for finding the temperature, where these properties influenced the rate of diffusion and effusion.
So, having an ability to compute molar mass not only aids in such calculations but also gives better insight into chemical substances and their behaviors.