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 at which the diffusion rate of \( \mathrm{N}_2 \) is \( 1.625 \) times that of \( \mathrm{SO}_2 \) is \( 373 \; \text{K} \).

Step 1: Understand Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \) where \( r_1 \) and \( r_2 \) are the rates of effusion or diffusion of the gases, and \( M_1 \) and \( M_2 \) are their molar masses respectively.

Step 2: Identify Given Values and Relation

We are given that the rate of diffusion of \( \mathrm{N}_2 \) is \( 1.625 \) times the rate of effusion of \( \mathrm{SO}_2 \). Thus, \( \frac{r_{\mathrm{N}_2}}{r_{\mathrm{SO}_2}} = 1.625 \). We also know that the rate of \( \mathrm{SO}_2 \) at \( 50^{\circ}C \) is required.

Step 3: Use Molar Mass in Graham's Law

The molar mass of \( \mathrm{N}_2 \) is \( 28 \; \text{g/mol} \) and for \( \mathrm{SO}_2 \) it is \( 64 \; \text{g/mol} \). Using Graham's Law: \( \frac{r_{\mathrm{N}_2}}{r_{\mathrm{SO}_2}} = \sqrt{\frac{64}{28}} \). Simplifying gives \( \sqrt{2.29} \approx 1.514 \). This relates the rates at standard temperature equations.

Step 4: Apply Temperature Correction

Since \( r_{\mathrm{N}_2} \) is \( 1.625 \) times that of \( r_{\mathrm{SO}_2} \), establish the following relation considering temperature: \( \frac{1.625}{1.514} = \sqrt{\frac{T_{\mathrm{N}_2}}{323}} \). Solving, \( 1.0735 = \sqrt{\frac{T_{\mathrm{N}_2}}{323}} \). Squaring both sides gives \( 1.1524 = \frac{T_{\mathrm{N}_2}}{323} \).

Step 5: Solve for Temperature

Rearrange the equation to solve for \( T_{\mathrm{N}_2} \): \( T_{\mathrm{N}_2} = 1.1524 \times 323 = 372.8 \), which rounds to \( 373 \; \text{K} \).

Key Concepts

Rate of Diffusion

In the world of gases, diffusion refers to how gas particles spread from an area of high concentration to an area of low concentration. This spread continues until an equilibrium is reached. Graham's Law provides a way to compare the diffusion rates of two different gases. It's important to know that the diffusion rate of a gas is influenced by its molar mass. In simpler terms:

• Gas particles with lighter molar mass will diffuse faster.
• Gas particles with heavier molar mass will diffuse more slowly.

Understanding this concept helps you predict and calculate the behavior of gases under different conditions. For instance, knowing that nitrogen gas (\(\mathrm{N}_2\)) diffuses faster than sulfur dioxide (\(\mathrm{SO}_2\)) because of their different molar masses, allows us to use Graham's Law to calculate their diffusion rates relative to each other.In our exercise, we saw that the rate of diffusion of \(\mathrm{N}_2\) was given as 1.625 times the rate of effusion of \(\mathrm{SO}_2\). Using Graham's Law, the rates can be expressed in terms of their molar masses, allowing students to successfully set up equations and find out at what temperature conditions the relationship holds true.

Temperature Correction in Gas Laws

Temperature plays a pivotal role in the behavior of gases. As gases warm up, their particles gain more energy, move faster, and consequently, the rate of their diffusion or effusion increases. Graham's Law shows this relationship in terms of diffusion rate changes with temperature.When applying Graham's Law, it is necessary to consider the effect of temperature correctly. The diffusion rate of gas is not only dependent on the molar mass but also on the temperature at which the diffusion takes place. So, given a scenario like in our exercise, where you have to calculate the temperature at which the rate of diffusion is a certain multiple of another, we adjust for temperature using this structured formula:

• \(\frac{r_1}{r_2} = \sqrt{\frac{T_1}{T_2}}\)

This becomes particularly useful in real-world applications such as calculating the temperatures at which specific industrial gas processes work optimally or ensuring correct gas ratios in experiments where temperature changes are involved.In the textbook exercise example, this temperature correction was necessarily applied to solve for the unknown temperature based on the conditions provided. It involved solving for the unknown temperature under the assumption that other factors remain constant.

Molar Mass Calculation

Molar mass is a central piece in the puzzle of understanding gas behavior. It is defined as the mass of one mole of a given substance, often using g/mol as the unit. Why does this matter for gas diffusion? Molar mass directly influences how quickly gas particles can move or spread, as dictated by Graham's Law.Calculating the molar mass of common gases like nitrogen (\(\mathrm{N}_2\)) or sulfur dioxide (\(\mathrm{SO}_2\)) is the starting point in understanding their rates of diffusion. For example:

• Nitrogen has a molar mass of 28 g/mol, as each nitrogen atom weighs about 14, and there are two in a molecule of \(\mathrm{N}_2\).
• Sulfur dioxide's molar mass is 64 g/mol, comprised of one sulfur (32 g/mol) and two oxygen atoms (16 g/mol each).

Once these molar masses are known, they can be placed into Graham's formula to compare rates of either effusion or diffusion and to derive further related insights in solving for other variables like temperature.In any chemical process or computation where gas interaction is of concern, such as in the exercise given, knowing the molar mass is fundamental. It allows accurate predictions and assessments of the behavior and performance of gases under specific conditions.