Question

Calculate relative rate of effusion of \(\mathrm{SO}_{2}\) to \(\mathrm{CH}_{4}\), if the mixture obtained by effusing out a : mixture with molar ratio \(\frac{n_{\mathrm{SO}_{2}}}{n_{\mathrm{CH}_{4}}}=\frac{8}{1}\) for three effusing steps: (a) \(2: 1\) (b) \(1: 4\) (c) \(1: 2\) (d) none of these

Short answer

None of the given effusing steps (a), (b), or (c) will result in the original mixture molar ratio of \( \frac{n_{\mathrm{SO}_{2}}}{n_{\mathrm{CH}_{4}}} = \frac{8}{1} \) after three steps of effusion.

Step-by-step solution

Understanding Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The relative rate of effusion between two gases can be given by the equation \( \frac{Rate_1}{Rate_2} = \sqrt{\frac{Molar \ Mass_2}{Molar \ Mass_1}} \), where Rate_1 and Rate_2 are the effusion rates of gas 1 and gas 2, respectively, and Molar Mass_1 and Molar Mass_2 are the molar masses of the gases.

Calculate Molar Masses of the Gases

The molar mass of \( \mathrm{SO}_{2} \) is \( 32.07 \times 1 + 16.00 \times 2 = 64.07 \ g/mol \). The molar mass of \( \mathrm{CH}_{4} \) is \( 12.01 \times 1 + 1.008 \times 4 = 16.04 \ g/mol \).

Calculate Relative Rate of Effusion Using Graham's Law

Applying Graham's Law, the relative rate of effusion of \( \mathrm{SO}_{2} \) to \( \mathrm{CH}_{4} \) can be calculated using the formula \( \frac{Rate_{\mathrm{SO}_{2}}}{Rate_{\mathrm{CH}_{4}}} = \sqrt{\frac{Molar \ Mass_{\mathrm{CH}_{4}}}{Molar \ Mass_{\mathrm{SO}_{2}}}} \). Plugging in the molar masses: \( \frac{Rate_{\mathrm{SO}_{2}}}{Rate_{\mathrm{CH}_{4}}} = \sqrt{\frac{16.04 \ g/mol}{64.07 \ g/mol}} \) which simplifies to \( \frac{Rate_{\mathrm{SO}_{2}}}{Rate_{\mathrm{CH}_{4}}} = \frac{1}{2} \) when simplified.

Apply the Mixture Molar Ratio to Determine the Effused Molar Ratios

The molar ratio of the mixture will affect the relative rates of effusion in the system. The original molar ratio is \( \frac{8}{1} \) in favor of \( \mathrm{SO}_{2} \). Each effusing step changes the ratio, eventually giving a new ratio after three steps. Since the given ratios for each step are not equivalent and do not accumulate to the original ratio, none of the options will result in the correct final mixture ratio.

Key concepts

Effusion Rate Calculation

The effusion rate of a gas reflects how fast the gas particles pass through a tiny orifice into a vacuum. A common method to calculate the effusion rate involves applying Graham's Law of Effusion, which outlines the relationship between the molar mass of a gas and its effusion rate. According to Graham's Law, the effusion rate is inversely proportional to the square root of its molar mass. Let's break down this principle with a straightforward example.

If we consider two gases, A and B, and their molar masses are M_A and M_B respectively, the law states that:\[\begin{equation}\frac{Rate_A}{Rate_B} = \sqrt{\frac{M_B}{M_A}}\end{equation}\]In the context of our exercise, we use this formula to find the relative rate of effusion between sulfur dioxide (\text{SO}_{2}) and methane (\text{CH}_{4}). After calculating their molar masses, we can determine how much faster or slower one gas effuses compared to the other. This concept is crucial for students preparing for competitive exams like JEE in Physical Chemistry, where understanding gas properties is essential.

Molar Mass

Molar mass serves as a bridge between the microscopic and macroscopic worlds. It is defined as the mass of one mole of a substance, and its common unit is grams per mole (g/mol). To calculate the molar mass of a compound like \text{SO}_{2} or \text{CH}_{4}, you need to sum up the molar masses of all atoms within the molecule.For \text{SO}_{2}, we calculate:\[\begin{equation}M_{\text{SO}_{2}} = (32.07 \times 1) + (16.00 \times 2) = 64.07 \ g/mol\end{equation}\]Similarly, for \text{CH}_{4}:\[\begin{equation}M_{\text{CH}_{4}} = (12.01 \times 1) + (1.008 \times 4) = 16.04 \ g/mol\end{equation}\]Understanding the concept of molar mass is vital for students, as it is used in stoichiometry to convert between mass and moles of a substance. Moreover, it's a crucial input when applying Graham's Law for effusion rate calculations.

Chemical Kinetics

Chemical kinetics is the branch of physical chemistry that studies reaction rates and the factors affecting them. While Graham's Law is not about chemical reactions, it's closely related to the principles of kinetics as it deals with the movement of molecules. Kinetics teaches us how various factors like temperature, pressure, concentration, and physical state of the reactants influence reaction rates.

Graham's Law provides insights into how molecular speeds (and hence effusion rates) are affected by the properties of the gases, such as their molar masses. In practical applications, understanding effusion is helpful in processes such as gas purification and isotope separation. For students tackling the JEE and other competitive examinations, grasping the fundamental concepts of kinetics offers a significant advantage, as kinetics questions often require integrating multiple chemistry concepts.

JEE Physical Chemistry

Physical Chemistry is an essential subject for the Joint Entrance Examination (JEE), which students need to clear for admission into engineering colleges in India. Within this subject, topics like gas laws, effusion, and the concept of molar mass play a pivotal role.

Competencies in these areas are critical because they form the basis of many more complex topics. An understanding of Graham's Law and chemical kinetics can lead to solving a wide array of physical chemistry problems. A solid foundation, along with practising problems related to effusion, will be beneficial for students aiming to ace the JEE. It should be noted that not only are these concepts important for entrance tests, but they also form the fundamental knowledge required for any student moving forward in the field of chemistry or engineering.