Problem 10 - A Complete Resource Book in Chemistry for JEE Main

Question

Which pair of the gases diffuses with the same rate at same temperature and pressure? (a) \(\mathrm{CO}\) and \(\mathrm{NO}\) (b) \(\mathrm{NO}_{2}\) and \(\mathrm{CO}_{2}\) (c) \(\mathrm{NH}_{3}\) and \(\mathrm{PH}_{3}\) (d) \(\mathrm{NO}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\)

Step-by-Step Solution

Answer

NO and C2H6 diffuse at the same rate.

Step 1: Understand the Problem

According to Graham's Law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass, which can be mathematically expressed as \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \). For two gases to diffuse at the same rate, their molar masses \(M_1\) and \(M_2\) must be equal.

Step 2: Calculate Molar Masses

Calculate the molar mass of each gas:- \( \text{CO} \) has a molar mass of \(12 + 16 = 28\) g/mol.- \( \text{NO} \) has a molar mass of \(14 + 16 = 30\) g/mol.- \( \text{NO}_2 \) has a molar mass of \(14 + 32 = 46\) g/mol.- \( \text{CO}_2 \) has a molar mass of \(12 + 32 = 44\) g/mol.- \( \text{NH}_3 \) has a molar mass of \(14 + 3 = 17\) g/mol.- \( \text{PH}_3 \) has a molar mass of \(31 + 3 = 34\) g/mol.- \( \text{C}_2\text{H}_6 \) has a molar mass of \(24 + 6 = 30\) g/mol.

Step 3: Compare Molar Masses

Compare the molar masses calculated in Step 2 to identify which pair has the same molar mass.- \( \text{CO} \) and \( \text{NO} \) molar masses are 28 and 30 g/mol respectively (not equal).- \( \text{NO}_2 \) and \( \text{CO}_2 \) molar masses are 46 and 44 g/mol respectively (not equal).- \( \text{NH}_3 \) and \( \text{PH}_3 \) molar masses are 17 and 34 g/mol respectively (not equal).- \( \text{NO} \) and \( \text{C}_2\text{H}_6 \) molar masses are both 30 g/mol.\( \text{NO} \) and \( \text{C}_2\text{H}_6 \) have equal molar masses, indicating they diffuse at the same rate.

Key Concepts

Molar Mass Calculation

Calculating molar mass is vital when working with gases. It assists in determining the rate at which a gas diffuses according to Graham's Law.
Molar mass is simply the mass of a given element in a compound represented in grams per mole (g/mol).
To calculate the molar mass, you need to sum the atomic masses of each atom in the molecule.

• For example, the molar mass of carbon monoxide (CO) is calculated by adding the atomic masses of carbon (12) and oxygen (16), giving a total of 28 g/mol.
• Similarly, the molar mass of nitrogen oxide (NO) is 14 (nitrogen) + 16 (oxygen) = 30 g/mol.

Using these calculations, different gas pairs can be compared to determine which would diffuse at similar rates under identical conditions of temperature and pressure.
This is crucial in many scientific applications and industrial processes.

Rate of Diffusion

Graham's Law of Diffusion discusses how different gases diffuse. The rate at which a gas diffuses is inversely proportional to the square root of the gas's molar mass.
This means lighter gases diffuse more rapidly than heavier ones. Graham's Law can be expressed mathematically:\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]where:

• \( r_1 \) and \( r_2 \) are the rates of diffusion for gases 1 and 2.
• \( M_1 \) and \( M_2 \) are the molar masses of the gases.

This formula helps calculate and predict how quickly a gas will diffuse under certain conditions.
When the molar masses of two gases are equal, they will diffuse at the same rate as was identified with \( NO \) and \( C_2H_6 \) in the exercise example.
Understanding diffusion rates is essential for research and applications in chemistry and environmental science.

Gas Comparison

When comparing gases, it's essential to consider their molar masses and how these affect their diffusion rates.
In the original exercise, we examined pairs of gases to determine which pair diffuses at the same rate.Key factors in comparing gases include:

• **Molar Mass Equality**: For two gases to diffuse at the same rate, they must have the same molar mass. For instance, in the example given, both \( NO \) and \( C_2H_6 \) have a molar mass of 30 g/mol.
• **Temperature and Pressure**: While the molar mass is a major factor in diffusion rates, temperature and pressure must remain constant to ensure accurate comparison results.
  These conditions help control the kinetic energy of the gases for proper diffusion comparison.

Gas comparison is especially valuable in fields like industrial chemistry, where selecting the right gas for a specific process can enhance efficiency and safety.