Problem 83 - General Chemistry: Principles and Modern Applications

Question

What are the ratios of the diffusion rates for the pairs of gases (a) \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2} ;\) (b) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{D}_{2} \mathrm{O}\) \(\left(\mathrm{D}=\text { deuterium, i.e., }_{1}^{2} \mathrm{H}\right) ;\) (c) \(^{14} \mathrm{CO}_{2}\) and \(^{12} \mathrm{CO}_{2}\) (d) \(^{235} \mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6} ?\)

Step-by-Step Solution

Answer

The ratios of the diffusion rates are: (a) 1.069 for N2:O2, (b) 1.054 for H2O:D2O, (c) 1 for ^14C:^12C, and (d) 1 for ^235U:^238U.

Step 1: Identify molecular weights

Calculate the relative atomic masses for all the given gases. They are calculated as N2 (28 g/mol), O2 (32 g/mol), H2O (18 g/mol), D2O (20 g/mol), ^{14}CO2 (44 g/mol), ^{12}CO2 (44 g/mol), ^{235}UF6 (352 g/mol) and ^{238}UF6 (352 g/mol).

Step 2: Use Graham's Law

Apply Graham's law of diffusion, which states that the rate of diffusion \(r\) of a gas is inversely proportional to the square root of its molar mass \(M\). It is expressed as \( r_a / r_b = \sqrt{M_b / M_a} \) where \(r_a\) and \(r_b\) are the rates of diffusion, and \(M_a\) and \(M_b\) are the molar masses.

Step 3: Calculate diffusion rates

Substitute the values of molar masses in Graham's law equation and compute the rates of diffusion for each pair: (a) N2 and O2: \( r_{N2} / r_{O2} = \sqrt{32g/mol / 28g/mol} = 1.069 \) (b) H2O and D2O: \( r_{H2O} / r_{D2O} = \sqrt{20g/mol / 18g/mol} = 1.054 \) (c) ^{14}CO2 and ^{12}CO2: \( r_{^14CO2} / r_{^12CO2} = \sqrt{44g/mol / 44g/mol} = 1 \) (d) ^{235}UF6 and ^{238}UF6: \( r_{^235UF6} / r_{^238UF6} = \sqrt{352g/mol / 352g/mol} = 1 \)

Key Concepts

Molecular Weights

Molecular weight, also known as molecular mass, is a crucial concept in chemistry. It is the sum of the atomic weights of all atoms in a molecule. For example, when given a molecule like \(\mathrm{N}_2\), the molecular weight is calculated by adding the atomic weights of two nitrogen atoms. Each nitrogen atom has an atomic weight of approximately 14 units, so the molecular weight of \(\mathrm{N}_2\) is 28 g/mol.
Understanding molecular weights helps in comparing different molecules and understanding how heavy or light they are relative to each other. This information is essential when studying properties such as diffusion.

• Molecular weight of \(\mathrm{H}_2\mathrm{O}\): 18 g/mol as it includes two hydrogen atoms (approximately 1 g/mol each) and one oxygen atom (16 g/mol).
• Molecular weight of \(\mathrm{D}_2\mathrm{O}\): 20 g/mol; deuterium (\( {}^2\mathrm{H} \)) is heavier than regular hydrogen, contributing approximately 2 g/mol each.

By knowing the molecular weights of substances, we can then apply various principles to determine their behavioral properties, such as how quickly they diffuse.

Rate of Diffusion

The rate of diffusion is an important principle when considering how gases spread and mix. Graham's Law provides a way to understand and calculate this rate. The law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Using Graham's law, we can compare how fast different gases will diffuse. In the case of \(\mathrm{N}_2\) and \(\mathrm{O}_2\), by calculating their diffusion rates, we can see that \(\mathrm{N}_2\) diffuses slightly faster as it has a lower molecular weight.

• Formula: \( \frac{r_a}{r_b} = \sqrt{\frac{M_b}{M_a}} \) allows us to calculate the relative rates where \(r\) represents rate of diffusion and \(M\) represents molar mass.
• Example: For \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{D}_2\mathrm{O}\), by using this formula, we discovered that \(\mathrm{H}_2\mathrm{O}\) diffuses faster than \(\mathrm{D}_2\mathrm{O}\) due to its lighter molar mass.

Therefore, understanding diffusion rates helps predict how gases will behave in different environments, whether they are confined spaces or open atmospheres.

Molar Mass

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance. Measured in grams per mole (g/mol), molar mass is derived from the atomic weights listed on the periodic table and provides a bridge between mass and number of particles.
When solving problems related to diffusion, knowing the molar mass of gases is essential. Molar mass plays a crucial role in determining how quickly a gas will dissipate in a given space. For instance, \(^{12}\mathrm{CO}_2\) and \(^{14}\mathrm{CO}_2\) have identical molar masses, leading to equivalent diffusion rates under similar conditions.

• Consistency: Molar mass remains constant for any given substance under standard conditions. This consistency is vital for accurate predictions and calculations in chemical reactions and processes.
• Application in Graham's Law: It's used to compare and calculate rates of diffusion, allowing us to observe how different gases perform.

This understanding connects the macroscopic properties of substances to their microscopic identities.