Question

Under identical conditions of pressure and temperature, \(4 \mathrm{~L}\) of gaseous mixture \(\left(\mathrm{H}_{2}\right.\) and \(\left.\mathrm{CH}_{4}\right)\) effuses through a hole in 5 min whereas \(4 \mathrm{~L}\) of a gas \(X\) of molecular mass 36 takes to \(10 \mathrm{~min}\) to effuse through the same hole. The mole ratio of \(\mathrm{H}_{2}: \mathrm{CH}_{4}\) in the mixture is : (a) \(1: 2\) (b) \(2: 1\) (c) \(2: 3\) (d) \(1: 1\)

Short answer

The mole ratio of H2 to CH4 in the mixture is 1:1.

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. This can be expressed as Rate of effusion ∝ 1/√Molar mass. For two different gases 'A' and 'B', the law can be applied as Rate(A)/Rate(B) = √(Molar Mass of B/Molar Mass of A).

Calculate the relative rate of effusion for the gas X

Given the molecular mass of gas X is 36 and it takes 10 min to effuse, compared to the mixture taking 5 min, the relative rate of effusion for gas X compared to the mixture is 5 min / 10 min = 1/2.

Apply Graham's Law to find the effective molecular mass of the mixture

Assuming the molar mass of gas X is 36 and using the relative rate of effusion (1/2), apply Graham's Law: Rate(mixture)/Rate(X) = √(Molar Mass of X/Molar Mass of mixture) => 1 / (1/2) = √(36/Molar Mass of mixture) => 2 = √(36/Molar Mass of mixture) => Molar Mass of mixture = 36 / 2² = 9.

Determine the molar mass and mole ratio of H2 and CH4 in the mixture

Let the mole fraction of H2 be x and that of CH4 be (1-x). The effective molar mass of the mixture will be the weighted average of their molar masses: Mixture Molar Mass = x(Molar Mass of H2) + (1-x)(Molar Mass of CH4). Using the molar masses of H2 (2 g/mol) and CH4 (16 g/mol): 9 = x(2) + (1-x)(16).

Solve for the mole fraction x

Solve the equation 9 = 2x + 16 - 16x to find the value of x: 9 = 16 - 14x => 14x = 16 - 9 => 14x = 7 => x = 7/14 => x = 1/2.

Calculate the mole ratio of H2:CH4

The mole ratio of H2 to CH4 will be x to (1-x): Mole ratio H2:CH4 = x:(1-x) = (1/2):(1-1/2) = 1/2:1/2 = 1:1. Thus, the correct mole ratio of H2 to CH4 in the mixture is 1:1.

Key concepts

Effusion Rate

The effusion rate of a gas is the measure of how quickly it escapes through a tiny hole into a vacuum. Consider it a race between gas particles to slide out of an opening; lighter and smaller particles tend to zip through faster than larger, heavier ones. This is essential in understanding Graham's Law, which states that the effusion rate of a gas is inversely proportional to the square root of its molar mass. This means that if you double the molar mass of a gas, its effusion rate decreases by a factor ofthe square root of two. Hence, lighter gases effuse more rapidly than heavier ones under the same conditions.

Using Graham's Law, you can compare the effusion rates of two different gases. Let's say we have Gas A and Gas B. If the molar mass of Gas A is less than that of Gas B, Gas A will effuse faster. This core idea helps in solving problems where you need to find unknown variables related to gases escaping through tiny holes by using the known effusion rates or times of different gases.

Molar Mass

The molar mass of a substance is the mass (usually in grams) of one mole of that substance. It is a critical parameter in Graham’s Law and various other chemistry problems, as it is directly linked to the molecular weight of the substance. For atoms or simple molecules, it corresponds to the sum of the atomic weights of all atoms in the molecule, as listed in the periodic table of elements. For example, the molar mass of water (H₂O) is approximately 18 g/mol, adding up the mass of 2 hydrogen atoms (about 1 g/mol each) and one oxygen atom (about 16 g/mol).

In the context of effusion, molar mass becomes a central factor in determining the effusion rate — a crucial part of Graham's Law. Understanding this relation allows one to make predictions about how quickly a gas will effuse, which is directly related to finding solutions such as mole ratio in gas mixtures.

Mole Ratio Calculation

Mole ratio calculations are essential for quantifying the proportions of different substances involved in chemical reactions. In the context of gases, mole ratios help describe the composition of gas mixtures, such as the example provided in the exercise with hydrogen (H₂) and methane (CH₄). Here, computing the mole ratio involves determining the relative amounts of gases in a mixture, which can be simplified to a mathematical fraction.

To find a mole ratio, you need the mole fractions of each component in the mixture, which can be deduced using reliable data. In the problem we're exploring, we use the effusion rates and molar masses of the gases in question to calculate the mole fraction, and thus the mole ratio. When you solve for these values, you gain a deeper understanding of the composition of the gas mixture — a key step in many scientific and engineering applications.

Gas Mixture

A gas mixture consists of two or more different gases occupying the same space. The behavior of gas mixtures is essential in fields like environmental science, chemical engineering, and physical chemistry. Each component in a gas mixture contributes to the overall properties of the mixture, such as pressure, volume, temperature, and effusion rate.

Understanding the behavior of gas mixtures can be quite complex because it involves multiple variables; however, certain laws, such as Dalton’s Law of Partial Pressures and Graham’s Law of Effusion, simplify these complexities. In the exercise at hand, we analyze a binary gas mixture containing H₂ and CH₄. By applying Graham’s Law, we predict how the mixture’s components behave during effusion. This insight is invaluable for tasks such as separating gases, designing chemical processes, and even for practical applications like breathing mixtures for diving.