Question

If 0.0129 mol of N2O4 effuses through a pinhole in a certain amount of time, how much NO would effuse in that same amount of time under the same conditions?

Short answer

Answer: Approximately 0.0223 mol of NO gas effuse in the same amount of time under the same conditions as 0.0129 mol of N2O4 gas.

Step-by-step solution

Determine the molar masses of the gases

For this problem, we need to know the molar masses of N2O4 and NO. The molar mass of an element is given by the product of the number of atoms (in the molecule) and the atomic mass. Nitrogen (N) has an atomic mass of 14 g/mol, and oxygen (O) has an atomic mass of 16 g/mol. For N2O4: Molar mass = (2 * 14) + (4 * 16) = 92 g/mol For NO: Molar mass = (14) + (16) = 30 g/mol

Use 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. Thus: Rate of effusion of N2O4 / Rate of effusion of NO = sqrt(Molar mass of NO) / sqrt(Molar mass of N2O4) We know the number of moles of N2O4 that effuses and the molar masses of both gases. We can represent the "rate of effusion * time" for each gas as the number of moles that effuses (since effusion is a continuous process, more moles escaping means a higher rate). 0.0129 mol of N2O4 / x mol of NO = sqrt(30 g/mol) / sqrt(92 g/mol)

Solve for the number of moles of NO

Now we will solve the equation to find the number of moles of NO: x mol of NO = 0.0129 mol of N2O4 * (sqrt(30 g/mol) / sqrt(92 g/mol)) = 0.0223 mol So, under the same conditions, approximately 0.0223 mol of NO would effuse in the same amount of time as 0.0129 mol of N2O4.

Key concepts

Molar Mass

The term molar mass is a crucial concept in chemistry, particularly when we discuss the effusion of gases. It is defined as the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). You can think of the molar mass as the 'weight' of one mole of atoms or molecules.

For example, when we calculate the molar mass of N₂O₄, we multiply the number of atoms of each element by the atomic mass of that element, then sum up these values. This calculation provides us with a molar mass of 92 g/mol for N₂O₄. In contrast, the molecule NO has a molar mass of 30 g/mol. Having this information is critical for applying Graham's law of effusion, as we'll see in the subsequent sections.

Rate of Effusion

The rate of effusion refers to how fast a gas escapes through a small opening, like a pinhole. According to Graham's law, the rate is inversely proportional to the square root of the gas's molar mass. This means lighter gases (with a lower molar mass) will effuse faster than heavier gases.

Here's how you can picture this: Imagine you have two balloons, one filled with hydrogen and the other with oxygen. If both have a tiny pinhole, hydrogen will escape quicker because it's lighter than the oxygen. This concept is essential when comparing different gases, as in our example of N₂O₄ and NO.

Effusion of Gases

When talking about the effusion of gases, we're referring to the process by which gas molecules move through a tiny opening from an area of higher pressure to an area of lower pressure. This concept is not just theoretical; it has practical applications. For instance, it helps explain how we can smell perfume across a room or how a gas leak can be detected.

Understanding the effusion process gets easier when we recall that gas molecules are constantly moving and bumping into each other. When they find a way to escape, like through the pinhole we've been discussing, they take it! The rate at which they can escape depends on their size (or molar mass) and speed, directly linking effusion to Graham's law.

Solving for Moles

Finally, let's delve into the process of solving for moles concerning effusion. In the context of Graham's law, 'moles' underpin the concept—it's the unit of measurement that quantifies the amount of substance. When we talk about the rate of effusion, we can equate this to the amount (in moles) of gas escaping over a period of time.

Therefore, to find out how many moles of one gas would effuse under the same conditions as another, we can set up a proportion based on their molar masses and known moles that effused, as seen in the exercise. In our example, knowing the amount of N₂O₄ that effused and using the ratios derived from Graham's law allowed us to calculate the moles of NO.