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Chapter 10: Problem 78

A sample of N2O effuses from a container in 42 seconds. How long will it take the same amount of gaseous I2 to effuse from the same container under identical conditions?

Short Answer

Expert verified
It will approximately take 100.8 seconds for the same amount of gaseous \( I_2 \) to effuse from the same container under identical conditions.

Step by step solution

01

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. Mathematically, it can be expressed as \( \dfrac{Rate_{1}}{Rate_{2}} = \sqrt{\dfrac{M_{2}}{M_{1}}} \), where \( Rate_{1} \) and \( Rate_{2} \) are the effusion rates of the two gases (with \( Rate \) being inversely proportional to the time of effusion), and \( M_{1} \) and \( M_{2} \) are the molar masses of the gases.
02

Calculate Molar Mass of N2O and I2

First, calculate the molar mass of dinitrogen monoxide (N2O) and molecular iodine (\( I_2\)). For N2O, it's \( 2 \times 14.01 + 1 \times 16.00 \) g/mol, which equals 44.02 g/mol. For I2, it's \( 2 \times 126.90 \) g/mol, which equals 253.8 g/mol.
03

Apply Graham's Law to Find the Time for I2 Effusion

Using Graham's Law, \( \dfrac{Time_{N2O}}{Time_{I2}} = \sqrt{\dfrac{M_{I2}}{M_{N2O}}} \). Plug in the known values: \( \dfrac{42 \: sec}{Time_{I2}} = \sqrt{\dfrac{253.8}{44.02}} \). Solve for \(Time_{I2}\) by cross-multiplying and squaring both sides to remove the square root. This will give you the time it takes for I2 to effuse under the same conditions.
04

Calculate the Time for I2 Effusion

To find \(Time_{I2}\), calculate \(\dfrac{42 \: sec}{Time_{I2}} = \dfrac{1}{\sqrt{\dfrac{253.8}{44.02}}} \) and then \(Time_{I2} = 42 \: sec \times \sqrt{\dfrac{253.8}{44.02}} \). Upon computation, \(Time_{I2} \) will be the time it takes for the same amount of gaseous I2 to effuse.
05

Perform the Calculation

Perform the final calculation by inserting the molar masses into the equation: \(Time_{I2} = 42 \: sec \times \sqrt{\dfrac{253.8}{44.02}} \approx 42 \: sec \times \sqrt{5.764} \approx 42 \: sec \times 2.40 \approx 100.8 \: sec\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Understanding the concept of molar mass is vital when working with Graham's Law as it relates directly to how fast a gas will effuse. Molar mass refers to the mass of one mole (approximately 6.022 x 1023 particles) of a substance and is generally expressed in grams per mole (g/mol).

Importance in Effusion

Molar mass influences the speed at which a gas can escape through a small hole, also known as effusion. Heavier gases with a higher molar mass will effuse more slowly than lighter gases with a lower molar mass because the heavier gas molecules have more mass and, consequently, less velocity under the same conditions.
Effusion Rate
The effusion rate of a gas describes how quickly it escapes from a container through a small hole into a vacuum. According to Graham's Law of Effusion, this rate is inversely proportional to the square root of the gas's molar mass. This principle makes it possible to compare the effusion rates of different gases under the same conditions.

Determining Effusion Rates

To determine effusion rates comparatively, we can use the relationship described in Graham's Law: \( \text{Effusion Rate (Gas 1)} / \text{Effusion Rate (Gas 2)} = \sqrt{\text{Molar Mass (Gas 2)} / \text{Molar Mass (Gas 1)}} \). Knowing the effusion rates allows chemists to predict how long it will take a gas to effuse and can be extremely useful in industrial processes and in the laboratory.
Gas Effusion
Gas effusion is the process by which gas molecules escape from a container with a pinhole or porous barrier into a lower pressure (often vacuum) environment. Effusion is particularly of interest in fields that encompass kinetic molecular theory and gaseous behavior.

Understanding Gas Effusion

In regards to Graham's Law, gas effusion is important as it is a practical application of the law. For example, when comparing the effusion of N2O and I2 as in the exercise solution, knowing their molar masses and the time one gas takes to effuse enables us to calculate the time it will take for the other gas to effuse, which is an essential skill in scientific experiments and industrial applications involving gases.

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