Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 153

In preparation for a combustion demonstration, a professor fills a balloon with equal molar amounts of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2},\) but the demonstration has to be postponed until the next day. During the night, both gases leak through pores in the balloon. If \(35 \%\) of the \(\mathrm{H}_{2}\) leaks, what is the \(\mathrm{O}_{2} / \mathrm{H}_{2}\) ratio in the balloon the next day?

Short Answer

Expert verified
The \(\mathrm{O}_{2} / \mathrm{H}_{2}\) ratio in the balloon the next day is 1.4.

Step by step solution

01

Determine Initial Moles of Gases

Initially, the balloon has equal molar amounts of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\), let's assume the amount of each gas is \(n\). So we have: \[n(\mathrm{H}_{2}) = n(\mathrm{O}_{2})\]
02

Calculate the Amount of \(\mathrm{H}_{2}\) Leaked

Given that \(35\%\) of \(\mathrm{H}_{2}\) leaks out, we calculate the amount of \(\mathrm{H}_{2}\) left. The amount of \(\mathrm{H}_{2}\) that leaks = \(0.35n\), so the remaining \(\mathrm{H}_{2}\) is:\[ n_{\mathrm{H}_{2\, remaining}} = n - 0.35n = 0.65n \]
03

Apply Graham’s Law of Effusion

Graham's law tells us that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Molar mass of \(\mathrm{H}_{2}\) \[M_{\mathrm{H}_{2}} = 2\, \mathrm{g/mol} \] and \(\mathrm{O}_{2}\) \[M_{\mathrm{O}_{2}} = 32\, \mathrm{g/mol} \] Using Graham's Law: \[ \frac{Rate_{\mathrm{H}_{2}}}{Rate_{\mathrm{O}_{2}}} = \sqrt{\frac{M_{\mathrm{O}_{2}}}{M_{\mathrm{H}_{2}}}} = \sqrt{\frac{32}{2}} = 4 \]
04

Calculate the Amount of \(\mathrm{O}_{2}\) Leaked

Since \(\mathrm{H}_{2}\) leaks at four times the rate of \(\mathrm{O}_{2}\), in the same time period, \(\mathrm{O}_{2}\) would have leaked: \[ \text{Amount of } \mathrm{O}_{2} \text{ leaked} = \frac{0.35n}{4} = 0.0875n \] So the amount of \(\mathrm{O}_{2}\) remaining is: \[ n_{\mathrm{O}_{2 \ remaining}} = n - 0.0875n = 0.9125n \]
05

Calculate the New \(\mathrm{O}_{2} / \mathrm{H}_{2}\) Ratio

Finally, the \(\mathrm{O}_{2} / \mathrm{H}_{2}\) ratio in the balloon the next day is: \[ \frac{n_{\mathrm{O}_{2 \ remaining}}}{n_{\mathrm{H}_{2 \ remaining}}} = \frac{0.9125n}{0.65n} = 1.4 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

rate of effusion
One of the key principles in understanding this problem is the rate of effusion. Effusion is the process where gas molecules escape through tiny openings. Graham's Law of Effusion helps us compare the rates at which different gases effuse. It states that the rate of effusion (\text{Rate}) of a gas is inversely proportional to the square root of its molar mass (\text{M}). The formula is: \[\text{Rate} \propto \frac{1}{\sqrt{\text{M}}}\] This means a lighter gas will effuse faster than a heavier gas. In our problem, hydrogen (\text{H}_2) effuses faster than oxygen (\text{O}_2) because it has a lower molar mass. Specifically, hydrogen effuses four times faster than oxygen.
molar mass
Understanding molar mass is critical in this exercise. Molar mass is the mass of one mole of a substance and is usually given in grams per mole (\text{g/mol}). For hydrogen (\text{H}_2), the molar mass is 2 \text{g/mol}, while for oxygen (\text{O}_2), it is 32 \text{g/mol}. This property is essential in applying Graham's Law of Effusion, as it helps us determine the rate at which each gas will effuse. Since hydrogen is much lighter than oxygen, it escapes from the balloon faster. To calculate the effusion rates, we use: \[ \frac{\text{Rate}_{\text{H}_2}}{\text{Rate}_{\text{O}_2}} = \sqrt{ \frac{\text{M}_{\text{O}_2}}{\text{M}_{\text{H}_2}} } = \sqrt{ \frac{32}{2} } = 4 \]
gas leakage
Gas leakage from the balloon is determined by the principle of effusion. Overnight, both hydrogen and oxygen leak through the pores of the balloon, but at different rates. According to the problem, 35% of hydrogen leaks out.Using Graham's Law and knowing that hydrogen effuses four times faster than oxygen, we can calculate that only 8.75% of oxygen leaks. Here's how: \[0.0875n = \frac{0.35n}{4}\] The remaining amounts of gas are calculated next. For hydrogen, 65% remains: \[ n_{\text{H}_2\text{ remaining}} = n - 0.35n = 0.65n \] For oxygen, around 91.25% remains: \[ n_{\text{O}_2\text{ remaining}} = n - 0.0875n = 0.9125n \]
chemical ratios
Finally, understanding chemical ratios is essential to solve the problem. Initially, the balloon has an equal molar amount (\text{n}) of each gas. After calculating the amounts that remain post-leakage, we find: \[ n_{\text{H}_2\text{ remaining}} = 0.65n \] \[ n_{\text{O}_2\text{ remaining}} = 0.9125n \] To find the new oxygen-to-hydrogen ratio: \[ \text{Ratio}_{\text{O}_2/\text{H}_2} = \frac{0.9125n}{0.65n} = 1.4 \] This means, after the leakage, there is a 1.4 times greater molar amount of oxygen compared to hydrogen in the balloon.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Will the volume of a gas increase, decrease, or remain unchanged with each of the following sets of changes? (a) The pressure is decreased from 2 atm to 1 atm, while the temperature is decreased from \(200^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C} .\) (b) The pressure is increased from 1 atm to 3 atm, while the temperature is increased from \(100^{\circ} \mathrm{C}\) to \(300^{\circ} \mathrm{C}\). (c) The pressure is increased from 3 atm to 6 atm, while the temperature is increased from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\). (d) The pressure is increased from 0.2 atm to 0.4 atm, while the temperature is decreased from \(300^{\circ} \mathrm{C}\) to \(150^{\circ} \mathrm{C}\).

Standard conditions are based on relevant environmental conditions. If normal average surface temperature and pressure on Venus are \(730 . \mathrm{K}\) and \(90 \mathrm{~atm},\) respectively, what is the standard molar volume of an ideal gas on Venus?

The lunar surface reaches \(370 \mathrm{~K}\) at midday. The atmosphere consists of neon, argon, and helium at a total pressure of only \(2 \times 10^{-14} \mathrm{~atm} .\) Calculate the \(\mathrm{rm}\) speed of each component in the lunar atmosphere. [Use \(R=8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})\) and express \(\mathscr{A}\) in \(\mathrm{kg} / \mathrm{mol} .]\)

Naturally occurring uranium ore is \(0.7 \%\) by mass fissionable \({ }^{235} \mathrm{U}\) and \(99.3 \%\) by mass nonfissionable \({ }^{238} \mathrm{U}\). For its use as nuclear reactor fuel, the amount of \({ }^{235} \mathrm{U}\) must be increased relative to the amount of \({ }^{238} \mathrm{U}\). Uranium ore is treated with fluorine to yield a gaseous mixture of \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\) that is pumped through a series of chambers separated by porous barriers; the lighter \({ }^{235} \mathrm{UF}_{6}\) molecules \((\mathscr{A}=349.03 \mathrm{~g} / \mathrm{mol}\) ) effuse through each barrier faster than molecules of \({ }^{238} \mathrm{UF}_{6}(\mathscr{A}=352.04 \mathrm{~g} / \mathrm{mol}),\) until the final mixture obtained is \(3-5 \%\) by mass \({ }^{235} \mathrm{UF}_{6} .\) This process generated \(33 \%\) of the world's enriched uranium in 2008 but has now been replaced with a less expensive centrifuge process. Calculate the ratio of the effusion rates of \({ }^{235} \mathrm{UF}_{6}\) to \({ }^{238} \mathrm{UF}_{6}\).

A mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{Kr}\) weighs \(35.0 \mathrm{~g}\) and exerts a pressure of 0.708 atm in its container. since \(\mathrm{Kr}\) is expensive, you wish to recover it from the mixture. After the \(\mathrm{CO}_{2}\) is completely removed by absorption with \(\mathrm{NaOH}(s),\) the pressure in the container is 0.250 atm. How many grams of \(\mathrm{CO}_{2}\) were originally present? How many grams of Kr can you recover?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free