Question

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

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

Step-by-step solution

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: \[n(\mathrm{H}_{2}) = n(\mathrm{O}_{2})\]

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:\[ n_{\mathrm{H}_{2\, remaining}} = n - 0.35n = 0.65n \]

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: \[ \frac{Rate_{\mathrm{H}_{2}}}{Rate_{\mathrm{O}_{2}}} = \sqrt{\frac{M_{\mathrm{O}_{2}}}{M_{\mathrm{H}_{2}}}} = \sqrt{\frac{32}{2}} = 4 \]

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: \[ n_{\mathrm{O}_{2 \ remaining}} = n - 0.0875n = 0.9125n \]

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: \[ \frac{n_{\mathrm{O}_{2 \ remaining}}}{n_{\mathrm{H}_{2 \ remaining}}} = \frac{0.9125n}{0.65n} = 1.4 \]

Key concepts

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.