Ethane burns in air to give H2 O and CO2 2 C2 H6(g)+7 O2(g) →4 CO2(g)+6 H2 O(g) (a) Four gases a

step 1 This is question 68 from chapter 10. And we have FAA burning to give water vapor and carbon diox

Ethane burns in air to give H2 O and CO2 2 C2 H6(g)+7 O2(g)...

Ethane burns in air to give $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{CO}_{2}$ $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.) (b) $\mathrm{A} 3.26-\mathrm{L}$ flask contains $\mathrm{C}_{2} \mathrm{H}_{6}$ at a pressure of $256 \mathrm{mm}$ Hg and a temperature of $25^{\circ} \mathrm{C}$ Suppose $\mathrm{O}_{2}$ gas is added to the flask until $\mathrm{C}_{2} \mathrm{H}_{6}$ and $\mathrm{O}_{2}$ are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of $\mathbf{O}_{2}$ and what is the total pressure in the flask?

Ethane burns in air to give $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{CO}_{2}$
$$
2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})
$$
(a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.)
(b) $\mathrm{A} 3.26-\mathrm{L}$ flask contains $\mathrm{C}_{2} \mathrm{H}_{6}$ at a pressure of $256 \mathrm{mm}$ Hg and a temperature of $25^{\circ} \mathrm{C}$ Suppose $\mathrm{O}_{2}$ gas is added to the flask until $\mathrm{C}_{2} \mathrm{H}_{6}$ and $\mathrm{O}_{2}$ are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of $\mathbf{O}_{2}$ and what is the total pressure in the flask?

Key Concepts

Ideal Gas Law and Dalton’s Law of Partial Pressures

The ideal gas law (PV = nRT) connects pressure, volume, temperature, and moles of gas, establishing that at constant temperature and volume, pressure is proportional to the number of moles. Dalton’s Law states that in a mixture of gases, the total pressure is the sum of the individual partial pressures, which is crucial when calculating the pressures of individual gases in a reaction mixture.

Stoichiometry in Gas-Phase Reactions

Stoichiometry involves using balanced chemical equations to relate the amounts (in moles) of reactants and products during a reaction. In gas-phase reactions, these mole ratios directly influence the relationships between the partial pressures of the gases when the reaction occurs at the same temperature and volume.

Relationship Between Molecular Mass and Molecular Speed

The kinetic energy of gas molecules is the same for all gases at the same temperature. Since kinetic energy is proportional to the square of the speed and inversely proportional to molecular mass, lighter molecules, having smaller masses, must move faster on average to have equivalent kinetic energies.

Kinetic Molecular Theory and Root-Mean-Square Speed

This concept explains how, at a given temperature, the average speed of gas molecules (measured as the root-mean-square speed) depends inversely on the square root of their molar mass. Thus, lighter molecules move faster than heavier ones, which is key when comparing the speeds of different gases.

Transcript

00:01 This is question 68 from chapter 10.

00:05 And we have faa burning to give water vapor and carbon dioxide.

00:12 So there's two parts to the question.

00:15 And a is rank all the gases in order of increasing rms speed.

00:29 So rms speed is a root mean square speed.

00:33 So the equation is rms.

00:36 Speed is equal square root of 3rt over the molar mass.

00:50 But for this, we can just compare it by looking at the molar mass, because a lower molar mass means it has a faster rms speed.

00:58 So water vapor has a molar mass of 18 grams per mole, the carbon dioxide that's 44 grams per mole.

01:17 Methane is 30 grams per mole and the 02 is 32 grams per mole so the largest one's going to have the smallest rm s speed so in that case that is the co2 followed by o2 then methane then water so water is going to have the biggest rms speed due to having the lowest smaller mass okay so b is asking or we're going to be adding o2 until the correct stoichiometric ratio.

02:05 So we're trying to say, what point is the partial pressure of oxygen and what is the total pressure? so we'll do some versions first.

02:16 We have 256 millimeters of mercury.

02:19 We're divided by 760 millimeters of mercury because we're going to use atmospheric pressure.

02:24 And what atmospheric pressure equals 760.

02:29 And this comes out to be 0 .336 and the next conversion will be temperature we have 25 degrees celsius to convert the kelvin add to 73 and this comes up to be 298 kelvin so now we have that we can solve for moles and we're going to be using the ideal gas law is pv is equal to nr2 okay, so moles n is equal to pv over r t.

03:24 So we can plug in the values for these.

03:31 So pressure, volume of 3 .26 liters with a gas constant, look at the temperature, and this comes out to be 0 .0448 moles of 0 .02...

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