The atmosphere slowly oxidizes hydrocarbons in a number of...

The atmosphere slowly oxidizes hydrocarbons in a number of steps that eventually convert the hydrocarbon into carbon dioxide and water. The overall reaction of a number of such steps for methane gas is $$ \begin{aligned} \mathrm{CH}_{4}(g)+5 \mathrm{O}_{2}(g)+5 \mathrm{NO}(g) & \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ +& 5 \mathrm{NO}_{2}(g)+2 \mathrm{OH}(g) \end{aligned} $$ Suppose that an atmospheric chemist combines $155 \mathrm{~mL}$ of methane at STP, 885 mL of oxygen at STP, and $55.5 \mathrm{~mL}$ of $\mathrm{NO}$ at STP in a $2.0 \mathrm{~L}$ flask. The flask is allowed to stand for several weeks at $275 \mathrm{~K}$. If the reaction reaches $90.0 \%$ of completion $(90.0 \%$ of the limiting reactant is consumed), what is the partial pressure of each of the reactants and products in the flask at $275 \mathrm{~K} ?$ What is the total pressure in the flask?

The atmosphere slowly oxidizes hydrocarbons in a number of steps that eventually convert the hydrocarbon into carbon dioxide and water. The overall reaction of a number of such steps for methane gas is
$$
\begin{aligned}
\mathrm{CH}_{4}(g)+5 \mathrm{O}_{2}(g)+5 \mathrm{NO}(g) & \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\
+& 5 \mathrm{NO}_{2}(g)+2 \mathrm{OH}(g)
\end{aligned}
$$
Suppose that an atmospheric chemist combines $155 \mathrm{~mL}$ of methane at STP, 885 mL of oxygen at STP, and $55.5 \mathrm{~mL}$ of $\mathrm{NO}$ at STP in a $2.0 \mathrm{~L}$ flask. The flask is allowed to stand for several weeks at $275 \mathrm{~K}$. If the reaction reaches $90.0 \%$ of completion $(90.0 \%$ of the limiting reactant is consumed), what is the partial pressure of each of the reactants and products in the flask at $275 \mathrm{~K} ?$ What is the total pressure in the flask?

Key Concepts

Reaction Yield and Percentage Completion

In practical scenarios, reactions often do not go to full completion. Percentage completion reflects the fraction of the limiting reactant that has reacted during the process. This concept is essential for adjusting stoichiometric calculations to account for incomplete reactions, thereby affecting the final amounts and pressures of reactants and products present in the system.

Partial Pressure and the Ideal Gas Law

For gases in a mixture, Dalton's Law of Partial Pressures states that the total pressure is the sum of the partial pressures of the individual gases. The ideal gas law (PV = nRT) relates the pressure, volume, temperature, and number of moles of a gas, allowing the calculation of partial pressures when the number of moles is known. This connection is critical when determining the behavior of the reaction mixture in a confined space, such as a flask.

Limiting Reactant Determination

In chemical reactions, the limiting reactant is the substance that is completely consumed first, dictating the extent to which the reaction can proceed. It is determined by comparing the mole ratios of the reactants as indicated by the balanced equation, ensuring that the reaction’s stoichiometric requirements are met. Once identified, this reactant’s consumption controls how many moles of products can form and the amounts of excess reactants remaining.

Stoichiometric Calculations

Stoichiometry involves the quantitative relationship between reactants and products in a balanced chemical equation. It allows one to convert between volumes (at STP) and moles, to determine how much of each species is consumed or produced. Using these mole relationships, one can calculate the changes in the number of moles of reactants and products as the reaction proceeds towards a given percentage of completion.

Transcript

00:01 This, unfortunately, is this an extremely long problem because of all the parts that it asks you to answer.

00:07 Not only is it a gas stoichiometry reaction, but it's also a limiting reactant reaction.

00:15 And when it's all completed, it wants you to calculate the total pressure inside of the entire flask, which requires you to calculate each individual pressure for all three reactants and all four products.

00:30 So let's get started with the information that we do know.

00:36 Here's the chemical reaction.

00:39 One mole methane reacts with 5 moles oxygen and 5 moles nitrogen monoxide to produce carbon dioxide, water, 5 moles nitrogen dioxide, and 2 moles oh.

00:54 We know that we initially combine 155 mil liters of methane, which is 0 .155 liters.

01:01 Because it's at stp, we can divide by 22 .4 liters to get the moles of methane.

01:06 Methane.

01:08 Once we have moles of methane, because it's carbon dioxide we're interested in, we'll calculate the moles of carbon dioxide that can be produced, recognizing that it is a one -to -one mole relationship, and we would get 6 .92 times 10 to negative 3 moles carbon dioxide if all the methane reacted.

01:28 We'll then do the same thing with the amount of the second product oxygen.

01:31 We have 885 mill liters or 0 .885.

01:41 I'm pretty sure i changed that point, yeah, 0 .885.

01:45 Because it's at stp, we'll divide by 22 .4 liters to get the moles of oxygen, and then recognize that the stoichiometry is 5 moles oxygen are required to just make 1 mole carbon dioxide.

01:58 So we would produce 7 .90 times 10 to negative 3 moles of carbon dioxide.

02:03 And the third reactant is nitrogen monoxide.

02:08 We have 55 .5 mill liters or 0 .555 liters of carbon monoxide, or nitrogen monoxide.

02:17 It's also at stp, so we'll divide by 22 .4 liters to get our moles.

02:22 Once we have moles of nitrogen monoxide, we recognize the 5 to 1 mole relationship.

02:29 So if all of the nitrogen monoxide were consumed, we would get 4 .96 times 10 to the negative 4 moles of carbon dioxide.

02:43 The least amount of carbon dioxide is produced from nitrogen monoxide, so that is our limiting reactant.

02:51 And if the reaction goes to completion, we would produce 4 .96 times 10 to the negative 4 moles of carbon dioxide.

02:58 However, it says the reaction only goes to 90 % completion.

03:07 So to calculate the pressure of carbon dioxide produced, we will only have 90 % of the maximum 4 .96 times 10 to negative 4 .9.

03:18 So now we'll go to the ideal gas law.

03:20 We'll take 90 % or 0 .9 of the 4 .9 times 10 to the negative, 4 .96 times 10 to negative 4 .9.

03:28 This will be the moles of carbon dioxide produced.

03:32 To calculate the pressure of carbon dioxide, we're using the ideal gas loss, so we'll take the moles multiplied by r, multiplied by the kelvin temperature, which is provided at 275 kelvin, divided by the volume.

03:48 The volume of the reaction is 2 liters, and we get 5 .03 times 10 to the negative 3 atmospheres of carbon dioxide that is produced when the reaction is produced when the reaction is 2 liters.

04:00 Reaction goes to completion.

04:03 Now that would be a long enough problem just by itself, but it wants us to do this for everything.

04:10 So what about water? we're also going to make some water at a 90 % completion.

04:18 So if we take 90 % of the amount of carbon dioxide that we're going to make, recognize the stoichiometry is one to one, we'll make the same number of moles of water, multiply that by r and t, and divide by the volume in liters, and we get the same amount of water as carbon dioxide as expected because of the one -to -one mole relationship.

04:43 Next, we'll look at nitrogen dioxide.

04:47 It turns out that we make five times the amount of nitrogen dioxide as carbon dioxide and water...

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