The emission of NO2 by fossil fuel combustion can be...

The emission of $\mathrm{NO}_{2}$ by fossil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture. The urea reduces NO (which oxidizes in air to form $\mathrm{NO}_{2}$ ) according to the reaction: $$ \begin{aligned} 2 \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(g)+4 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightarrow 4 \mathrm{~N}_{2}(g) \\ &+2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ Suppose that the exhaust stream of an automobile has a flow rate of $2.55 \mathrm{~L} / \mathrm{s}$ at $655 \mathrm{~K}$ and contains a partial pressure of $\mathrm{NO}$ of $12.4$ torr. What total mass of urea is necessary to react completely with the NO formed during $8.0$ hours of driving?

The emission of $\mathrm{NO}_{2}$ by fossil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture. The urea reduces NO (which oxidizes in air to form $\mathrm{NO}_{2}$ ) according to the reaction:
$$
\begin{aligned}
2 \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(g)+4 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightarrow 4 \mathrm{~N}_{2}(g) \\
&+2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)
\end{aligned}
$$
Suppose that the exhaust stream of an automobile has a flow rate of $2.55 \mathrm{~L} / \mathrm{s}$ at $655 \mathrm{~K}$ and contains a partial pressure of $\mathrm{NO}$ of $12.4$ torr. What total mass of urea is necessary to react completely with the NO formed during $8.0$ hours of driving?

Key Concepts

Reaction Stoichiometry

Chemical reaction stoichiometry involves the quantitative relationships between reactants and products in a balanced chemical equation. It allows one to determine how many moles of a given reactant are required to completely react with another substance based on the molar ratios derived from the balanced equation.

Ideal Gas Law

The ideal gas law (PV = nRT) connects the pressure, volume, moles, and temperature of a gas. This fundamental relationship is used to calculate the number of moles in a given volume of gas under specific conditions, which is essential for determining reactant quantities in reactions involving gases.

Partial Pressure

Partial pressure refers to the pressure contributed by an individual gas in a mixture of gases. According to Dalton's law, the total pressure is the sum of the partial pressures of all individual gases, and understanding this concept is key to determining the mole fraction and the number of moles of a specific gas in a mixture.

Flow Rate and Time Integration

Flow rate combined with time integration is used to calculate the total volume of gas processed over a period. By knowing the rate of gas flow and the duration, one can determine the total amount of reactant involved, which is crucial for scaling chemical reactions and calculating reagent requirements.

Transcript

00:01 Hi guys, i want to stop this question.

00:05 The emission of nitrogen, no2 by fossil fuel combustion can be prevented by injecting urea into the combustion mixture.

00:21 If we have this conversion mixture, nitrogen oxide gas plus oxygen gas, so we need to get no2 gas.

00:38 We're going to get n -o -2 gas and then we try to avoid this try to avoid this n -o -2 so we don't want this n -o -2 so the best way to do is to use this urea co n -h2 gas plus n0 gas plus o2 gas.

01:23 So the product will not contain o2, the product will rather be n2 instead of an o2 plus co2 plus water.

01:43 So we can help us all that, all the products are gas.

01:55 This is for, so this is what the question is all about.

02:02 And the question says the flow risk of the exhaust steam the suppose that the exhaust stream of an automobile as a flow rate of 2 .55 later 2 .55 later per seconds as temperature of 655 carving and the partial pressure of no2 is this is this 12.

02:50 So we want to find the mass of urea, mass of c -o -n -a -2, that is urea.

03:08 That is all we're trying to find.

03:14 Mass of urea that is necessary to react completed with the endo form during eight hours.

03:24 The time duration is eight hours.

03:30 So i can convert eight hours to minutes.

03:36 You know i have eight hours one hour to 60 minutes and six minutes six one minute so if i multiply eight okay it gives me three six zero zero times eight that's good a big two eight seconds two eight eight eight zero zero second so the volume will be 2 .55 later per second the total volume during this time of esos during this time will be 2 .55 letter per second time 2880 second we have later that is 7 3 440 seconds seconds we divide second we have later that is 7 3 440 later so now we have the volume also to be 9 -180 later temperature to be 65550 and then the partial pressure of no2 the pressure of no equals 12 point 4 tall so 12 .4 tall i can confirm that one to at most by dividing by 760 0 .0 .0163 atmosphere.

06:20 Now i want to find the mole of no using gas of pv equals nrt because this is not the total we want to find if we have the partial pressure of no we can get the mole of no so i'm only getting the mole of no so nno will be partial pressure of no pv divided by rt so i get my n nl because my p my p is my piece then 2 to 1 my p is 0 .0 163 atmosphere my v we got v to be 73440 73440 73440 7 3 440 later r is 0 .0 .082 the atmosphere park heaven panel and then my temperature is 655 kervi.

08:04 Okay, i use my calculator.

08:07 .0 -163 multiplied by 73440 divided by 1084 divided by 1082...

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