Nitrogen dioxide dimerizes to give dinitrogen tetroxide: 2 NO2(g) →N2 O4(g) . At 298 K, 9.66 g o

Nitrogen dioxide dimerizes to give dinitrogen tetroxide: 2...

Nitrogen dioxide dimerizes to give dinitrogen tetroxide: 2 $\mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .$ At $298 \mathrm{K}, 9.66 \mathrm{g}$ of an $\mathrm{NO}_{2} / \mathrm{N}_{2} \mathrm{O}_{4}$ mixture exerts a pressure of 0.487 $\mathrm{atm}$ in a volume of 6.51 $\mathrm{L}$ . What are the mole fractions of the two gases in the mixture?

Nitrogen dioxide dimerizes to give dinitrogen tetroxide: 2 $\mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .$ At $298 \mathrm{K}, 9.66 \mathrm{g}$ of an $\mathrm{NO}_{2} / \mathrm{N}_{2} \mathrm{O}_{4}$ mixture exerts a pressure of 0.487 $\mathrm{atm}$ in a volume of 6.51 $\mathrm{L}$ .
What are the mole fractions of the two gases in the mixture?

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in which the rates of the forward and reverse reactions are equal, leading to constant concentrations of all species in the system. Understanding equilibrium is crucial for predicting the composition of reacting systems and determining the extent of a reaction.

Ideal Gas Law

The ideal gas law (PV = nRT) relates the pressure, volume, temperature, and number of moles in a gas. It is a fundamental tool in thermodynamics for computing the total number of moles in a gaseous mixture when pressure, volume, and temperature are known, making it essential for problems involving gas calculations.

Mole Fraction

Mole fraction is the ratio of the number of moles of a component to the total number of moles in a mixture. It provides a dimensionless way to express the composition of mixtures, which simplifies the analysis of how different species contribute to the overall properties of the system.

Dimerization

Dimerization is a process in which two identical molecules combine to form a dimer. This reaction is a specific type of association reaction and plays an important role in modifying the composition and physical properties of the system under study, particularly in equilibrium scenarios.

Transcript

00:03 All right, so in this question, we have two no2 yields, n204 is our reaction.

00:11 And we are absolutely going to take advantage of the fact that the n204 is just double of the n02.

00:19 There aren't any other stoichiometric fractions to worry about.

00:23 And the reason that's going to be super useful to us is because what we can do in this question to figure out what the mole fraction of each of these gases is.

00:33 Is we can compare how many moles we would expect to have if it was all no2 or all n204 might actually even be easier, versus how many moles of gas we actually have according to our ideal gas law.

00:53 So we know that any moles, if we assume that this is all n204, let's go ahead and start with that.

01:02 So if we have 9 .66 grams of this mixture, and these are in the same proportion, if it were all n204, n204 has a molar mass of n is 14, so 28, 0 is 16, so 64 plus 28, 64 plus 28 is 92 grams per mole.

01:35 Our 9 .66 and divide by 92 grams per mole, then what we get is 0 .105 moles if it's 100 % in 204.

01:58 That's just based on the mass.

02:01 So that means that once we solve for the number of moles we have according to the ideal gas law, assuming that we have more than 0 .105, which we should, that means that anything above and beyond the point 105 is the result of n204 splitting into 2n02s.

02:22 So let's go ahead and apply that ideal gas law now.

02:27 We have pv equals nrt.

02:30 We're trying to solve for n.

02:32 So pv over rt equals n.

02:38 And in this example, our pressure is 0 .487.

02:47 Our volume is 6 .51...

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