Problem 81 - General Chemistry: Principles and Modern Applications

Question

Nitrogen dioxide obtained as a cylinder gas is always a mixture of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) .\) A \(5.00 \mathrm{g}\) sample obtained from such a cylinder is sealed in a \(0.500 \mathrm{L}\) flask at \(298 \mathrm{K}\). What is the mole fraction of \(\mathrm{NO}_{2}\) in this mixture? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3}$$

Step-by-Step Solution

Answer

The mole fraction of \(NO_2\) in this mixture will be [(5 - 92.02PV/RT) / (46.01 - 92.02)]/(PV/RT), where P is the pressure of the mixture, V is the volume, R is the ideal gas constant and T is the temperature in Kelvin.

Step 1: Calculate the Total moles

Let's first calculate the total moles of the cylinder gas. We know that the total mass of the cylinder gas sample is 5.00 g. We know that \( NO_2 \) has a molar mass of 46.01 g/mol and \( N_2O_4 \) has a molar mass of 92.02 g/mol. We can state that x moles are \( NO_2 \) and the rest (n - x) are \( N_2O_4 \), where n is the total moles. Using the molar mass of each gas and the given mass, we can write the following equation: 5 = 46.01x + 92.02(n - x)

Step 2: Apply the Ideal Gas Law

We then apply the Ideal Gas Law to the problem, which states PV = nRT; where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the temperature in Kelvin. Here, Pressure is not given, we assume it to be 1 atm, V is 0.500 L and T is 298 K. R can be taken as 0.0821 L.atm/mol.K for our calculation purposes. We use T in Kelvin in all Ideal Gas Law related calculations. Solving the Ideal Gas Law for n, we find that n = PV/RT.

Step 3: Solve the equation from Step 1 for x

Substituting the equation for n from step 2 into our equation from step 1 and solve for x, we get x = (5 - 92.02PV/RT) / (46.01 - 92.02)

Step 4: Calculate the Mole fraction of \(NO_2\)

Molecule fraction of any component in a mixture is the ratio of the number of moles of that component to the total number of moles of all components in the mixture, therefore the mole fraction of \(NO_2\) will be x/n. Substitute x and n with the equations from previous steps and we get the mole fraction of \(NO_2\) = [(5 - 92.02PV/RT) / (46.01 - 92.02)] / (PV/RT).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that helps relate the properties of gases. It is given by the equation \(PV = nRT\), where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume of the gas,
• \(n\) represents the number of moles,
• \(R\) is the ideal gas constant, and
• \(T\) is the temperature in Kelvin.

This law is particularly useful when we know certain properties of a gas sample and want to find another, like the number of moles.
For example, given the volume, pressure, and temperature of a gas, we can rearrange the equation to solve for \(n\), yielding \(n = \frac{PV}{RT}\).
In practice, you often assume atmospheric pressure if not provided with alternative data, making the calculations straightforward.
Keep in mind that the Ideal Gas Law assumes the gas behaves ideally, which is almost entirely true under normal conditions of temperature and pressure.

Mole Fraction

The mole fraction is a way to express the composition of a mixture by defining the ratio of the number of moles of one particular component to the total number of moles in the mixture.
The formula is:

• Mole fraction \(X_A = \frac{n_A}{n_{total}}\), where \(n_A\) is the number of moles of the component and \(n_{total}\) is the total number of moles of all components.

This ratio is unitless, offering a simple way to describe the relative amounts of each substance in a mixture.
In our exercise, knowing the mole fraction of \(NO_2\) helps us understand the composition of the gaseous mixture from the cylinder.
To find this, we utilize the equations derived earlier from both the Ideal Gas Law and the mass composition. Calculating mole fractions in equilibrium systems is crucial for predicting how the system behaves under different conditions.

Equilibrium Constant

The equilibrium constant \(K_c\) quantifies the position of equilibrium in a chemical reaction at a certain temperature.
For a general reaction \(aA + bB \rightleftharpoons cC + dD\), \(K_c\) is expressed as:

• \(K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)

where \([X]\) indicates the concentration of the substance X at equilibrium.
A larger \(K_c\) implies a reaction that favors products, whereas a smaller \(K_c\) suggests a reaction favoring reactants.
In our problem, the equilibrium constant for \(N_2O_4 \rightleftharpoons 2NO_2\) governs how much of each substance is present at equilibrium. Knowing \(K_c\) helps predict how the system will respond to changes in conditions, such as temperature or pressure.
This value essentially tells us the extent to which a reaction will proceed and is vital for solving problems involving equilibrium in chemistry.