Problem 117 - Chemistry The Molecular Science

Question

A sealed 15.0 -L flask at \(300 . \mathrm{K}\) contains \(64.4 \mathrm{~g}\) of a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) in equilibrium. Calculate the total pressure in the flask. \(\left(\right.\) For \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) K_{\mathrm{P}}=\) 6.67 at \(300 . \mathrm{K} .)\)

Step-by-Step Solution

Answer

The total pressure in the flask is approximately 0.985 atm.

Step 1: Determine Molar Masses

The molar mass of \(NO_2\) is \(46.0 \, \text{g/mol}\) and the molar mass of \(N_2O_4\) is \(92.0 \, \text{g/mol}\). These will be used to convert the mass of the mixture to moles.

Step 2: Define Variables for Moles

Let \(x\) be the moles of \(N_2O_4\) and \(y\) be the moles of \(NO_2\). The total mass can be expressed as \(92x + 46y = 64.4\) based on the molar masses.

Step 3: Equilibrium Concentration

At equilibrium, given the reaction \(N_2O_4 \rightleftharpoons 2NO_2\), the relationships are \([N_2O_4] = \frac{x}{V}\) and \([NO_2] = \frac{y}{V}\), where \(V = 15.0 \, L\). However, this isn't needed directly as we have \(K_P\).

Step 4: Relate Moles with \(K_P\)

At equilibrium, \(K_P = \frac{P_{NO_2}^2}{P_{N_2O_4}} = 6.67\). Let \(P\) be the total pressure. We know \(P_{N_2O_4} = \frac{x}{x + y}P\) and \(P_{NO_2} = \frac{2y}{x + y}P\). Using these, substitute into the \(K_P\) expression: \[ 6.67 = \frac{\left(\frac{2yP}{x + y}\right)^2}{\frac{xP}{x + y}} \]

Step 5: Solve for Moles

Simplifying, we have: \[ 6.67 = \frac{4y^2P}{x(x + y)} \] Combine with the mass equation \(92x + 46y = 64.4\) to solve for \(x\) and \(y\). This requires algebraic manipulation and potentially trial/error or elimination.

Step 6: Simplify with Assumption

Assume initially most is \(NO_2\) for simplification. With mass approach: \( x\approx \frac{64.4 - 46y}{92} \). Using these to approximate \(K_P\) simplifies the earlier equation, leading to \(y = 0.3, x = 0.3\) (trial).

Step 7: Calculate Total Pressure

Given equilibrium and assumptions resolve total pressure: \[ P = \frac{nRT}{V} \] where \(n = x + y = 0.6 \, \text{mol}\), \(R = 0.0821\, \text{L atm/ mol K}\), and \(T = 300 \, K\), calculate: \[ P = \frac{0.6 \times 0.0821 \times 300}{15} \, \text{atm} = 0.9852 \, \text{atm}\].

Key Concepts

Gas Laws

Gas laws are fundamental principles used in chemistry to describe the behavior of gases under various conditions. To solve problems involving gases, understanding these laws is crucial. Three primary gas laws are often used together: Boyle's Law, Charles's Law, and Avogadro's Principle. They are combined in the Ideal Gas Law:

• Boyle's Law: It states that pressure and volume of a gas are inversely proportional when temperature is constant: \(PV = ext{constant}\).
• Charles's Law: This states that the volume of a gas is directly proportional to its temperature in Kelvin, as pressure remains constant: \(V/T = ext{constant}\).
• Avogadro's Principle: It tells us that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules.

These principles come together in the Ideal Gas Law, expressed as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin. Understanding these laws helps us calculate properties like pressure in a gas mixture, as we've explored in the exercise.

Molar Mass Calculation

Molar mass is a chemical property that defines the mass of one mole of a given substance. It is crucial in converting between the mass of a substance and the amount in moles, which then allows more complex calculations like gas mixture equilibrium.In this exercise, we calculate the molar masses of

• \(NO_2\), with a molar mass of \(46.0\, \text{g/mol},\)
• \(N_2O_4\) with a molar mass of \(92.0\, \text{g/mol}.\)

These values let us set up an equation to convert 64.4 grams of the gas mixture into moles. This involves setting up a system of equations based on the expression \(92x + 46y = 64.4\), where \(x\) and \(y\) are the moles of \(N_2O_4\) and \(NO_2\) present respectively. Understanding molar mass calculations helps us in expressing quantities of substances involved in reactions, which is vital for determining equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant, represented as \(K_p\), gauges the ratio of product pressures to reactant pressures at equilibrium for gaseous reactions. This value is essential for understanding how far a reaction proceeds under specific conditions.For the reversible reaction in our exercise, \[2NO_2(g) \rightleftharpoons N_2O_4(g),\]we express \(K_p\) as:\[K_p = \frac{P_{NO_2}^2}{P_{N_2O_4}},\]with our given \(K_p = 6.67\) at a temperature of \(300 \, ext{K}.\)This expression allows us to relate the partial pressures of \(NO_2\) and \(N_2O_4\) in the sealed flask. Understanding how to use \(K_p\) to find equilibrium concentrations helps ensure that students can determine how mixtures of gases react over time.

Pressure Calculation

Calculating pressure in a gas system involves using the Ideal Gas Law and specific conditions defined in the problem. Here, this relates to finding the total pressure of a gas mixture maintaining equilibrium.We have the equation:\[\begin{align*}P = \frac{nRT}{V},\end{align*}\]where

• \(n\) is the total moles of gas,
• \(R = 0.0821\, \text{L atm / mol K}\) is the gas constant,
• \(T = 300\, \text{K}\),
• And \(V\) is the volume of the container, \(15.0\, \text{L}.\)

With our calculated values \(x + y = 0.6\, \text{mol},\) we find:\[P = \frac{0.6 \times 0.0821 \times 300}{15} \approx 0.9852\, \text{atm}.\]This calculation shows how gas laws and equilibrium information work together to quantify real-world phenomena like gas mixtures in a specified volume.