Question

Air (21 percent O \(_{2}, 79\) percent \(\mathrm{N}_{2}\) ) is heated to \(5400 \mathrm{R}\) at a pressure of 1 atm. Determine the equilibrium composition, assuming that only \(\mathrm{O}_{2}, \mathrm{N}_{2}, \mathrm{O},\) and NO are present. Is it realistic to assume that no N will be present in the final equilibrium mixture?

Short answer

Question: Determine the equilibrium composition of an air mixture containing 21% O₂ and 79% N₂ heated to 5400 R at a pressure of 1 atm, assuming that only O₂, N₂, O, and NO are present. Answer: The equilibrium composition can be found by following the outlined steps, which involve determining the initial moles, writing equilibrium constant expressions, setting up a system of equations to solve for equilibrium mole fractions, and checking if it's realistic to assume no N at equilibrium. Upon solving the system of equations for x and y representing the moles of O and NO formed, equilibrium mole fractions can be calculated for all species (O, N₂, O₂, and NO) in the mixture.

Step-by-step solution

Determine the Initial Moles of Each Species

We can assume that initially, we have 1 mole of air in the system containing 0.21 moles of O\(_2\) and 0.79 moles of N\(_2\). There are no moles of O and NO initially. Initial moles: - O\(_2\): 0.21 - N\(_2\): 0.79 - O: 0 - NO: 0

Write Equilibrium Constant Expressions for Possible Reactions

We will consider two reactions: 1. O\(_2\) dissociates into 2O (Reaction 1) 2. N\(_2\) reacts with O\(_2\) to form 2NO (Reaction 2) The equilibrium constant expressions for these reactions are as follows: For Reaction 1: \(K_1 = \frac{[O]^2}{[O_2]}\) For Reaction 2: \(K_2 = \frac{[NO]^2}{[N_2][O_2]}\)

Set Up System of Equations to Solve for Equilibrium Mole Fractions

Let x and y represent the moles of O and NO formed at equilibrium, respectively. Then, the moles of O\(_2\) and N\(_2\) at equilibrium will be (0.21 - x) and (0.79 - y), respectively. The mole fractions of the species at equilibrium can be written as: - \(X_{O} = \frac{x}{1+x+y}\) - \(X_{NO} = \frac{y}{1+x+y}\) - \(X_{O_2} = \frac{0.21-x}{1+x+y}\) - \(X_{N_2} = \frac{0.79-y}{1+x+y}\) Now, substitute these expressions into the equilibrium constant expressions for K_1 and K_2, and solve the system of equations for x and y: \(K_1 = \frac{x^2}{0.21-x}\) and \(K_2 = \frac{y^2}{(0.79-y)(0.21-x)}\)

Solve the System of Equations for x and y

To solve the system of equations, we can first look up the values of K_1 and K_2 at 5400 R. Then, use numerical methods or software to find the values of x and y that satisfy both equations.

Determine the Equilibrium Composition

Use the values of x and y obtained in Step 4 to calculate the equilibrium mole fractions of each species: - \(X_{O} = \frac{x}{1+x+y}\) - \(X_{NO} = \frac{y}{1+x+y}\) - \(X_{O_2} = \frac{0.21-x}{1+x+y}\) - \(X_{N_2} = \frac{0.79-y}{1+x+y}\)

Check If It's Realistic to Assume No N at Equilibrium

To check if it's realistic to assume that no N is present in the final equilibrium mixture, we can look at the equilibrium mole fractions for N. If the mole fraction for N is negligible compared to other species, it can be considered realistic to assume that no N is present at equilibrium.