Problem 59 - Chemistry

Question

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ What is the partial pressure of NO in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of 0.80 and \(0.20 \mathrm{atm},\) respectively?

Step-by-Step Solution

Answer

In order to find the partial pressure of NO in equilibrium with N2 and O2, we need to set up a table representing the initial pressures, change in pressures, and equilibrium pressures for all species. Then, we can write the equilibrium constant expression (Kp) for the reaction and substitute the equilibrium pressures from the table. Solving for the change in pressure (x), we find x ≈ 0.036. The equilibrium partial pressure of NO is given by P_NO = 2x, which is approximately 0.072 atm.

Step 1: Write down the given information

We are given the following information: - Initial Pressure of N2 = 0.80 atm - Initial Pressure of O2 = 0.20 atm - The equilibrium constant, Kp = 0.050 - Temperature = 2200°C The balanced chemical equation is: \[ N_{2}(g) + O_{2}(g) \rightleftharpoons 2 NO(g) \]

Step 2: Set up a table to represent initial, change, and equilibrium pressures

Now, we will represent the initial pressures (I), the change in pressures (C), and the equilibrium pressures (E) using a table: | | N2 | O2 | 2NO | |---|----|----|----| | I | 0.80 atm | 0.20 atm | 0 atm | | C | -x | -x | +2x | | E | 0.80-x | 0.20-x | 2x | In this table, 'x' represents the change in pressure for each of the species as the reaction moves towards equilibrium.

Step 3: Write down the equilibrium constant expression (Kp)

Now, we'll write the equilibrium constant expression for the given reaction: \[ K_p = \frac{[NO]^2_{eq}}{[N_2]_{eq}[O_2]_{eq}} \] In terms of partial pressures, the expression becomes: \[ K_p = \frac{P_{NO}^2}{P_{N_2} P_{O_2}} \]

Step 4: Substitute equilibrium pressures from the table into the Kp expression

Now, we will substitute the equilibrium pressures from the table (row E) into the Kp expression: \[ 0.050 = \frac{(2x)^2}{(0.80-x)(0.20-x)} \]

Step 5: Solve for x

Next, we need to solve the equation for x: \[ 0.050 = \frac{4x^2}{(0.80-x)(0.20-x)} \] Multiplying both sides by (0.80-x)(0.20-x): \[ 0.050(0.80-x)(0.20-x) = 4x^2 \] Expanding and simplifying: \[ 0.16 - 0.20x - 0.80x + x^2 = 4x^2 \] Rearranging terms: \[ 3x^2 + x^2 - 1x - 0.16 = 0 \] Solving the quadratic equation: From the quadratic formula, we know that the solutions for x are \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] a = 4, b = 1, c = -0.16 is for our equation, and after solving for x, we obtain 2 potential solutions: - x1 ≈ 0.036 - x2 ≈ -1.103 Since we cannot have negative pressure, we will accept the positive value of x: x = 0.036

Step 6: Calculate the equilibrium partial pressure of NO

Finally, we can find the equilibrium partial pressure of NO by substituting the value of x back into the equilibrium pressure expression: \[ P_{NO} = 2x \] \[ P_{NO} = 2(0.036) \] \[ P_{NO} ≈ 0.072 \, atm \] So, the equilibrium partial pressure of NO is ≈ 0.072 atm.

Key Concepts

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \( K_p \), is a fundamental concept in chemistry that quantifies the ratio of the concentration or pressure of products to reactants at equilibrium. For gases, it is expressed in terms of partial pressures. For the reaction \( \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g) \), the equilibrium constant expression is:

• \( K_p = \frac{P_{NO}^2}{P_{N_2} \cdot P_{O_2}} \)

This equation shows how \( K_p \) relates the partial pressures of nitrogen, oxygen, and nitric oxide at equilibrium. Understanding \( K_p \) helps us predict how far a reaction proceeds before reaching equilibrium. A small \( K_p \) value, such as 0.050, indicates that at equilibrium, the concentration of the products (NO) is relatively low compared to the reactants (\( N_2 \) and \( O_2 \)). Thus, knowing \( K_p \) allows chemists to calculate the extent of chemical reactions under given conditions.

Partial Pressure

Partial pressure is an essential term when dealing with gas-phase reactions. It represents the pressure a particular gas exerts in a mixture, behaving as if it occupied the entire volume alone. Each component in a gaseous mixture has its partial pressure, and these pressures add up to the total pressure. In our reaction:

• \( N_2(g) + O_2(g) \rightleftharpoons 2 \mathrm{NO}(g) \)

we start with the initial partial pressures: \( 0.80 \text{ atm} \) for \( N_2 \) and \( 0.20 \text{ atm} \) for \( O_2 \). The partial pressure of NO at equilibrium can be calculated using the changes in pressure as shown in the step-by-step solution. By setting up an ICE (Initial, Change, Equilibrium) table, we calculate the shifts in each gas's pressure as the system reaches equilibrium. For NO, its equilibrium partial pressure is determined by the formula \( P_{NO} = 2x \), where \( x \) is the change in pressure during the reaction process.

Chemical Reactions

Chemical reactions involve the transformation of substances, breaking bonds in reactants and forming new bonds in products. In the reaction \( \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) \), reactant molecules collide and interact under high temperatures, such as \( 2200^{\circ} \text{C} \), to form products. This particular reaction is reversible, indicated by the equilibrium sign \( \rightleftharpoons \), meaning that it can proceed in both forward and backward directions until equilibrium is reached.
Reversible reactions are characterized by their ability to reach a state where the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. The study of these conditions is crucial because it allows chemists to manipulate conditions to favor the formation of products or reactants, depending on the desired outcome.