Problem 53 - Chemistry An Atoms First Approach

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 atm, respectively?

Step-by-Step Solution

Answer

The partial pressure of NO in equilibrium with N2 and O2, given their initial pressures of 0.80 atm and 0.20 atm, respectively, at a temperature of 2200°C and \(K_p = 0.050\), is approximately 0.028 atm.

Step 1: Identify the Balanced Equation and Equilibrium Constant

We are given the balanced equation: \[\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\] At 2200°C, the equilibrium constant \(K_p\) is given as 0.050.

Step 2: Define the Initial and Change in Pressures

The initial pressures of N2 and O2 are given as 0.80 atm and 0.20 atm, respectively. Initially, there is no NO present. Let's denote the change in pressure for each component as follows: - N2: -x (pressure decreases as it reacts) - O2: -x (pressure decreases as it reacts) - NO: +2x (pressure increases as it forms)

Step 3: Write Equilibrium Pressures

Using the expressions in Step 2, we can write the equilibrium pressures for each component: - N2: 0.80 - x - O2: 0.20 - x - NO: 2x

Step 4: Set Up the Equilibrium Constant Expression

Now, let's set up the equilibrium constant expression using the equilibrium constant value and the pressures for each component: \[K_p = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_2][\mathrm{O}_2]}\] Substitute the values of the equilibrium pressures: \[0.050 = \frac{(2x)^2}{(0.80 - x)(0.20 - x)}\]

Step 5: Solve for x

Solve the above equation for the value of x: \[0.050 = \frac{4x^2}{(0.80 - x)(0.20 - x)}\] It's a quadratic equation, and solving it using any preferred method (such as factoring, quadratic formula, or numerically) gives: x ≈ 0.014

Step 6: Determine the Equilibrium Pressure of NO

With the value of x, we can now find the equilibrium pressure for NO: Partial pressure of NO = 2x = 2 × 0.014 ≈ 0.028 atm Therefore, the partial pressure of NO in equilibrium with N2 and O2 is approximately 0.028 atm.

Key Concepts

Partial Pressure

Partial pressure is a concept used when dealing with gases that are present in a mixture. It helps us understand how each gas contributes to the total pressure exerted by the entire gaseous mix.
Partial pressure is defined as the pressure a gas would exert if it were alone in the container, at the same temperature as the mixture.

• Individual Contributions: Each gas in a mixture contributes its own portion to the total pressure. This individual pressure is the gas's partial pressure.
• Additive Nature: The sum of all individual partial pressures equals the total pressure of the gas mixture. This is known from Dalton's Law of Partial Pressures.

In our exercise, we are asked to determine the partial pressure of nitrogen monoxide (NO) when nitrogen (N2) and oxygen (O2) gases react in a container. Initially, these gases have partial pressures of 0.80 atm and 0.20 atm, respectively. As the reaction reaches equilibrium, the individual pressures change, leading to the formation of NO. Partial pressure calculation allows us to assess these changes precisely.

Chemical Equilibrium

Chemical equilibrium is an important concept when studying chemical reactions, especially reversible ones. It denotes the state where the reaction rates of the forward and reverse reactions are equal.
At this point, the concentrations of reactants and products remain constant over time.

• Dynamic Process: Though it seems static, equilibrium is dynamic, with continuous forward and reverse processes happening simultaneously.
• Influence of Temperature: Temperature changes can shift the balance, altering equilibrium concentrations.
• Equilibrium Constant (\( K_p \)): This constant is specific to a particular reaction at a specific temperature, representing the ratio of product pressures to reactant pressures at equilibrium.

For the reaction given in the exercise \( \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g) \), we use the equilibrium constant \( K_p \) which is given as 0.050 at 2200°C. This allows us to understand the relationship and calculate how much NO is formed when equilibrium is achieved.

Quadratic Equation

In chemical equilibrium problems, sometimes solving for unknowns leads us to quadratic equations. These equations have a particular format and require specific methods for their solutions.

• Form: A typical quadratic equation is in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.
• Solve Methods: Quadratics can be solved in various ways including factoring, using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), or numerically using tools like calculators.
• Applications in Chemistry: When creating expressions for equilibrium due to the change in pressures, the mathematical relationships end in a quadratic equation, needing solutions for further calculations.

In our solution, we form a quadratic equation to determine the change in pressure, \( x \), by substituting into its equilibrium expression:\[ 0.050 = \frac{4x^2}{(0.80 - x)(0.20 - x)} \]Solving gives \( x \approx 0.014 \), leading to the calculated partial pressure of NO. Understanding quadratic equations aids in accurately predicting these equilibrium situations in chemistry.