Calculating Partial Pressure Of NO At Equilibrium Kp = 1.0
Introduction
Understanding chemical equilibrium is fundamental in chemistry, especially when dealing with gaseous reactions. Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the concentrations of reactants and products. In gaseous reactions, we often use partial pressures to describe the amounts of reactants and products. The equilibrium constant, denoted as Kp, is a crucial parameter that relates the partial pressures of the gases at equilibrium. This article delves into how to determine the partial pressure of nitric oxide (NO) at equilibrium given the partial pressures of nitrogen (N₂) and oxygen (O₂) and the equilibrium constant Kp for the reaction: N₂ + O₂ ⇌ 2NO.
The concepts of partial pressure and equilibrium constants are essential for understanding this problem. Partial pressure refers to the pressure exerted by an individual gas in a mixture of gases. According to Dalton's Law of Partial Pressures, the total pressure of a gas mixture is the sum of the partial pressures of each component gas. The equilibrium constant Kp is defined in terms of partial pressures and provides a quantitative measure of the extent to which a reaction proceeds to completion. For the given reaction, the expression for Kp is:
Kp = (P(NO)²) / (P(N₂) * P(O₂))
This expression indicates that Kp is the ratio of the square of the partial pressure of NO to the product of the partial pressures of N₂ and O₂ at equilibrium. Knowing the value of Kp and the partial pressures of N₂ and O₂, we can calculate the partial pressure of NO at equilibrium. This calculation involves setting up an equilibrium table (ICE table) and solving for the unknown partial pressure using the Kp expression. Understanding these principles is crucial for solving equilibrium problems in gaseous systems.
Background: Equilibrium and Partial Pressures
To accurately determine the partial pressure of NO at equilibrium, a firm grasp of equilibrium principles and partial pressures is essential. Chemical equilibrium is a dynamic state where the forward and reverse reaction rates are equal, resulting in constant concentrations of reactants and products. This state does not mean the reaction has stopped; rather, the rates of formation and consumption of reactants and products are balanced. Understanding this dynamic nature is crucial for solving equilibrium problems.
Partial pressure, as defined by Dalton's Law of Partial Pressures, is the pressure exerted by a single gas in a mixture of gases. The total pressure of the mixture is the sum of the partial pressures of all the gases present. In the context of gaseous reactions, partial pressures are used to express the concentrations of reactants and products. The partial pressure of a gas is directly proportional to its mole fraction in the mixture and the total pressure. This relationship is expressed as:
Pᵢ = xᵢ * P_total
where Pᵢ is the partial pressure of gas i, xᵢ is the mole fraction of gas i, and P_total is the total pressure of the mixture. For reactions at equilibrium, the partial pressures of reactants and products are related by the equilibrium constant Kp. The equilibrium constant is a temperature-dependent value that indicates the extent to which a reaction proceeds to completion. For the general gaseous reaction:
aA + bB ⇌ cC + dD
The expression for Kp is:
Kp = (P(C)ᶜ * P(D)ᵈ) / (P(A)ᵃ * P(B)ᵇ)
where P(A), P(B), P(C), and P(D) are the partial pressures of the gases A, B, C, and D at equilibrium, and a, b, c, and d are their respective stoichiometric coefficients. For the specific reaction N₂ + O₂ ⇌ 2NO, the Kp expression is:
Kp = (P(NO)²) / (P(N₂) * P(O₂))
Understanding these foundational concepts of equilibrium and partial pressures is crucial for setting up and solving equilibrium problems, including determining the partial pressure of NO at equilibrium.
Problem Setup
To determine the partial pressure of NO at equilibrium, we start by outlining the given information and the reaction equation. We are given the partial pressures of N₂ and O₂ at equilibrium as 4 torr and 16 torr, respectively. The equilibrium constant Kp for the reaction is given as 1.0. The balanced chemical equation for the reaction is:
N₂ (g) + O₂ (g) ⇌ 2NO (g)
The equilibrium constant expression for this reaction, in terms of partial pressures, is:
Kp = (P(NO)²) / (P(N₂) * P(O₂))
We are tasked with finding the partial pressure of NO at equilibrium, which we will denote as P(NO). To solve this, we can use an ICE (Initial, Change, Equilibrium) table to systematically track the changes in partial pressures as the reaction reaches equilibrium. The ICE table helps organize the information and makes it easier to apply the equilibrium constant expression.
- Initial (I): This row represents the initial partial pressures of the reactants and products. We are given the initial partial pressures of N₂ and O₂ as 4 torr and 16 torr, respectively. We will assume the initial partial pressure of NO is 0 torr since it is not specified otherwise.
- Change (C): This row represents the change in partial pressures as the reaction proceeds to equilibrium. Let's denote the change in partial pressure of N₂ and O₂ as -x, since they are reactants and their partial pressures will decrease. The change in partial pressure of NO will be +2x, since it is a product and its partial pressure will increase, and the stoichiometric coefficient for NO is 2.
- Equilibrium (E): This row represents the partial pressures of the reactants and products at equilibrium. These are the sums of the initial partial pressures and the changes. So, the equilibrium partial pressures are P(N₂) = 4 - x, P(O₂) = 16 - x, and P(NO) = 2x.
By setting up the ICE table, we can now express the equilibrium partial pressures in terms of a single variable, x, and use the Kp expression to solve for x. This value will then allow us to determine the partial pressure of NO at equilibrium.
Solving for Equilibrium Partial Pressure of NO
Now that we have set up the ICE table and expressed the equilibrium partial pressures in terms of x, we can use the equilibrium constant expression to solve for the value of x. Recall the balanced chemical equation and the Kp expression:
N₂ (g) + O₂ (g) ⇌ 2NO (g)
Kp = (P(NO)²) / (P(N₂) * P(O₂))
We are given that Kp = 1.0, P(N₂) = 4 - x, P(O₂) = 16 - x, and P(NO) = 2x. Substituting these values into the Kp expression, we get:
- 0 = (2x)² / ((4 - x) * (16 - x))
Simplifying the equation, we have:
- 0 = 4x² / (64 - 20x + x²)
To solve for x, we can multiply both sides by the denominator to get:
64 - 20x + x² = 4x²
Rearranging the terms, we obtain a quadratic equation:
3x² + 20x - 64 = 0
This quadratic equation can be solved using the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
where a = 3, b = 20, and c = -64. Plugging in these values, we get:
x = (-20 ± √(20² - 4 * 3 * (-64))) / (2 * 3)
x = (-20 ± √(400 + 768)) / 6
x = (-20 ± √1168) / 6
x = (-20 ± 34.176) / 6
We have two possible solutions for x:
x₁ = (-20 + 34.176) / 6 ≈ 2.363
x₂ = (-20 - 34.176) / 6 ≈ -9.029
Since partial pressures cannot be negative, we discard the negative value and take x ≈ 2.363. Now we can find the partial pressure of NO at equilibrium using P(NO) = 2x:
P(NO) = 2 * 2.363 ≈ 4.726 torr
Rounding this value to the nearest whole number, we get approximately 4.7 torr. This is very close to 4 torr, so the closest answer is 4 torr.
Conclusion
In conclusion, we have successfully determined the partial pressure of NO at equilibrium for the given reaction N₂ + O₂ ⇌ 2NO. By utilizing the principles of chemical equilibrium, partial pressures, and the equilibrium constant Kp, we were able to systematically solve the problem. We started by setting up the ICE table to track the changes in partial pressures of the reactants and products as the reaction reached equilibrium. This table allowed us to express the equilibrium partial pressures in terms of a single variable, x. The Kp expression was then used to form a quadratic equation, which we solved using the quadratic formula. After discarding the negative root as physically irrelevant, we found the value of x and calculated the partial pressure of NO at equilibrium.
The calculated partial pressure of NO at equilibrium was approximately 4.726 torr, which we rounded to 4 torr. This value corresponds to option C in the given choices. The process involved a clear understanding of the relationships between partial pressures, equilibrium constants, and the stoichiometry of the reaction. This method can be applied to a variety of similar problems involving gaseous equilibria.
Understanding these equilibrium calculations is crucial in various chemical applications, including industrial processes, environmental chemistry, and research. The ability to predict the partial pressures of reactants and products at equilibrium allows for optimization of reaction conditions and yields. By mastering these concepts, one can gain a deeper understanding of chemical reactions and their behavior under different conditions.
Therefore, the partial pressure of NO at equilibrium, given the initial conditions and the equilibrium constant, is approximately 4 torr. This detailed step-by-step solution demonstrates the importance of understanding and applying chemical equilibrium principles to solve practical problems.
