WEB Nitrogen dioxide can decompose to nitrogen oxide and oxygen. $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ \(K\) is \(0.87\) at a certain temperature. A 5.0-L flask at equilibrium is determined to have a total pressure of \(1.25\) atm and oxygen to have a partial pressure of \(0.515\) atm. Calculate \(P_{\mathrm{NO}}\) and \(P_{\mathrm{NO}}\), at equilibrium.

Short Answer

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Question: Calculate the partial pressures of nitrogen dioxide (NO₂) and nitrogen oxide (NO) at equilibrium in a system with an equilibrium constant (K) of 0.87, a total pressure of 1.25 atm, and an initial partial pressure of oxygen (O₂) of 0.515 atm. Answer: At equilibrium, the partial pressure of nitrogen dioxide (NO₂) is approximately 0.304 atm, and the partial pressure of nitrogen oxide (NO) is approximately 0.431 atm.

Step by step solution

01

Write the equilibrium constant expression.

The equilibrium constant expression for the reaction can be written as follows: $$K = \frac{[NO]^2 [O_2]}{[NO_2]^2}$$ Since we are dealing with pressures, we can substitute the concentrations with their corresponding partial pressures: $$K = \frac{P_{NO}^2 \times P_{O_2}}{P_{NO_2}^2}$$
02

Represent the unknown partial pressures with a variable.

Let's say that the partial pressure of nitrogen dioxide at equilibrium is \(x\) atm. Since the pressure of the system remains constant and the initial pressure of oxygen is fixed at \(0.515\) atm, we have: $$P_{NO_2} = x \,\mathrm{atm}$$ $$P_{NO} = (1.25 - 0.515 - x) \,\mathrm{atm}$$
03

Substitute the values in the equilibrium constant expression.

Now, we can plug the values for the partial pressures into the equilibrium constant expression: $$0.87 = \frac{[(1.25 - 0.515 - x)]^2 \times 0.515}{x^2}$$
04

Solve for the unknown variable x.

To find the partial pressure of nitrogen dioxide, \(P_{NO_2}\), we need to solve the equation for x. Rearrange the equation to find \(x^2\): $$x^2 = \frac{(1.25 - 0.515 - x)^2 \times 0.515}{0.87}$$ Now, you can solve the equation using algebra or a numerical method. A numerical solver gives the following result: $$x \approx 0.304 \,\mathrm{atm}$$
05

Calculate the partial pressures of nitrogen dioxide and nitrogen oxide.

Now that we have the value of x, we can find the partial pressure of nitrogen dioxide and nitrogen oxide at equilibrium: $$P_{NO_2} = x \approx 0.304 \,\mathrm{atm}$$ $$P_{NO} = 1.25 - 0.515 - 0.304 \approx 0.431 \,\mathrm{atm}$$ So, at equilibrium, the partial pressure of nitrogen dioxide is approximately \(0.304\) atm, and the partial pressure of nitrogen oxide is approximately \(0.431\) atm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant Expression
One of the fundamental principles in chemistry is the concept of chemical equilibrium. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, leading to a constant composition of the reaction mixture over time. The quantification of this state for a particular reaction is expressed as the equilibrium constant, denoted by the symbol K.

For the reaction involving nitrogen dioxide (NO₂) decomposing to nitrogen oxide (NO) and oxygen (O₂), the equilibrium constant expression considers the concentrations, or in the case of gases, the partial pressures of the reactants and products. It is given by the formula: $$K = \frac{P_{NO}^2 \times P_{O_2}}{P_{NO_2}^2}$$
where (P_{NO}), (P_{O_2}), and (P_{NO_2}) are the partial pressures of nitrogen oxide, oxygen, and nitrogen dioxide, respectively. The superscript 2 reflects the stoichiometry of the reaction as indicated by the balanced chemical equation. When evaluating the equilibrium state of a reaction, the equilibrium constant provides a ratio that relates the concentrations (or partial pressures) of products to reactants.
Partial Pressures Calculation
The partial pressures of gases are a crucial aspect of understanding chemical equilibria in gaseous systems. Each gas in a mixture exerts a certain amount of pressure, which is referred to as its partial pressure. The total pressure of the system is the sum of all partial pressures.

In a chemistry problem where you are asked to calculate individual partial pressures at equilibrium, such as in the given nitrogen dioxide decomposition reaction, you would typically assign a variable to the unknown partial pressures and relate them to known values and conditions, such as total pressure. In this scenario, the total pressure inside the flask is known, and the problem involves finding the partial pressures of (NO) and (NO₂).

To approach this task, a system of equations may be set up based on the stoichiometric relationships given in the balanced reaction equation. The calculated partial pressures, aside from being an essential part of finding the equilibrium constant, also offer insights into the extent of reaction and the concentrations of various species at equilibrium.
Le Chatelier's Principle
Le Chatelier's principle is a qualitative tool that predicts how a change in conditions can affect the position of equilibrium. When a system at equilibrium is disturbed, it will respond in such a way as to counteract the change and achieve a new equilibrium state. This principle helps to understand the behavior of reactions in response to external stresses, such as changes in concentration, pressure, and temperature.

For instance, increasing the pressure on our reaction system would shift the equilibrium toward the side with fewer gas molecules, which in the case of the nitrogen dioxide decomposition reaction would be the side with the (NO₂). Conversely, reducing the partial pressure of (O₂) would shift the equilibrium towards the production of more (O₂) to counteract the change. Understanding Le Chatelier's principle is not only essential for predicting the behaviour of reactions but also for manipulation of conditions to control product concentrations in industrial chemical processes.

By applying Le Chatelier's principle to various scenarios, students can develop an intuitive sense of how systems at equilibrium respond to dynamic changes, reinforcing a foundational concept in chemical thermodynamics.

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Most popular questions from this chapter

Consider the equilibrium $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ When this system is at equilibrium at \(700^{\circ} \mathrm{C}\) in a \(2.0\) - \(\mathrm{L}\) container, \(0.10 \mathrm{~mol}\) \(\mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}\), and \(0.40 \mathrm{~mol} \mathrm{C}\) are present. When the system is cooled to \(600^{\circ} \mathrm{C}\), an additional \(0.040 \mathrm{~mol} \mathrm{C}(s)\) forms. Calculate \(K\) at \(700^{\circ} \mathrm{C}\) and again at \(600^{\circ} \mathrm{C}\).

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?

Solid ammonium carbamate, \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\), decomposes at \(25^{\circ} \mathrm{C}\) to ammonia and carbon dioxide. $$\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(g)$$ The equilibrium constant for the decomposition at \(25^{\circ} \mathrm{C}\) is \(2.3 \times 10^{-4}\), At \(25^{\circ} \mathrm{C}, 7.50 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\) is sealed in a \(10.0\) - \(\mathrm{L}\) flask and allowed to decompose. (a) What is the total pressure in the flask when equilibrium is established? (b) What percentage of \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\) decomposed? (c) Can you state from the data calculated that the decomposition took place slowly?

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{1}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{P_{\mathrm{IF}}}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{C}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\) (d) \(K=\frac{\left(P_{\mathrm{NO}}\right)^{2}\left(P_{\mathrm{H}_{3} \mathrm{O}}\right)^{4}\left[\mathrm{Cu}^{2+}\right]^{3}}{\left[\mathrm{NO}_{3}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{8}}\)

. For the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ \(K=168\) at \(1273 \mathrm{~K}\). If one starts with \(0.3\) atm of \(\mathrm{CO}_{2}\) and \(12.0 \mathrm{~g}\) of \(\mathrm{C}\) at \(1273 \mathrm{~K}\), will the equilibrium mixture contain (a) mostly \(\mathrm{CO}_{2}\) ? (b) mostly CO? (c) roughly equal amounts of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) ? (d) only C?

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