At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ If \(2.0\) moles of \(\mathrm{CO}_{2}\) is initially placed into a \(5.0-\mathrm{L}\) vessel, calculate the equilibrium concentrations of all species.

Short Answer

Expert verified
At equilibrium, the concentrations are approximately: [CO₂] ≈ 0.3764 M, [CO] ≈ 0.0236 M, and [O₂] ≈ 0.0118 M.

Step by step solution

01

Write the expression for the equilibrium constant Kc

First, we need to write down the expression for the equilibrium constant, Kc, for the given reaction: 2 CO₂(g) ⇌ 2 CO(g) + O₂(g) The Kc expression is: $$ K_c = \frac{[\mathrm{CO}]^2[\mathrm{O_2}]}{[\mathrm{CO_2}]^2} $$ We are given the value of Kc as \(2.0 \times 10^{-6}\).
02

Set up an ICE table

An ICE table is useful for keeping track of the initial concentrations (I), changes in concentrations (C), and equilibrium concentrations (E) for the reactants and products in the reaction. | | CO₂ | CO | O₂ | |----------|-------------|-----------|-----------| | Initial | \(\frac{2}{5}\) | 0 | 0 | | Change | -2x | +2x | +x | | Equil. | \(\frac{2}{5}-2x\) | 2x | x | The initial concentration of CO₂ is calculated as follows: \[\frac{\text{Moles}}{\text{Volume}} = \frac{2.0}{5.0} = 0.4\text{ M}\] At equilibrium, the new concentrations are as described in the table.
03

Substitute the equilibrium concentrations into the Kc expression

Now, we will substitute the equilibrium concentrations of each species into the expression for Kc: \(2.0 \times 10^{-6} = \frac{(2x)^2 \cdot x}{(0.4 - 2x)^2}\)
04

Solve for x

To solve for x, we will simplify the equation and solve for x: \( 2.0 × 10^{-6} = \frac{4x^3}{(0.4 - 2x)^2} \) Now multiply both sides by \((0.4 - 2x)^2\): \( (0.4 - 2x)^2 \cdot 2.0 × 10^{-6} = 4x^3 \) Solving for x in the above equation can be rather tedious or involve an approximation to find the value of x (e.g., neglecting small terms). For the sake of simplicity, we can use a numerical method or calculator to find the value of x, which turns out to be approximately \( x \approx 0.0118\text{ M}\).
05

Calculate the equilibrium concentrations

Now that we have the value of x, we can plug it back into the ICE table to find the molar concentrations of all substances at equilibrium. CO₂: \(0.4 - 2x \approx 0.4 - 2(0.0118) \approx 0.3764\text{ M}\) CO: \(2x \approx 2(0.0118) \approx 0.0236\text{ M}\) O₂: \(x \approx 0.0118\text{ M}\) So, at equilibrium, the concentrations are: [CO₂] ≈ 0.3764 M; [CO] ≈ 0.0236 M; [O₂] ≈ 0.0118 M.

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Key Concepts

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

Chemical Equilibrium
The concept of chemical equilibrium is foundational to understanding how reactions proceed in chemistry. At a basic level, equilibrium represents a state in which the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentration of reactants and products remains constant over time, although not necessarily equal.

When a reaction reaches this state, it is said to be 'at equilibrium.' Importantly, reaching equilibrium does not mean that the reaction has stopped; rather, reactants continue to form products, and products continue to form reactants at equal rates. Understanding this dynamic allows us to predict how various changes to the system, such as altering concentrations, temperature, or pressure, could shift the equilibrium position, a concept often described by Le Chatelier's Principle.
ICE Table
An ICE Table, standing for Initial, Change, and Equilibrium, is a strategic tool used to track the changes in concentrations of reactants and products during a dynamic chemical process. It is particularly useful in visualizing how equilibrium is established in a reaction system.

The 'Initial' section of the table lists the starting concentrations of reactants and products. Often, as in the original exercise, the initial concentration of products is zero in a reaction that starts only with reactants. The 'Change' section expresses the amounts by which the reactants and products will increase or decrease as the system approaches equilibrium. This is symbolized with a variable, often 'x,' which signifies this shift. Finally, the 'Equilibrium' section reveals the concentrations of reactants and products once the system has reached its equilibrium state. By setting up an ICE table, students can systematically calculate unknown concentrations at equilibrium using the equilibrium constant of the reaction.
Equilibrium Constant Calculations
The equilibrium constant, represented by K or Kc for concentrations, is a numerical value that conveys the ratio of the concentration of products to the concentration of reactants at equilibrium for a reversible chemical reaction. Its definition is specific to the reaction at hand, derived from the balanced chemical equation.

For the example given, the equilibrium constant expression is \[ K_c = \frac{[\mathrm{CO}]^2[\mathrm{O_2}]}{[\mathrm{CO_2}]^2} \]. Calculating these constants involves algebraic manipulation where we substitute equilibrium concentrations from the ICE table into the Kc expression. Solving for the remaining variables allows us to determine the equilibrium concentrations of all species in the reaction. The difficulty often lies in solving a cubic or a polynomial equation, which may require approximation techniques or numeric methods for solutions. Understanding these calculations helps to forecast how the system would respond under different conditions, which is extremely pertinent in chemical industries and research.

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Most popular questions from this chapter

Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet- proof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right)\), an extremely poisonous gas. Phosgene decomposes by the reaction $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an jnitial pressure of \(1.0\) atm decomposes, calculate the equilibrium pressures of all species.

What will happen to the number of moles of \(\mathrm{SO}_{3}\) in equilibrium with \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) in the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ in each of the following cases? a. Oxygen gas is added. b. The pressure is increased by decreasing the volume of the reaction container. c. In a rigid reaction container, the pressure is increased by adding argon gas. d. The temperature is decreased (the reaction is endothermic). e. Gaseous sulfur dioxide is removed.

Consider the following statements: "Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)\), for which at equilibrium \([\mathrm{A}]=2 M\) \([\mathrm{B}]=1 M\), and \([\mathrm{C}]=4 M .\) To a \(1-\mathrm{L}\) container of the system at equilibrium, you add 3 moles of \(\mathrm{B}\). A possible equilibrium condition is \([\mathrm{A}]=1 M,[\mathrm{~B}]=3 M\), and \([\mathrm{C}]=6 M\) because in both cases \(K=2 . "\) Indicate everything that is correct in these statements and everything that is incorrect. Correct the incorrect statements, and explain.

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calcu- late the value of \(K\) for the reaction.

Old-fashioned "smelling salts" consist of ammonium carbonate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammonium carbonate $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

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