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For the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ an analysis of an equilibrium mixture is performed. At a particular temperature, it is found that \([\mathrm{CO}]=0.11 M,\left[\mathrm{O}_{2}\right]=0.055 M,\) and \(\left[\mathrm{CO}_{2}\right]=1.4 M .\) Calculate \(K\) for the reaction.

Short Answer

Expert verified
The equilibrium constant (K) for the reaction \(2\mathrm{CO}_{2}(g) \rightleftharpoons 2\mathrm{CO}(g)+\mathrm{O}_{2}(g)\) is approximately \(3.40\times10^{-4}\) when the equilibrium concentrations are \([\mathrm{CO}]=0.11\,M\), \([\mathrm{O}_2]=0.055\,M\), and \([\mathrm{CO}_2]=1.4\,M\).

Step by step solution

01

Recall the formula for the equilibrium constant (K)

When a chemical reaction is at equilibrium, the concentrations of the species are related by the equilibrium constant (K). In general, for a reaction of the form: $$ a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D} $$ The equilibrium constant K is given by: $$ K = \frac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b} $$ In our case, the reaction is: $$ 2\mathrm{CO}_2(g) \rightleftharpoons 2\mathrm{CO}(g) + \mathrm{O}_2(g) $$ So, we'll have: $$ K = \frac{[\mathrm{CO}]^2 [\mathrm{O}_2]}{[\mathrm{CO}_2]^2} $$
02

Plug in the given equilibrium concentrations

We are given the equilibrium concentrations as \([\mathrm{CO}] = 0.11\,M\), \([\mathrm{O}_2] = 0.055\,M\), and \([\mathrm{CO}_2] = 1.4\,M\). Plug these values into the expression for K: $$ K = \frac{(0.11)^2\times 0.055}{(1.4)^2} $$
03

Calculate the equilibrium constant (K)

Now, perform the calculations to find K: $$ K = \frac{0.0121\times 0.055}{1.96} $$ $$ K = \frac{0.0006655}{1.96} \sim 3.40\times10^{-4} $$ Thus, the equilibrium constant, K, for the given reaction is approximately \(3.40\times10^{-4}\).

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Key Concepts

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

Equilibrium Constant
When a chemical reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction, it is said to be at dynamic equilibrium. At this point, the concentration of the reactants and products remains constant over time, although they are not necessarily equal. The equilibrium constant, denoted as K, is a numerical value that expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their coefficients in the balanced chemical equation.

The equilibrium constant provides insight into the position of equilibrium and the extent of the reaction. A high value of K (significantly larger than 1) indicates that, at equilibrium, the reaction mixture is predominantly products. Conversely, a low value of K (significantly smaller than 1) suggests that the reactants are favored. The precise value of K is determined by the specific reaction conditions, such as temperature. It's important to note that K only changes with temperature; it is not affected by changes in concentration or pressure.
Equilibrium Concentrations
Understanding the concept of equilibrium concentrations is vital for grasping the nature of chemical equilibrium. These are the concentrations of the reactants and products in a reaction mixture when the reaction has reached equilibrium. To calculate the equilibrium constant, you need the concentrations of each species at equilibrium.

The given equilibrium concentrations enable us to apply the equilibrium constant expression. Substituting the equilibrium concentrations into the equilibrium constant expression allows us to solve for the numerical value of K. In practical terms, knowing the equilibrium concentrations can inform us about the efficiency of a reaction and predict how the system will respond to external disturbances, according to Le Châtelier's Principle.
Chemical Reaction Equilibrium
Grasping the principle of chemical reaction equilibrium is essential for understanding how reactions proceed in nature and industry. It is the state in a chemical reaction where the rate of the forward reaction equals the rate of the backward reaction, causing no net change in the concentration of reactants and products. However, this does not imply that the reaction has stopped; rather, the reactants and products are being created and consumed at an equal rate.

Chemical reaction equilibrium is dynamic and can be shifted by changes in concentration, pressure, or temperature, as stated by Le Châtelier's Principle. This principle helps to predict how the equilibrium position will shift if external conditions are altered. For instance, increasing the concentration of products typically pushes the reaction towards the reactants to re-establish equilibrium. Depending on external conditions and the nature of the reaction, equilibrium can favor either the reactants or the products to varying degrees.

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Most popular questions from this chapter

Suppose the reaction system $$ \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ has already reached equilibrium. Predict the effect of each of the following changes on the position of the equilibrium. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Any liquid water present is removed from the system. b. \(\mathrm{CO}_{2}\) is added to the system by dropping a chunk of dry ice into the reaction vessel. c. The reaction is performed in a metal cylinder fitted with a piston, and the piston is compressed to decrease the total volume of the system. d. Additional \(\mathrm{O}_{2}(g)\) is added to the system from a cylinder of pure \(\mathrm{O}_{2}\).

Calcium sulfate, \(\mathrm{CaSO}_{4}\), is only soluble in water to the extent of approximately 2.05 \(\mathrm{g} / \mathrm{L}\) at \(25 \quad \mathrm{C} .\) Calculate \(K_{\mathrm{sp}}\) for calcium sulfate at \(25 \mathrm{C}\)

What is the special name given to the equilibrium constant for the dissolving of an ionic solute in water?

Most hydroxides are not very soluble in water. For example, \(K_{\mathrm{sp}}\) for nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2},\) is \(2.0 \times 10^{-15}\) at 25 . \(\mathrm{C}\). How many grams of nickel(II) hydroxide dissolve per liter at 25 ' \(\mathrm{C}\) ?

As you learned in Chapter 7 , most metal hydroxides are sparingly soluble in water. Write balanced chemical equations describing the dissolving of the following metal hydroxides in water. Write the expression for \(K_{\mathrm{sp}}\) for each process. a. \(\mathrm{Cu}(\mathrm{OH})_{2}(s)\) b. \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) c. \(\mathrm{Ba}(\mathrm{OH})_{2}(s)\) d. \(\operatorname{Sn}(\mathrm{OH})_{2}(s)\)

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