For the reaction \(4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g)\) at \(25^{\circ} \mathrm{C}, K_{\mathrm{eq}}=0.150 .\) What is the equilibrium concentration of \(\mathrm{NO}_{2}(g)\) if \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.300 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=1.20 \mathrm{M} ?\)

Short Answer

Expert verified
The equilibrium concentration of \(\mathrm{NO}_{2}(g)\) is approximately \(0.84 \mathrm{M}\).

Step by step solution

01

1. Write the expression for the reaction quotient (Q):

Using the balanced chemical equation, we can write the equilibrium expression (Q) as: \(Q = \frac{[\mathrm{N}_{2} \mathrm{O}_{5}]^{2}}{[\mathrm{NO}_{2}]^{4} [\mathrm{O}_{2}]} \)
02

2. Use the given K_eq value to set up an equation to solve for the equilibrium concentration of NO_2 (x):

At equilibrium, we know that Q = K_eq. We can now use the given K_eq value and the given equilibrium concentrations to set up an equation and solve for the equilibrium concentration (x) of \(\mathrm{NO}_{2}(g)\). \(\frac{(0.300)^{2}}{(x)^{4}(1.20)}=0.150 \)
03

3. Solve for x:

Now, we'll solve for x by first isolating x^4 on one side of the equation. \((x)^{4}= \frac{(0.300)^{2}}{1.20 \times 0.150} \) \((x)^{4}= \frac{0.09}{0.18} \) \((x)^{4}= 0.5 \) Now, calculate \(\mathrm{NO}_2\) concentration by taking the fourth root of both sides: \(x = \sqrt[4]{0.5} \) The equilibrium concentration of \(\mathrm{NO}_{2}(g)\) is \(x \approx 0.84 \mathrm{M}\).

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Key Concepts

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

Chemical Equilibrium
When a chemical reaction occurs, the reactants transform into products, and in some cases, these products can react to form the reactants again. A state of chemical equilibrium is established in a closed system when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant over time. However, even though the concentrations are constant, it should be noted that the reactions are still occurring in both directions.

Understanding equilibrium is crucial because it helps us predict the extent of a reaction and the concentrations of various compounds at equilibrium. Factors such as temperature, pressure, and the presence of catalysts can shift this equilibrium, as described by Le Châtelier's principle.
Reaction Quotient (Q)
The reaction quotient (Q) is a measure of the relative amounts of products and reactants present during a reaction at a given moment. It provides a snapshot of the reaction's current state and is defined by the same expression as the equilibrium constant, but for non-equilibrium conditions. This means that if you're given a reaction in progress and know the concentrations of all the reactants and products, you can calculate Q.

The value of Q can tell us the direction in which the reaction will proceed to reach equilibrium. If Q < Keq, the forward reaction will be favored, and more products will form until equilibrium is reached. Conversely, if Q > Keq, the reaction will proceed in the reverse direction.
Equilibrium Constant (K_eq)
The equilibrium constant (Keq) is a number that expresses the ratio of the concentration of the products to the concentration of the reactants at equilibrium, with each raised to the power of their coefficients from the balanced chemical equation. The magnitude of Keq can indicate the extent to which a reaction will proceed. A high value (much greater than 1) implies a reaction that favors the formation of products, while a low value (much less than 1) means that the reactants are favored.

In practice, you can calculate Keq if you know the concentrations of the reactants and products at equilibrium. Conversely, if you know Keq and the concentrations of either the reactants or the products, you can determine the other set of concentrations at equilibrium, as illustrated in the textbook exercise.
Balancing Chemical Equations
Balancing chemical equations is the process of ensuring that the number of atoms for each element is the same on both sides of the chemical equation. It is a fundamental principle, based on the law of conservation of mass, stating that matter cannot be created or destroyed in an isolated system.

To balance an equation, we adjust the coefficients in front of the chemical formulas to make sure that the atoms are balanced. These coefficients are also essential when determining the reaction quotient and equilibrium constant because they factor into the powers to which the concentrations are raised in the mathematical expressions.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using the coefficients from the balanced equations to calculate the quantities of reactants and products.

In a reactant-limited reaction, stoichiometry can tell us how much product can be produced from a given amount of reactants. Likewise, if only products' concentrations are known, stoichiometry allows us to deduce the quantities of the reactants involved. When dealing with a reaction at equilibrium or any chemical reaction, understanding stoichiometry is crucial to converting between moles, mass, and particles, and interpreting the quantifiable aspects of reactions.

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Most popular questions from this chapter

Suppose a reaction is at equilibrium and you then disturb the equilibrium by adding reactants. What happens to the value of \(K_{\text {eq }}\) ? Explain your answer.

Some liquid water is placed in a room-temperature container and sealed. After some time, the level of the liquid drops by a small amount. After that, the water level remains constant.

Diamond and graphite are two forms of elemental carbon. Under the appropriate conditions they will be in equilibrium with each other: \(C_{\text {diamond }} \rightleftarrows C_{\text {graphite }}\) If graphite is subjected to very high pressure and temperature, it will convert into the diamond form. (a) Is the above equilibrium reaction exothermic or endothermic? Explain how you know. (b) Which form, graphite or diamond, has the higher density? (Hint: Think about what increasing the pressure of a gas does to its density. It works the same for the solid and liquid phases as well.)

Indicate with an arrow the direction of the equilibrium shift and predict what will happen to the amount of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (increases, decreases, unchanged, need more information) when the following stresses are applied to the following exothermic reaction: \(2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g)\) The reaction is cooled. \(\mathrm{H}_{2}\) gas is added. \(\mathrm{H}_{2} \mathrm{O}\) is removed. Volume is reduced. A catalyst is added. Fe is added while the reaction temperature is increased.

(a) How would you prepare a saturated aqueous solution of copper(I) iodide at \(25^{\circ} \mathrm{C}\) ? (b) What is the mass in milligrams of CuI in \(400.0 \mathrm{~mL}\) of the saturated solution? (Hint: Begin with a \(K_{\mathrm{sp}}\) value from Table 14.1.) (c) Suppose you add some CuI* to this saturated solution, where \(\mathrm{I}^{*}\) is a radioactive form of iodide ion. A student says, "Because the solution is already saturated, the added CuI* won't dissolve and there's no danger of getting any radioactive iodide ion in solution." What is wrong with his thinking?

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