For the reaction: $$ 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g) $$ \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 M\), calculate the equilibrium \(\mathrm{O}_{3}\) concentration.

Short Answer

Expert verified
The equilibrium concentration of \(\mathrm{O}_3\) is approximately \(2.07 \times 10^{-6} M\).

Step by step solution

01

Write the equilibrium expression

Based on the given reaction, the equilibrium expression is as follows: \[K = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3}\] Now, we substitute the known values and solve for the unknown concentration.
02

Substitute given values into the equation

We know the equilibrium constant and the concentration of \(\mathrm{O}_2\). Substitute these values into the equation: \[ 1.8 \times 10^{-7} = \frac{[\mathrm{O}_3]^2}{(0.062)^3} \]
03

Solve for the equilibrium concentration of \(\mathrm{O}_3\)

To find the concentration of \(\mathrm{O}_3\), we first need to isolate it by multiplying both sides of the equation by \((0.062)^3\): \[ [\mathrm{O}_3]^2 = 1.8 \times 10^{-7} \times (0.062)^3 \] Calculate the right-hand side of the equation: \[ [\mathrm{O}_3]^2 \approx 4.29 \times 10^{-11} \] Now, take the square root of both sides to find the equilibrium concentration of \(\mathrm{O}_3\): \[ [\mathrm{O}_3] = \sqrt{4.29 \times 10^{-11}} \] \[ [\mathrm{O}_3] \approx 2.07 \times 10^{-6} M \] So, the equilibrium concentration of \(\mathrm{O}_3\) is approximately \(2.07 \times 10^{-6} M\).

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is fundamental when studying reactions and their behaviors. It's a state in which both reactants and products have no net change over time. This doesn't mean the reactants stop turning into products or vice versa; rather, the rate at which the reactants produce products is equal to the rate at which products revert back to reactants. This dynamic state might seem static, but it's actually a balance of ongoing processes.

Imagine a crowded dance floor where the number of people entering is the same as those leaving; while people are always moving in and out, the total number doesn't change – that's like chemical equilibrium.
Equilibrium Constant (K)
The equilibrium constant, symbolized by 'K', is a snapshot of a reaction's balance at a specific temperature. It tells us the ratio of product concentrations to reactant concentrations at equilibrium, raised to the power of their coefficients from the balanced equation. The value of 'K' is essential for chemists to understand how much product can form under given conditions and to predict the direction of the reaction.

For instance, a large 'K' suggests that at equilibrium, the reaction favors the formation of products, whereas a small 'K' implies that reactants are preferred. Remember, 'K' is temperature-dependent; if the temperature changes, so does the 'K' value. By knowing the equilibrium constant, chemists can forecast the composition of a reaction mixture when equilibrium is reached.
Equilibrium Expression
To converse with chemists, you need to speak the language, and the equilibrium expression is part of that lexicon. It mathematically conveys the ratio of products to reactants at equilibrium, in the form of concentrations and partial pressures, depending on the state of the reactants and products involved (aqueous or gaseous).

For a general reaction like \(aA + bB \rightleftharpoons cC + dD\), the equilibrium expression is \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\), where each letter represents the substances involved, and the brackets denote their molarity. The coefficients from the balanced reaction equation become the exponents in this expression. Importantly, solids and pure liquids are left out of this equation, as their concentrations don't change during the reaction.
Le Chatelier's Principle
Le Chatelier's principle is like the universe's way of maintaining balance. It states that if a dynamic equilibrium is disturbed, the system will adjust itself to counteract the disturbance and restore equilibrium. Picture a seesaw where kids are constantly adjusting their positions to keep it level; that's what reactions do, according to this principle.

Changes can be in the form of concentration, pressure, volume, or temperature. For example, increasing the concentration of a reactant will push the equilibrium toward making more products. If you increase the pressure, the equilibrium will shift to the side with fewer gas molecules. Heating or cooling a system will also cause shifts, depending on whether the reaction is endothermic or exothermic. Le Chatelier's principle helps to predict how changes will affect the position of equilibrium, enabling chemists to optimize reaction conditions for the desired outcome.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}} \approx 1 \times 10^{-31}\) for the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ a. Calculate the concentration of \(\mathrm{NO}\), in molecules \(/ \mathrm{cm}^{3}\), that can exist in equilibrium in air at \(25^{\circ} \mathrm{C}\). In air, \(P_{\mathrm{N}_{2}}=0.8 \mathrm{~atm}\) and \(P_{\mathrm{o},}=0.2\) atm. b. Typical concentrations of NO in relatively pristine environments range from \(10^{8}\) to \(10^{10}\) molecules \(/ \mathrm{cm}^{3}\). Why is there a discrepancy between these values and your answer to part a?

At a particular temperature, a 3.0-L flask contains \(2.4\) moles of \(\mathrm{Cl}_{2}, 1.0\) mole of \(\mathrm{NOCl}\), and \(4.5 \times 10^{-3}\) mole of NO. Calculate \(K\) at this temperature for the following reaction: $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$

Consider the reaction $$ \mathrm{P}_{4}(g) \longrightarrow 2 \mathrm{P}_{2}(g) $$ where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{~K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{~K}\), the equilibrium mixture of \(\mathrm{P}_{4}(\mathrm{~g})\) and \(\mathrm{P}_{2}(g)\) has a total pressure of \(1.00 \mathrm{~atm} .\) Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g)\) : $$ \mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g) $$ A \(2.00-\mathrm{g}\) sample of \(\mathrm{XY}\) (molar mass \(=165 \mathrm{~g} / \mathrm{mol}\) ) is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\). The pressure is held constant at \(0.967\) atm. As \(X Y\) begins to dissociate, the piston moves until \(35.0\) mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of \(25^{\circ} \mathrm{C}\).

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) c. \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) d. \(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\)

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