Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation \(\mathrm{KI}(a q)+\mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{KI}_{3}(a q)\) give the same expression for the reaction quotient. \(\mathrm{KI}_{3}\) is composed of the ions \(K^{+}\) and \(I_{3}^{-}\)

Short Answer

Expert verified
The reaction quotient expressions for the complete chemical equation, total ionic equation, and net ionic equation are equivalent after accounting for the cancellation of spectator ions.

Step by step solution

01

Write the Complete Chemical Equation

Write down the complete balanced chemical equation for the reaction. In this case, the equation is given: \(\mathrm{KI}(aq) + \mathrm{I}_{2}(aq) \rightleftharpoons \mathrm{KI}_{3}(aq)\).
02

Identify Ions in Solution

Identify the ions that each compound dissociates into in an aqueous solution. KI dissociates into \(K^{+}\) and \(I^{-}\), while \(\mathrm{I}_{2}\) stays as a molecule in the solution, and \(\mathrm{KI}_{3}\) dissociates into \(K^{+}\) and \(I_{3}^{-}\) ions.
03

Write the Total Ionic Equation

The total ionic equation shows all the ions present in the reaction. For this reaction, it is: \(K^{+}(aq) + I^{-}(aq) + \mathrm{I}_{2}(aq) \rightleftharpoons K^{+}(aq) + I_{3}^{-}(aq)\).
04

Write the Net Ionic Equation

Cancel the spectator ions from both sides of the equation to get the net ionic equation. The spectator ion here is \(K^{+}\), which gives us the net ionic equation: \(I^{-}(aq) + \mathrm{I}_{2}(aq) \rightleftharpoons I_{3}^{-}(aq)\).
05

Write the Expression for the Reaction Quotient

For each equation (complete, total ionic, and net ionic), write the expression for the reaction quotient \(Q\). This is done by dividing the product of the concentrations of the products by the product of the concentrations of reactants. Complete chemical equation \(Q\) expression: \(\frac{[\mathrm{KI}_{3}]}{[\mathrm{KI}][\mathrm{I}_{2}]}\). Total ionic equation \(Q\) expression: \(\frac{[K^{+}][I_{3}^{-}]}{[K^{+}][I^{-}][\mathrm{I}_{2}]}\). Net ionic equation \(Q\) expression: \(\frac{[I_{3}^{-}]}{[I^{-}][\mathrm{I}_{2}]}\). Note that the \(K^{+}\) ion cancels out in the total ionic equation.
06

Compare the Reaction Quotient Expressions

Compare the expressions for the reaction quotient \(Q\) from the complete, total ionic, and net ionic equations. After canceling out the \(K^{+}\) ions in the total ionic equation, it becomes apparent that all three expressions are equivalent because they all simplify to the same form.

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Key Concepts

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

Chemical Equation
Understanding the chemical equation involves recognizing how substances react to form new substances. Chemical equations provide a succinct way to represent these reactions. Each chemical equation consists of reactants (substances that start the reaction) and products (substances that are produced by the reaction). For educational clarity, let's consider the provided equation:
In the exercise, the balanced chemical equation is \[\begin{equation}\mathrm{KI}(aq) + \mathrm{I}_{2}(aq) \rightleftharpoons \mathrm{KI}_{3}(aq)\end{equation}\]which shows potassium iodide reacting with iodine to form triiodide. This equation communicates not only the substances involved but also their states; in this case, 'aq' denotes that the substances are aqueous, or dissolved in water. It's important for students to acknowledge that the law of conservation of mass is upheld in chemical equations: the number of atoms of each element is the same on both sides of the reaction.
Reaction Quotient (Q)
The reaction quotient () is a measure of the relative amounts of products and reactants present during a chemical reaction at a given moment. It is calculated similarly to the equilibrium constant, but for a reaction that has not necessarily reached equilibrium. \[\begin{equation}Q = \frac{[Products]}{[Reactants]}\end{equation}\]In the context of the given exercise, for the complete chemical equation, total ionic equation, and net ionic equation is used to show the state of the reaction. When equals the equilibrium constant (), the reaction is at equilibrium and no net change occurs in the concentration of reactants and products. By demonstrating that all expressions for are equivalent, it illustrates the consistency of the reaction's progress, regardless of how the reaction is represented.
Total Ionic Equation
The total ionic equation provides an in-depth view of the chemical reaction by showing all the ions present when the reactants are dissolved in aqueous solutions. This representation is vital for understanding reactions in an ionic context. For the given exercise, the total ionic equation is:\[\begin{equation}K^{+}(aq) + I^{-}(aq) + \mathrm{I}_{2}(aq) \rightleftharpoons K^{+}(aq) + I_{3}^{-}(aq)\end{equation}\]This equation demonstrates the reaction on an ionic level, highlighting which compounds dissociate into ions in solution. It’s an intermediate step between the complete chemical equation and the net ionic equation, helping students grasp the reaction's nuance.
Spectator Ions
Spectator ions are ions that do not change during the course of a reaction; they appear unchanged on both the reactant and the product sides of a chemical equation. These ions are not directly involved in the formation of the product and are therefore 'spectators' to the reaction. In our example, the potassium ion () is a spectator ion, as it remains unchanged from the reactants to the products:\[\begin{equation}K^{+}(aq) \text{ (reactants) } \rightarrow K^{+}(aq) \text{ (products) }\end{equation}\]Identifying and eliminating spectator ions when transitioning from the total ionic equation to the net ionic equation simplifies the reaction and focuses on the species that actually undergo chemical change.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain unchanged over time. It's a dynamic state where reactions continue to occur, but without net change in concentration of reactants and products. This concept is significant when discussing reversible reactions, such as the one in the exercise:\[\begin{equation}\mathrm{KI}(aq) + \mathrm{I}_{2}(aq) \rightleftharpoons \mathrm{KI}_{3}(aq)\end{equation}\]At equilibrium, the reaction quotient () equals the equilibrium constant (), signifying no further change in the system. Understanding equilibrium is fundamental in predicting how the system will respond to changes, as outlined by Le Chatelier's principle, making it a key concept in chemistry.

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Most popular questions from this chapter

Write the mathematical expression for the reaction quotient, \(Q_{c}\), for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)=2 \mathrm{NO}_{2}(g)\) (d) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{NH}_{4} \mathrm{Cl}(s)=\mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\) (f) \(2 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightleftharpoons 2 \mathrm{PbO}(s)+4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) (g) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)=2 \mathrm{H}_{2} \mathrm{O}(l)\) (h) \(S_{8}(g) \rightleftharpoons 8 S(g)\)

The equilibrium constant \(\left(K_{c}\right)\) for this reaction is 5.0 at a given temperature. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) (a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of \(\mathrm{CO}, 0.30\) mol of water vapor, and \(0.90 \mathrm{mol}\) of \(\mathrm{H}_{2}\) in a liter. How many moles of \(\mathrm{CO}_{2}\) were there in the equilibrium mixture? (b) Maintaining the same temperature, additional \(\mathrm{H}_{2}\) was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of \(\mathrm{CO}, 0.30\) mol of water vapor, and 1.2 mol of \(\mathrm{H}_{2}\) in a liter. How many moles of \(\mathrm{CO}_{2}\) were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.

Calcium chloride 6-hydrate, \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\), dehydrates according to the equation \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{CaCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{P}=5.09 \times 10^{-44} \mathrm{at} 25^{\circ} \mathrm{C}\) What is the pressure of water vapor at equilibrium with a mixture of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CaCl}_{2}\) at \(25^{\circ} \mathrm{C} ?\)

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, \(S O_{3}\), from sulfur dioxide, \(S O_{2},\) and oxygen, \(O_{2}\), shown here. At high temperatures, the rate of formation of \(S O_{3}\) is higher, but the equilibrium amount (concentration or partial pressure) of \(\mathrm{SO}_{3}\) is lower than it would be at lower temperatures. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases? (b) Is the reaction endothermic or exothermic?

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each? (a) \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \quad \Delta H=92 \mathrm{kJ}\) (b) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H=181 \mathrm{kJ}\) (c) \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g) \quad \Delta H=-285 \mathrm{kJ}\) (d) \(\mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CaCO}_{3}(s) \quad \Delta H=-176 \mathrm{kJ}\)

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