Problem 85
Question
The following reaction in aqueous solution has \(K_{\mathrm{c}}=\) \(1 \times 10^{-85}\) at a temperature of \(25^{\circ} \mathrm{C}\). \(7 \mathrm{IO}_{3}^{-}(a q)+9 \mathrm{H}_{2} \mathrm{O}+7 \mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{I}_{2}(a q)+5 \mathrm{H}_{5} \mathrm{IO}_{6}(a q)\) What is the equilibrium law for this reaction?
Step-by-Step Solution
Verified Answer
The equilibrium law is \( K_{\mathrm{c}} = \frac{[\mathrm{I}_{2}] \times [\mathrm{H}_{5}\mathrm{IO}_{6}]^{5}}{[\mathrm{IO}_{3}^{-}]^{7} \times [\mathrm{H}^{+}]^{7}} \).
1Step 1: Write the General Form of the Equilibrium Law
For a general reaction, the equilibrium law is expressed as the product of the concentrations of the products raised to their stoichiometric coefficients, divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.
2Step 2: Identify Reactants and Products
In the given reaction, the reactants are \( \mathrm{IO}_{3}^{-}(aq) \) and \( \mathrm{H}^{+}(aq) \) and the products are \( \mathrm{I}_{2}(aq) \) and \( \mathrm{H}_{5} \mathrm{IO}_{6}(aq) \) .
3Step 3: Insert the Reactants and Products into the Equilibrium Law
The equilibrium law for the reaction is given by:\[ K_{\mathrm{c}} = \frac{[\mathrm{I}_{2}]^{1} \times [\mathrm{H}_{5}\mathrm{IO}_{6}]^{5}}{[\mathrm{IO}_{3}^{-}]^{7} \times [\mathrm{H}^{+}]^{7}} \]Note that the stoichiometric coefficients become the exponents in the equilibrium expression and the concentration of water is not included because it is the solvent.
Key Concepts
Equilibrium LawEquilibrium Constant (Kc)StoichiometryReaction Quotient
Equilibrium Law
To get a grasp on the concept of equilibrium in chemistry, picture a seesaw that is perfectly balanced. This is similar to a chemical reaction that has reached a state where the rate of the forward reaction is equal to the rate of the reverse reaction. In this balanced state, the concentrations of reactants and products remain constant over time.
The equilibrium law provides a mathematical way to express this balance. It states that at chemical equilibrium, the ratio of the product of the concentrations of the products, each raised to the power of their stoichiometric coefficient, to the product of the concentrations of the reactants, also raised to their stoichiometric coefficients, is a constant. This constant is known as the equilibrium constant, denoted as Kc for reactions in solution.
Understanding this law is important because it describes how different concentrations of substances influence the position of equilibrium. It is essential for predicting how the system will react when conditions change, such as adjusting concentrations or changing the temperature (note that Kc is temperature-dependent).
The equilibrium law provides a mathematical way to express this balance. It states that at chemical equilibrium, the ratio of the product of the concentrations of the products, each raised to the power of their stoichiometric coefficient, to the product of the concentrations of the reactants, also raised to their stoichiometric coefficients, is a constant. This constant is known as the equilibrium constant, denoted as Kc for reactions in solution.
Understanding this law is important because it describes how different concentrations of substances influence the position of equilibrium. It is essential for predicting how the system will react when conditions change, such as adjusting concentrations or changing the temperature (note that Kc is temperature-dependent).
Equilibrium Constant (Kc)
Diving deeper into the equilibrium constant (Kc), it acts like a snapshot of a reaction's balance point. It is a number that provides a measure of the degree to which a reaction has proceeded at equilibrium and is determined by the specific reaction at a given temperature.
The Kc value is unitless, despite being calculated from concentrations (usually expressed in moles per liter). This is because the units typically cancel out when taking the ratio as prescribed by the stoichiometry of the reaction. A high Kc value indicates that at equilibrium, the products dominate. Conversely, a low Kc value, like the extremely small one given in the original exercise (\(1 \times 10^{-85}\)), suggests the reactants are favored at equilibrium.
When interpreting Kc, it's also important to realize that if you reverse the reaction, the value of the Kc inverts. If you multiply the coefficients in the balanced equation, then the Kc is raised to the power of the factor you multiplied by.
The Kc value is unitless, despite being calculated from concentrations (usually expressed in moles per liter). This is because the units typically cancel out when taking the ratio as prescribed by the stoichiometry of the reaction. A high Kc value indicates that at equilibrium, the products dominate. Conversely, a low Kc value, like the extremely small one given in the original exercise (\(1 \times 10^{-85}\)), suggests the reactants are favored at equilibrium.
When interpreting Kc, it's also important to realize that if you reverse the reaction, the value of the Kc inverts. If you multiply the coefficients in the balanced equation, then the Kc is raised to the power of the factor you multiplied by.
Stoichiometry
Stoichiometry is the term that refers to the quantitative relationship between the reactants and products in a chemical reaction. These relationships are derived from the balanced chemical equation, providing crucial information such as the proportions in which substances react and form.
For practical purposes, stoichiometry is the chemist's tool for recipe making at the molecular level. It is used to calculate quantities such as how much reactant is needed to produce a desired amount of product. In terms of equilibrium, stoichiometry determines the exponents in the equilibrium law expression. Each substance's stoichiometric coefficient in the balanced equation becomes the power to which its concentration is raised in the equilibrium expression.
For practical purposes, stoichiometry is the chemist's tool for recipe making at the molecular level. It is used to calculate quantities such as how much reactant is needed to produce a desired amount of product. In terms of equilibrium, stoichiometry determines the exponents in the equilibrium law expression. Each substance's stoichiometric coefficient in the balanced equation becomes the power to which its concentration is raised in the equilibrium expression.
Reaction Quotient
The reaction quotient (Q) is a close cousin of the equilibrium constant (Kc). It is calculated in the same way as Kc, using the current concentrations of the reactants and products involved in the reaction at any point in time, not just at equilibrium.
If we were to draw an analogy, think of Q as a real-time scoreboard during a game, whereas Kc is the final score once the game has ended and is tied. Just like following a sport, comparing Q to Kc can tell us which way the reaction will shift to reach equilibrium. If Q is less than Kc, the reaction proceeds forward, producing more products. If Q is greater than Kc, the reaction will go in reverse, producing more reactants. When the reaction is at equilibrium, Q equals Kc, signaling that the concentrations of reactants and products have stabilized at their equilibrium values.
If we were to draw an analogy, think of Q as a real-time scoreboard during a game, whereas Kc is the final score once the game has ended and is tied. Just like following a sport, comparing Q to Kc can tell us which way the reaction will shift to reach equilibrium. If Q is less than Kc, the reaction proceeds forward, producing more products. If Q is greater than Kc, the reaction will go in reverse, producing more reactants. When the reaction is at equilibrium, Q equals Kc, signaling that the concentrations of reactants and products have stabilized at their equilibrium values.
Other exercises in this chapter
Problem 82
The reaction \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) has a \(K_{\mathrm{c}}=\) \(2.0 \times 10^{9}\) at \(25^{\circ} \math
View solution Problem 83
The reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) has \(K_{\mathrm{P}}=0.140\) at \(25^{\circ} \mathrm{C}\). In a reacti
View solution Problem 87
At \(27^{\circ} \mathrm{C}, K_{\mathrm{P}}=1.5 \times 10^{18}\) for the reaction $$ 3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}
View solution Problem 88
Consider the equilibrium \(2 \mathrm{NaHSO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g)\) How w
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